Contents

(Summer 2001)

(Final)

Final Exam Math 3720 Summer, 2001

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vv = (1, 0, \tm2)$ and $\vw = (\tm1, 2, 3).$

[3] Calculate $\vv\cdot\vw.$

[3] Calculate $\Vert\vv\Vert.$

[3] Find a unit vector in the same direction as $\vv.$

[4] Find a $2\times 2$ matrix $A = \mat(2,2)[](\(a_{{}_{11}}\) & \(a_{{}_{12}}\)\\ \(a_{{}_{21}}\) & \(a_{{}_{22}}\))$ such that $\mat(2,2)[](\(a_{{}_{11}}\) & \(a_{{}_{12}}\)\\ \(a_{{}_{21}}\) & \(a_{{}_{22}}\)) \mat(2,2)[](\(a\) & \(b\)\\ \(c\) & \(d\))$ $=$ $\mat(2,2)[](\(c\) & \(d\)\\ \(a - c\) & \(d - b\))$ for any $2\times 2$ matrix $\mat(2,2)[](\(a\) & \(b\)\\ \(c\) & \(d\)).$

[4] Express the following matrix as a product of a lower triangular matrix and an upper triangular matrix (i.e., $A = LU$).

$A = \mat(3,3)[](1 & 2 & 3\\ -1 & 1 & 0\\ 0 & 6 & -2)$

[4] Use your answers from part (a) to solve $A\vx= \vb$ where $\vb = \cvt[3 -3 -16 ].$

[3] Let $A = \mat(3,3)[](-7 & 3 & 1\\ 2 & 0 & 3\\ 1 & 2 & -7).$ Is $A$ symmetric?

[3] Prove that if $A$ is invertible, then $\tran A;$ is invertible. Hint: Show that $\invp{\tran A;};$ $=$ $\tranp{\inv A;};.$

For each of the following, determine whether or not the given matrix is invertible.

[3] $\mat(2,2)[]({\SMALL\(\fric<3,5>\)} & {\SMALL\(\fric<2,5>\)}\\ -{\SMALL\(\fric<1,5>\)} & {\SMALL\(\fric<1,5>\)})$

[4] $\mat(4,4)[](1 & 2 & -1 & 4\\ 0 & -3 & 7 & 18\\ 2 & 4 & 7 & 12\\ 1 & -1 & 6 & 22)$

For each of the following, you are given a matrix $A$ and a vector $\vb$. Find

1. The particular solution to $A\vx = \vb.$

2. The special solutions to $A\vx = \vzero.$

3. The complete solution to $A\vx = \vb.$

4. The null space of $A.$

[4] $A = \mat(3,3)[](1 & 0 & 2\\ 0 & 1 & 7\\ 0 & 0 & -4)$ and $\vb = \cvt[1 13 -8 ]$

[5] $A = \mat(4,4)[](1 & 2 & 1 & -2\\ 2 & -1 & 1 & 1\\ 4 & 3 & 3 & 3)$ and $\vb = \cvt[5 3 -5 ]$

Let $A = \mat(4,4)[](1 & 2 & 1 & -2\\ 2 & -1 & 1 & 1\\ 4 & 3 & 3 & 3)$. Note that $A$ is the same matrix that you worked with in the previous problem.

[3] Describe $C(A).$

[3] Describe $C(\tran A;).$

[3] Describe $N(\tran A;).$

Calculate the determinant of each of the following.

[3] $A = \mat(3,3)[](2 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0)$

[3] $B = \mat(3,3)[](-4 & 0 & 1\\ -3 & 2 & 0\\ 1 & 0 & 0)$

[3] $C = \mat(4,4)[](0 & 1 & 0 & 0\\ 0 & 1 & 0 & -1\\ 0 & 27 & 3 & -5\\ 2 & 15 & -8 & 0)$

[4] Let $A = \mat(2,2)[](2 & 4\\ 2 & 4)$. Find the eigenvlaues and eigenvectors of $A$. If possible, diagonalize $A.$

Let $\vV$ be the set of vectors in $\Rss3$ whose components sum to 0.

[3] Prove that $\vV$ is a subspace of $\Rss 3 .$

[3] Find a basis for $\vV.$

[4] Let $\vV$ and $\vW$ be vector spaces and $T:\vV\to\vW$ be a linear transformation. Prove that $\ker(T)$ is a subspace of $\vV.$

Let $\vV$ be $\Rss3$ with basis $\{(1,0,0), (1,2,0), (1,2,3)\}$ and let $\vW$ be $\Rss3$ with basis $\{(1,0,0), (0,1,0), (0,0,1)\}$. Define $T:\vV\to\vW$ by $T(\vv) = \tm\vv.$

[3] Prove that $T$ is a linear transformation.

[4] Find the transformation matrix for $T.$

[3] Let $\vv = (3,4,3)$. Use the transformation matrix to find $T(\vv)$.

[3] If possible, give an example of a linear transformation $T:\Rss2 \to \Rss2$ such that $T\left(\cv[1 2 ]\right) = \cv[0 1 ]$ and $T\left(\cv[-2 -4 ]\right) = \cv[1 3 ].$ If this is not possible, explain why it is not.

Answer the following as true or false (write the entire word). If the statement is true, then prove it. If the statement is false, then give a counter example.

[4] The set $\{(1, 0, 1), (\tm1, 0, 0), (1, 2, 3)\}$ spans $\Rss 3 .$

[4] If $A$ and $B$ are both $n\times n$ matrices, then $\det(AB) = \det(BA).$

[4] A $2\times 2$ matrix has 2 distinct eigenvalues if and only if it has two pivots.

(Spring 2009)

(Quizzes)

[2]( Solve each of the following.)

[2]

$x_{{}_{1}}$	$+$	$3x_{{}_{2}}$	=	$4$
$x_{{}_{1}}$	$-$	$x_{{}_{2}}$	=	$4$

[3]

$3x_{{}_{1}}$	$+$	$x_{{}_{2}}$	$-$	$x_{{}_{3}}$	=	$-2$
$x_{{}_{1}}$	$+$	$4x_{{}_{2}}$	$+$	$2x_{{}_{3}}$	=	$-2$
$x_{{}_{1}}$	$+$	$x_{{}_{2}}$	$+$	$x_{{}_{3}}$	=	0

[2]( For each of the following, determine whether or not $A\vx$ $=$ $\vb$ is consistent. If so, find all solutions. If not, explain why it is not.)

[3] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 3 & -1 & 4\\ -1 & 1 & 2 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -1\\ 9\\ -5 \end{tabular}\hspace{-4pt}\right]$

[3] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 2 & 1 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ 1\\ 1 \end{tabular}\hspace{-4pt}\right]$

[3] Show that if $A$ $=$ $\tran A;,$ then $A$ is square.

[3] Let $A$ and $B$ be symmetric matrices. Show that $AB$ is symmetric if and only if
$AB$ $=$ $BA.$

[2] Let $A = \left[\hspace{-4pt} \begin{tabular}{rr} \vspace{-12pt}\\ 1 & 3 \\ 1 & -2 \end{tabular}\hspace{-4pt}\right]$ and $B = \left[\hspace{-4pt} \begin{tabular}{rr} \vspace{-12pt}\\ 13 & -1 \\ -7 & -1 \end{tabular}\hspace{-4pt}\right]$. Find a matrix $X$ such that $AX$ $=$ $B.$

[3] Use the LU factorization to solve the following.

$2x$	$+$	$3y$	=	-5
-$x$	$+$	$2y$	=	-8

(Page 81: 12) [3] Let $A$ and $B$ be $n\times n$ matrices and $M$ be a block matrix of the form $M = \left[\hspace{-4pt} \begin{tabular}{rr} \vspace{-12pt}\\ A & 0 \\ 0 & B \end{tabular}\hspace{-4pt}\right]$. Prove that if $A$ is singular, then $M$ is singular.

[2]( Calculate the following determinants.)

[2] $\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 3 & 4 \end{tabular}\hspace{-4pt}\right\vert$

[3] $\left\vert\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 2 ... ... 0 & 0 & 0 & -4 & 6\\ 0 & 0 & 0 & 0 & 2 \end{tabular}\hspace{-4pt}\right\vert$

[3] Let $A$ be an $n\times n$ matrix with determinant $6$. Also, let $P$ be the $n\times n$ matrix formed by interchanging two rows of the $n\times n$ identity matrix. Calculate $\vert PA\vert.$

[3] Let $A$ be an $n\times n$ matrix where $n$ is odd. Show that $A^{2}$ $\ne$ $-I.$

(Page 122: 7) [3] Show that the zero element in a vector space is unique.

[3] Let $\vV$ be a vector space and $\vv\in\vV$. Show that if $\vv + \vw$ $=$ $\vzero,$ then $\vw$ $=$ $\tm\vv.$

(Page 122: 9) [6] Let $\vV$ be a vector space and let $\vx\in\vV$. Show that:

$\beta\vzero$ $=$ $\vzero$ for each scalar $\beta.$

If $\alpha\vx$ $=$ $\vzero,$ then either $\alpha$ $=$ 0 or $\vx$ $=$ $\vzero.$

[1] Let $\vV$ be a vector space and let $S,T\subseteq \vV$ such that $T\subseteq$span$(S)$ and $S\subseteq$span$(T)$. Prove that span$(S) =$   span$(T).$

[1] Page 131: 18.

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[1] Page 145: 17.

(Page 167: 1) [6] For the following matrix, find a basis for the row space, a basis for the column space, and a basis for the null space.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -3 & 1 & 3 & 4 \\ 1 & 2 & -1 & -2\\ -3 & 8 & 4 & 2 \end{tabular}\hspace{-4pt}\right]$

[4] Let $\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}},\vv_{{}_{4}}\in\Rss3$. Is it possible for the vectors to be linearly independent? Must they span $\Rss3 ?$

[3] Let $\vv\in\Rss n$ and define $T_{{}_{\vv}}: \Rss n \to \R$ by $T_{{}_{\vv}}(\vx)$ $=$ $\vv\cdot\vx$. Show that $T_{{}_{\vv}}$ is a linear transformation.

[3] Let $\vV$ and $\vW$ be vector spaces and $T:\vV\to\vW$ be a linear transformation. Show that the kernel of $T$ is a subspace of $\vW.$

[3] Show that the composition of linear transformations is a linear transformation.

(Exam 1)

Exam 1 Math 3720 Spring 2009

[2]( For each of the following, determine whether or not $A\vx$ $=$ $\vb$ is consistent. If so, find all solutions. If not, explain why it is not.)

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 2 & -1\\ 1 & 3\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 7\\ 0\\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & -1 & 7\\ 1 & 3 & 0\\ \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 3\\ 0 & 1 & -1\\ \end{tabular}\hspace{-4pt}\right]$

$\vx$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 3 \\ -1\\ \end{tabular}\hspace{-4pt}\right]$

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 1 & 4\\ 1 & 1 & 5\\ 7 & -1 & 5 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 7\\ 7\\ 11 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 4 & 7\\ 1 & 1 & 5 & 7\\ 7 & -1 & 5 & 11\\ \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\\ \end{tabular}\hspace{-4pt}\right]$

$\vx$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 1\\ 1\\ \end{tabular}\hspace{-4pt}\right]$

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 2 & -1 & 2\\ -1 & 2 & 3\\ 2 & 1 & -2\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 6\\ 3\\ 7\\ 3\\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 6\\ 2... ...2 & 3\\ -1 & 2 & 3 & 7\\ 2 & 1 & -2 & 3\\ \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 0... ... 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{tabular}\hspace{-4pt}\right]$

$0 + 0 + 0$ $\ne$ $1$

Inconsistent

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1... ...& 0 & 1 & 0 & 2\\ 1 & 0 & 0 & 2 & 1 & 0\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 9\\ 12\\ 4\\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrrrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1 & ... ...& 0 & 2 & 12\\ 1 & 0 & 0 & 2 & 1 & 0 & 4\\ \end{tabular}\hspace{-4pt}\right]$

$\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 2 & ... ... & -2 & -4\\ 0 & 0 & 1 & -4 & -2 & 3 & 9\\ \end{tabular}\hspace{-4pt}\right]$

Lead Variables: $x_{{}_{1}},$ $x_{{}_{2}},$ $x_{{}_{3}}$

Free Variables: $x_{{}_{4}},$ $x_{{}_{5}},$ $x_{{}_{6}}$

Solutions: $\left\{ \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ \(4 - 2x_{{}_{4... ...tabular}\hspace{-4pt}\right]: x_{{}_{4}}, x_{{}_{5}}, x_{{}_{6}} \in\R \right\}$

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 2 & 1 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ 1\\ 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 1 & 1\\ 1 & 2 & 1 & 1\\ 1 & 2 & 1 & 1\\ \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{tabular}\hspace{-4pt}\right]$

Lead Variables: $x_{{}_{1}}$

Free Variables: $x_{{}_{2}},$ $x_{{}_{3}}$

Solutions: $\left\{ \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ \(1 - 2x_{{}_{2... ...\) \\ \end{tabular}\hspace{-4pt}\right]: x_{{}_{2}}, x_{{}_{3}} \in\R \right\}$

[5] Show that if $A$ and $B$ are $n\times n$ invertible matrices, then $AB$ is invertible.

Note that

$(AB)(\inv B;\inv A;)$ $=$ $A(B\inv B;)\inv A;$ $=$ $A(I)\inv A;$ $=$ $(AI)\inv A;$ $=$ $A\inv A;$ $=$ $I$

and

$(\inv B;\inv A;)(AB)$ $=$ $B(A\inv A;)\inv B;$ $=$ $B(I)\inv B;$ $=$ $(BI)\inv B;$ $=$ $B\inv B;$ $=$ $I.$

Therefore, $\inv B;\inv A;$ $=$ $\inv(AB);.$

[5] Show that if $A$ $=$ $\tran A;,$ then $A$ is square.

Suppose that $A$ is an $m\times n$ matrix. Then $\tran A;$ is an $n\times m$ matrix. Since $A$ $=$ $\tran A;,$ $A$ is both an is an $m\times n$ matrix and an $n\times m$ matrix. Therefore, $m$ $=$ $n$ and $A$ is square.

[5] Give an example that illustrates that matrix multiplication is not commutative.

[5] Show that the matrix $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 1 & 0\\ \end{tabular}\hspace{-4pt}\right]$ has no inverse.

Let $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ a & b\\ c & d\\ \end{tabular}\hspace{-4pt}\right]$ be any $2\times 2$ matrix and consider $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ a & b\\ c & d\\ \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 1 & 0\\ \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ b & 0\\ 0 & 0\\ \end{tabular}\hspace{-4pt}\right]$ $\ne$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ \end{tabular}\hspace{-4pt}\right].$

[5] Let $A$ and $B$ be symmetric matrices. Show that $AB$ is symmetric if and only if
$AB$ $=$ $BA.$

First suppose that $AB$ is symmetric. Then $AB$ $=$ $\tran(AB);$ $=$ $\tran B;\tran A;$ $=$ $BA.$

Now suppose that $AB$ $=$ $BA$. Then $\tran(AB);$ $=$ $\tran B;\tran A;$ $=$ $BA$ $=$ $AB.$

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

For each of the following, determine whether or not $A\vx$ $=$ $\vb$ is consistent. If so, find all solutions. If not, explain why it is not.

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 2 & -1\\ 1 & 3\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 7\\ 0\\ \end{tabular}\hspace{-4pt}\right]$ [5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 1 & 4\\ 1 & 1 & 5\\ 7 & -1 & 5 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 7\\ 7\\ 11 \end{tabular}\hspace{-4pt}\right]$

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 2 & -1 & 2\\ -1 & 2 & 3\\ 2 & 1 & -2\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 6\\ 3\\ 7\\ 3\\ \end{tabular}\hspace{-4pt}\right]$

[5] $A = \left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1... ...& 0 & 1 & 0 & 2\\ 1 & 0 & 0 & 2 & 1 & 0\\ \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 9\\ 12\\ 4\\ \end{tabular}\hspace{-4pt}\right]$ [5] $A = \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & 2 & 1\\ 1 & 2 & 1 \end{tabular}\hspace{-4pt}\right],$ $\vb=\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ 1\\ 1 \end{tabular}\hspace{-4pt}\right]$

[5] Show that if $A$ and $B$ are $n\times n$ invertible matrices, then $AB$ is invertible.

[5] Show that if $A$ $=$ $\tran A;,$ then $A$ is square.

[5] Give an example that illustrates that matrix multiplication is not commutative.

[5] Show that the matrix $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 1 & 0\\ \end{tabular}\hspace{-4pt}\right]$ has no inverse.

[5] Let $A$ and $B$ be symmetric matrices. Show that $AB$ is symmetric if and only if
$AB$ $=$ $BA.$

(Exam 2)

Exam 2 Math 3720 Spring 2009

[5] Find elementary matrices whose product is $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 0 & 2 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right].$

Let $E_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ $E_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 2 & 1 \end{tabular}\hspace{-4pt}\right],$ $E_{{}_{3}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right],$ then $E_{{}_{3}}$ $E_{{}_{2}}$ $E_{{}_{1}}$ $=$ $A.$

[5] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 3 & 4 \end{tabular}\hspace{-4pt}\right]$ and give the $LU$ factorization of $A.$

Let $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ -3 & 1 \end{tabular}\hspace{-4pt}\right],$ $L$ $=$ $\inv E;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 3 & 1 \end{tabular}\hspace{-4pt}\right],$ and $U$ $=$ $EA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & -2 \end{tabular}\hspace{-4pt}\right]$. Then $A$ $=$ $LU.$

[5] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 1 & 2 & -1\\ 3 & -1 & 1 \end{tabular}\hspace{-4pt}\right],$ $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ -1 & 1 & 0\\ -7 & 4 & 1 \end{tabular}\hspace{-4pt}\right],$ $L$ $=$ $\inv E;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 1 & 1 & 0\\ 3 & -4 & 1 \end{tabular}\hspace{-4pt}\right],$ $U$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & -2\\ 0 & 0 & -10 \end{tabular}\hspace{-4pt}\right],$ and $\vb$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ -5\\ -3 \end{tabular}\hspace{-4pt}\right].$ Note that $A$ $=$ $LU$. Use the $LU$ factorization of $A$ to solve $A\vx$ $=$ $\vb.$

[2]()

$A\vx$ $=$ $\vb$

$LU\vx$ $=$ $\vb$

$U\vx$ $=$ $\inv L;\vb$

$U\vx$ $=$ $E\vb$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & -2\\ 0 & 0 & -10 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ \(x_{{}_{1}}\)\\ \(x_{{}_{2}}\)\\ \(x_{{}_{3}}\) \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ -1 & 1 & 0\\ -7 & 4 & 1 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ -5\\ -3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & -2\\ 0 & 0 & -10 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ \(x_{{}_{1}}\)\\ \(x_{{}_{2}}\)\\ \(x_{{}_{3}}\) \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1\\ -6\\ -30 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ \(x_{{}_{1}}\)\\ \(x_{{}_{2}}\)\\ \(x_{{}_{3}}\) \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -2\\ 0\\ 3 \end{tabular}\hspace{-4pt}\right]$

[5] Find the inverse of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

Note that the row operations needed to transform $A$ to $I$ are subtracting row two from row one and subtracting row three from row two. Therefore, $\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

[5] Use the above answer to find a matrix $X$ such that $AX$ $=$ $B$ where $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{tabular}\hspace{-4pt}\right].$

$X$ $=$ $\inv A;B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -3 & -3 & -3\\ -3 & -3 & -3\\ 7 & 8 & 9 \end{tabular}\hspace{-4pt}\right]$

[5] Multiply. Hint: Use block multiplication.

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1 & 0... ...& 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4 & 5... ...& 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 2 & 4 & 6 & 8 & 1... ...& 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{tabular}\hspace{-4pt}\right]$

[3]( Calculate the determinants of the following matrices.)

[5] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -2\\ -1 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -2\\ -1 & 4 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $2$

[5] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 3 & -2\\ 0 & -5 & 3\\ 1 & -1 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 3 & -2\\ 0 & -5 & 3\\ 1 & -1 & 4 \end{tabular}\hspace{-4pt}\right\vert$

$=$ $1\cdot\tm17 + 1\cdot\tm1$

$=$ $\tm18$

[5] $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4 & 5... ...& 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{tabular}\hspace{-4pt}\right]$

$\left\vert\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4... ...& 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{tabular}\hspace{-4pt}\right\vert$ $=$ 0

[5] Given that $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & 0 & 1 & 4\\... ...2\\ 1 & 0 & -1 & -10\\ 7 & 5 & 1 & -8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $34,$ $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 0 & 1 & 4\\... ...2\\ 3 & 0 & -1 & -10\\ -1 & 5 & 1 & -8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $58,$ $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & 2 & 1 & 4\\... ...\\ 1 & 3 & -1 & -10\\ 7 & -1 & 1 & -8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm216,$

$\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & 0 & 2 & 4\\... ...2\\ 1 & 0 & 3 & -10\\ 7 & 5 & -1 & -8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $336,$ and $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & 0 & 1 & 2\\... ...& 1\\ 1 & 0 & -1 & 3\\ 7 & 5 & 1 & -1 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm38,$ solve $A\vx$ $=$ $\vb$ where $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & 0 & 1 & 4\\ 3... ... & -2\\ 1 & 0 & -1 & -10\\ 7 & 5 & 1 & -8 \end{tabular}\hspace{-4pt}\right]$

and $\vb$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 2\\ 1\\ 3\\ -1 \end{tabular}\hspace{-4pt}\right].$ By Cramer's rule, $\vx$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-10pt}\\ \(\fric<29,17>\)\\ \vs... ...-6pt}\\ \(\tm\fric<19,17>\) \vspace{3pt}\\ \end{tabular}\hspace{-4pt}\right].$

(Exam 3)

[5] Show that the zero element of a vector space is unique.

Suppose that $\vzero_{{}_{1}}$ and $\vzero_{{}_{2}}$ are both zeros of a vector space. Then $\vzero_{{}_{1}}$ $=$ $\vzero_{{}_{1}} + \vzero_{{}_{2}}$ $=$ $\vzero_{{}_{2}}.$

[5] Let $P$ be the vector space consisting of all polynomials. Find a basis for $P.$

$\{1,x,\x2 , \x3 , \x4 , \ldots\}$

[5] Suppose that $\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}}\in\Rss3$ are linearly independent and $A$ is a $3\times3$ invertible matrix. Show that $A\vv_{{}_{1}},A\vv_{{}_{2}},A\vv_{{}_{3}}$ are linearly independent.

Suppose that $\alpha_{{}_{1}},\alpha_{{}_{2}},\alpha_{{}_{3}}\in\R$ such that $\alpha_{{}_{1}}A\vv_{{}_{1}} + \alpha_{{}_{2}}A\vv_{{}_{2}} + \alpha_{{}_{3}}A\vv_{{}_{3}}$ $=$ $\vzero$. Then we have

$\alpha_{{}_{1}}\left(A\vv_{{}_{1}}\right) + \alpha_{{}_{2}}\left(A\vv_{{}_{2}}\right) + \alpha_{{}_{3}}\left(A\vv_{{}_{3}}\right)$ $=$ $\vzero$

$A\left(\alpha_{{}_{1}}\vv_{{}_{1}}\right) + A\left(\alpha_{{}_{2}}\vv_{{}_{2}}\right) + A\left(\alpha_{{}_{3}}\vv_{{}_{3}}\right)$ $=$ $\vzero$

$A\left(\alpha_{{}_{1}}\vv_{{}_{1}} + \alpha_{{}_{2}}\vv_{{}_{2}} + \alpha_{{}_{3}}\vv_{{}_{3}}\right)$ $=$ $\vzero$.

Since $A$ is invertible, $\alpha_{{}_{1}}\vv_{{}_{1}} + \alpha_{{}_{2}}\vv_{{}_{2}} + \alpha_{{}_{3}}\vv_{{}_{3}}$ $=$ $\vzero$. This implies that $\alpha_{{}_{1}}$ $=$ $\alpha_{{}_{2}}$ $=$ $\alpha_{{}_{3}}$ $=$ 0 since $\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}}$ are linearly independent. Therefore, $Av_{{}_{1}},Av_{{}_{2}},Av_{{}_{3}}$ are linearly independent.

Let $\vV$ $=$ $\Rss2\times2$ be the vector space of all $2\times 2$ matrices. For each of the following, determine whether or not the given set is a subspace of $\vV.$

[5] $\left\{A\in\Rss2\times2 :AB = BA\text{ for all } B\in\Rss2\times2 \right\}$

This is a subspace of $\Rss2\times2 .$

Let $\vW$ $=$ $\left\{A\in\Rss2\times2 :AB = BA\text{ for all } B\in\Rss2\times2 \right\}$. Suppose that $C,D\in\vW$ and $\alpha\in\R$. To see that $C + D\in\vW,$ let $M\in\Rss2\times2$ and consider $M(C + D)$ $=$ $MC + MD$ $=$ $CM + DM$ $=$ $(C + D)M$ and so $C + D\in\vW$. Also, note that $M(\alpha C)$ $=$ $\alpha(MC)$ $=$ $\alpha(CM)$ $=$ $(\alpha C)M$. Therefore, $\vW$ is a subspace of $\vV.$

[5] The set of all invertible $2\times 2$ matrices.

This is not a subspace of $\Rss2\times2 .$

Let $\mathcal{I}$ be the set of all invertible $2\times 2$ matrices. Note that $I,\tm I\in\mathcal{I}$ and $I + \tm I\notin\mathcal{I}$. Therefore, $\mathcal{I}$ is not a subspace of $\vV.$

[5] Recall that the vector space $P_{{}_{n}}$ consists of all polynomials with degree less than $n$ and has dimension $n$. Suppose that $n\geq 1$ and let $D$ $=$ $\{f':f\in P_{{}_{n}}\}$. Show that $D$ is a subspace of $P_{{}_{n}}$. What is the dimension of $D?$

Recall from calculus that if $f,g\in P_{{}_{n}}$ and $\alpha\in\R,$ then $(f + g)' = f' + g' \in P_{{}_{n}}$ and $(\alpha f)' = \alpha f'\in P_{{}_{n}}$.

Since the derivative of an $m^{\text{th}}$ degree polynomial is $m - 1,$ the dimension of $D$ is $n - 1$.

[5] Let $\vu_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 2 \\ -1 \end{tabular}\hspace{-4pt}\right],$ $\vu_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 2 \\ 5 \end{tabular}\hspace{-4pt}\right],$ $\vv_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -1 \\ 6 \end{tabular}\hspace{-4pt}\right],$ and $\vv_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 3 \end{tabular}\hspace{-4pt}\right]$. Give the transition matrix from the basis $\{\vu_{{}_{1}},\vu_{{}_{2}}\}$ to the basis $\{\vv_{{}_{1}},\vv_{{}_{2}}\}.$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 2 & 2 \\ -1 & 5 \end{tabular}\hspace{-4pt}\right]$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -1 & 1 \\ 6 & 3 \end{tabular}\hspace{-4pt}\right]$. Then the transition matrix is $\inv B;A$.

[5] State the Rank Plus Nullity Theorem.

Find the row space, column space, and null space of each of the following.

[5] $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -3 & 1 & 3 & 4 \\ 1 & 2 & -1 & -2\\ -3 & 8 & 4 & 2 \end{tabular}\hspace{-4pt}\right]$

Using row operations the matrix can be transformed to $\left[\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & -1 & -2\\ 0 & 7 & 0 & -2\\ 0 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right].$

So the row space and column space both have dimension $3$ and the null space has dimension $1.$

$C\left(\tran A;\right)$ $=$ span$\left( \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 2 \\ -1... ...ace{-12pt}\\ 0 \\ 0 \\ 1 \\ 0 \end{tabular}\hspace{-4pt}\right] \right)$

$C\left(A\right)$ $=$ $\Rss3$

$N\left(A\right)$ $=$ span$\left( \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 10 \\ 2 \\ 0 \\ 7 \end{tabular}\hspace{-4pt}\right] \right)$

[5] $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 1\\ 0 & 2 & 1\\ 1 & 1 & 1\\ 3 & 1 & 6 \end{tabular}\hspace{-4pt}\right]$

Using row operations the matrix can be transformed to $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right].$

So the row space and column space both have dimension $3$ and the null space has dimension $0.$

$C\left(\tran A;\right)$ $=$ $\Rss3$

$C\left(A\right)$ $=$ span$\left( \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 0 \\ 1... ...ace{-12pt}\\ 1 \\ 1 \\ 1 \\ 6 \end{tabular}\hspace{-4pt}\right] \right)$

$N\left(A\right)$ $=$ $\left\{ \left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 0 \\ 0 \\ 0 \end{tabular}\hspace{-4pt}\right] \right\}$

(Final)

Final Exam Math 3720 Spring 2009

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vv$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 0 \\ -2 \end{tabular}\hspace{-4pt}\right]$ and $\vw$ $=$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -1 \\ 2 \\ 3 \end{tabular}\hspace{-4pt}\right].$

[2] Calculate $\vv\cdot\vw.$

[2] Calculate $\Vert\vv\Vert.$

[2] Find a unit vector in the same direction as $\vv.$

[3] Find a $2\times 2$ matrix $A = \mat(2,2)[](\(a_{{}_{11}}\) & \(a_{{}_{12}}\)\\ \(a_{{}_{21}}\) & \(a_{{}_{22}}\))$ such that $\mat(2,2)[](\(a_{{}_{11}}\) & \(a_{{}_{12}}\)\\ \(a_{{}_{21}}\) & \(a_{{}_{22}}\)) \mat(2,2)[](\(a\) & \(b\)\\ \(c\) & \(d\))$ $=$ $\mat(2,2)[](\(c\) & \(d\)\\ \(a - c\) & \(d - b\))$ for any $2\times 2$ matrix $\mat(2,2)[](\(a\) & \(b\)\\ \(c\) & \(d\)).$

[3] Express the following matrix as a product of a lower triangular matrix and an upper triangular matrix (i.e., $A = LU$).

$A = \mat(3,3)[](1 & 2 & 3\\ -1 & 1 & 0\\ 0 & 6 & -2)$

[3] Use your answers from part (a) to solve $A\vx= \vb$ where $\vb = \cvt[3 -3 -16 ].$

[2] Prove that if $A$ is invertible, then $\tran A;$ is invertible. Hint: Show that $\invp{\tran A;};$ $=$ $\tranp{\inv A;};.$

For each of the following, determine whether or not the given matrix is invertible.

[3] $\mat(2,2)[]({\SMALL\(\fric<3,5>\)} & {\SMALL\(\fric<2,5>\)}\\ -{\SMALL\(\fric<1,5>\)} & {\SMALL\(\fric<1,5>\)})$

[2] $\mat(4,4)[](1 & 2 & -1 & 4\\ 0 & -3 & 7 & 18\\ 2 & 4 & 7 & 12\\ 1 & -1 & 6 & 22)$

For each of the following, find the row space, column space, and null space of the given matrix.

[3] $A = \mat(3,3)[](1 & 0 & 2\\ 0 & 1 & 7\\ 0 & 0 & -4)$

[3] $A = \mat(4,4)[](1 & 2 & 1 & -2\\ 2 & -1 & 1 & 1\\ 4 & 3 & 3 & 3)$

[2] Let $A = \mat(2,2)[](2 & 4\\ 2 & 4)$. Find the eigenvalues and eigenvectors of $A$. If possible, diagonalize $A.$

Let $\vV$ be the set of vectors in $\Rss3$ whose components sum to 0.

[2] Prove that $\vV$ is a subspace of $\Rss 3 .$

[2] Find a basis for $\vV.$

Let $\vV$ be $\Rss3$ with basis $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 0 \\ 0 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 2 \\ 0 \end{tabular}\hspace{-4pt}\right],$ and $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 2 \\ 3 \end{tabular}\hspace{-4pt}\right]$ and $\vW$ be $\Rss3$ with basis $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 0 \\ 0 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 0 \\ 1 \\ 0 \end{tabular}\hspace{-4pt}\right],$ and $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 0 \\ 0 \\ 1 \end{tabular}\hspace{-4pt}\right].$ Define $T:\vV\to\vW$ by $T(\vv) = \tm\vv.$

[2] Prove that $T$ is a linear transformation.

[2] Find the transformation matrix for $T.$

[2] Let $\vv = (3,4,3)$. Use the transformation matrix to find $T(\vv)$.

[3] Let $\vV$ and $\vW$ be vector spaces and $T:\vV\to\vW$ be a linear transformation. Prove that $\ker(T)$ is a subspace of $\vV.$

[3] If possible, give an example of a linear transformation $T:\Rss2 \to \Rss2$ such that $T\left(\cv[1 2 ]\right) = \cv[0 1 ]$ and $T\left(\cv[-2 -4 ]\right) = \cv[1 3 ].$ If this is not possible, explain why it is not.

Answer the following as true or false (write the entire word). If the statement is true, then prove it. If the statement is false, then give a counter example.

[2] The set $\left\{\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 1 \\ 0 \\ 1... ...} \vspace{-12pt}\\ 1 \\ 2 \\ 3 \end{tabular}\hspace{-4pt}\right]\right\}$ spans $\Rss 3 .$

[2] If $A$ and $B$ are both $n\times n$ matrices, then $\det(AB) = \det(BA).$

[2] A $2\times 2$ matrix has 2 distinct eigenvalues if and only if it has two pivots.

(Spring 2012) (Quizzes)

[1] Let $\vu$ $=$ $\cv[1 2 ]$ and $\vv$ $=$ $\cv[2 \tm1 ].$ Express $\cv[6 2 ]$ as a linear combination of $\vu$ and $\vv.$

$2\vu + 2\vv$ $=$ $\cv[6 2 ]$

(Page 26: 31) [1] Let $A$ $=$ $(1,1,\tm1),$ $B$ $=$ $(\tm3,2,\tm2),$ and $C$ $=$ $(2,2,\tm4)$. Prove that $\triangle ABC$ is a right-angled triangle.

[1] Solve the following system.

$2x$	$-$	$y$	$+$	$z$	$=$	3
$x$	$+$	$y$	$+$	$z$	$=$	$\tm$2
$x$	$+$	$y$	$-$	$z$	$=$	0

[2]()

$R_{{}_{1}} + R_{{}_{3}}$

$3x$ $=$ $3$

$x$ $=$ $1$

$R_{{}_{2}} - R_{{}_{3}}$

$2z$ $=$ $\tm2$

$z$ $=$ $\tm 1$

$x + y - z$ $=$ 0

$1 + y + 1$ $=$ 0

$y$ $=$ $\tm2$

[1] Suppose that $S$ is a set of vectors and $\vv\in$span$(S)\setminus S$. Let $T$ $=$ $S\cup\{\vv\}$. Prove that $T$ is a linearly dependent set.

Suppose that $S$ $=$ $\left\{ \jsub(\vv,1), \jsub(\vv,2), \ldots, \jsub(\vv,n) \right\}$. Since $\vv\in$span$(S),$ there exist $\jsub(a,1), \jsub(a,2), \ldots \jsub(a,n)\in\R$ such that $\vv$ $=$ $\jsub(a,1)\jsub(\vv,1) + \jsub(a,2)\jsub(\vv,2) + \cdots + \jsub(a,n)\jsub(\vv,n)$. Then $\jsub(a,1)\jsub(\vv,1) + \jsub(a,2)\jsub(\vv,2) + \cdots + \jsub(a,n)\jsub(\vv,n) - \vv$ $=$ $\vzero$. Since $\tm 1\ne 0,$ $T$ is linearly dependent.

[2]( For each of the following, determine whether the vectors are linearly dependent or linearly independent.)

[1] $\cvt[1 3 2 ],$ $\cvt[\tm4 0 3 ],$ $\cvt[0 1 5 ]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 3 & 2\\ -4 & 0 & 3\\ 0 & 1 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

Rank: 3

Linearly independent.

[1] $\cvt[1 1 1 ],$ $\cvt[2 0 3 ],$ $\cvt[5 1 7 ]$

Note that $2\cvt[2 0 3 ] + \cvt[1 1 1 ]$ $=$ $\cvt[5 1 7 ].$

Linearly dependent.

[1] Suppose that $\vu,\vv,\vw\in\Rss 3$ are linearly independent. Show that . Hint: Suppose that there is a vector $\vx\in \Rss 3$ such that $\vx\notin$   span$(\{\vu,\vv,\vw\})$.

Suppose that $\vx\notin$   span$(\{\vu,\vv,\vw\})$. Then $\{\vu,\vv,\vw,\vx\}$ is a linearly independent set. This is a contradiction since any set of four vectors in $\Rss3$ must be linearly dependent.

[2] Find the $\tran P;LU$ factorization of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 4 & 3 & 7\\ 2 & 1 & 3\\ -2 & -1 & -2 \end{tabular}\hspace{-4pt}\right].$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 3\\ 0 & 0 ... ...- \tm2R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} - \tm1R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & -1 & -2\\ 0 & ... ...\tiny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{1}}\)\\ \tiny\(R_{{}_{2}}\) \end{tabular}$

$P$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

$PA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & -1 & -2\\ 4 & 3 & 7\\ 2 & 1 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & -1 & -2\\ 0 & ... ...}_{2}} - \tm2R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - \tm1R_{{}_{1}}\) \end{tabular}$

$U$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & -1 & -2\\ 0 & 1 & 3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$L$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ -2 & 1 & 0\\ -1 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\tran P;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 4 & 3 & 7\\ 2 & 1 & 3\\ -2 & -1 & -2 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 0 & 0 ... ...}\\ -2 & -1 & -2\\ 0 & 1 & 3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[2] Find the row space, column space, and null space of the following matrix.

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 1 & 2 & 1 & -1\\ 0 & 1 & 2 & 0 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0... ...4,3>\)}\\ 0 & 0 & 1 & {\tiny\(\fric<2,3>\)} \end{tabular}\hspace{-4pt}\right]$

row$(A)$ $=$ span$\left(\colvec[](1\\ 0\\ 0\\ 1), \colvec[](1\\ 2\\ 1\\ -1), \colvec[](0\\ 1\\ 2\\ 0)\right)$

col$(A)$ $=$ $\Rss3$

null$(A)$ $=$ span$\left(\colvec[](-1\\ \tiny {\(\fric<4,3>\)}\\ \tiny {\(\tm\fric<2,3>\)}\\ 1)\right)$

[2]( Calculate the following determinants.)

(Page 280: 7) [1] $\left\vert\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 5 & 2 & 2\\ -1 & 1 & 2\\ 3 & 0 & 0 \end{tabular}\hspace{-4pt}\right\vert$

[1] $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\... ...1 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 \end{tabular}\hspace{-4pt}\right\vert$

[2] Let $A$ $=$ $\twomat[1 2 0 -1 ]$. Find a diagonal matrix $D$ and an invertible matrix $P$ such that $\inv P;AP$ $=$ $D.$

$D$ $=$ $\twomat[1 0 0 -1 ]$

$P$ $=$ $\twomat[1 -1 0 1 ]$

[1] Find an orthogonal basis for $\Rss 2$ that contains the vector $\cv[1 1 ].$

$\left\{\cv[1 1 ],\cv[1 -1 ]\right\}$

(Exam 1)

[2]( Let $\vv$ $=$ $\cvt[2 \tm3 3 ]$ and $\vw$ $=$ $\cvt[1 \tm1 2 ].$ )

[2] Calculate $\vv\cdot\vw.$

$\vv\cdot\vw$ $=$ $11$

[2] Find $\cos\theta$ where $\theta$ is the angle between $\vv$ and $\vw.$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<11,2\sqrt{33}>$

[2] Find comp$_{{}_{\vv}}\vw.$

comp$_{{}_{\vv}}\vw$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert>$ $=$ $\fric<11,\sqrt{22}>$

[2] Find proj$_{{}_{\vv}}\vw.$

proj$_{{}_{\vv}}\vw$ $=$ $\fricp<\vv\cdot\vw,\Vert\vv\Vert^{2}> \vv$ $=$ $\fric<1,2>\vv$

[2] Suppose that $\vv$ and $\vw$ are vectors such that $\Vert\vv\Vert$ $=$ $3$ and $\Vert\vw\Vert$ $=$ $2$. Is it possible that $\vv\cdot\vw$ $=$ $\tm7?$

No. According to the Cauchy-Schwarz-Buniakowsky Inequality, $\vert\vv\cdot\vw\vert$ $\leq$ $\Vert\vv\Vert\Vert\vw\Vert$ $=$ $6.$

[2] Give the parametric equations of the line containing the points $(\tm1,2,1)$ and $(2,1,3).$

[2]()

$\vd$ $=$ $\cvt[3 \tm1 2 ]$

$x$ $=$ $2 + 3t$

$y$ $=$ $1 - t$

$z$ $=$ $3 + 2t$

[2] Give the parametric equations of the plane containing the points $(1,\tm1,2),$ $(2,3,\tm1),$ and $(1,0,1).$

[3]()

$\vu$ $=$ $\cvt[1 4 \tm3 ]$

$\vv$ $=$ $\cvt[0 1 \tm1 ]$

$x$ $=$ $1 + r$

$y$ $=$ $0 + 4r + s$

$z$ $=$ $1 - 3r - s$

[2] Find the point of intersection of the line and the plane.

[3]()

$2 + 3t$ $=$ $1 + r$

$1 - t$ $=$ $4r + s$

$3 + 2t$ $=$ $1 - 3r - s$

$\tm r + 3t$ $=$ $\tm 1$

$\tm4r - s - t$ $=$ $\tm 1$

$3r + s + 2t$ $=$ $\tm2$

$\tm r + 3t$ $=$ $\tm 1$

$4r + s + t$ $=$ $1$

$3r + s + 2t$ $=$ $\tm2$

[2]()

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -1 & 0 & 3 & -1 \\ 4 & 1 & 1 & 1\\ 3 & 1 & 2 & -2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -1 & 0 & 3 & -1 \\ 1 & 0 & -1 & 3\\ 3 & 1 & 2 & -2 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -1 & 0 & 3 & -1 \\ 0 & 0 & 2 & 2\\ 0 & 1 & 11 & -5 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} + 3R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & -3 & 1 \\ 0 & 0 & 1 & 1\\ 0 & 1 & 11 & -5 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \tiny\(\tm R_{{}_{1}}\)\\ \tiny\(\fric<1,2>R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & -3 & 1\\ 0 & 1 & 11 & -5\\ 0 & 0 & 1 & 1\\ \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}}\)\\ \end{tabular}$

$t$ $=$ $1$

$s$ $=$ $\tm16$

$r$ $=$ $4$

Therefore, the point of intersection is $(5,0,5).$

[2]( Find all solutions (if any) of the following systems of linear equations.)

[2]

$2x$	$+$	$y$	$+$	$3z$	=	1
$$	$$	$y$	$-$	$z$	=	2
$2x$	$+$	$3y$	$+$	$z$	=	5

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 3 & 1\\ 0 & 1 & -1 & 2\\ 2 & 3 & 1 & 5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 3 & 1\\ 0 & 1 & -1 & 2\\ 0 & 2 & -2 & 4 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 3 & 1\\ 0 & 1 & -1 & 2\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 2R_{{}_{2}}\) \end{tabular}$

Free variable: $z$

Solution: $\left(\tm\fric<1,2>,2,0\right)$

$\cvt[\(x\) \(y\) \(z\) ]$ $=$ $\cvt[{\tiny\(\tm\frac{1}{2}\)} 2 0 ] + t\cvt[\(\tm2\) \(1\) 1 ]$

[2]

$x$	$$	$$	$+$	$2z$	=	1
$3x$	$+$	$y$	$-$	$z$	=	2
$5x$	$+$	$y$	$+$	$3z$	=	3

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1\\ 3 & 1 & -1 & 2\\ 5 & 1 & 3 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1\\ 3 & 1 & -1 & 2\\ 2 & 0 & 4 & 1 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{2}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1\\ 3 & 1 & -1 & 2\\ 0 & 0 & 0 & -1 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 2R_{{}_{1}}\) \end{tabular}$

No solution.

(Exam 2)

[10] Find $\tran A;$ if $A$ $=$ $\threematb[1 2 3 4 5 6 7 8 9].$

$\tran A;$ $=$ $\threematb[1 4 7 2 5 8 3 6 9]$

[10] Let $A$ $=$ $\threematb[1 -1 0 5 2 -3 0 1 0],$ $B$ $=$ $\threematb[-1 1 0 -5 -2 3 0 -1 0],$ $C$ $=$ $\threematb[3 0 2 1 1 1 4 4 2]$ and calculate $CA + CB$. Hint: There is an easy way to do this. Look closely at $A$ and $B.$

$CA + CB$ $=$ $C(A + B)$ $=$ $CO$ $=$ $O$

[10] Suppose that $A$ and $B$ are invertible matrices. Prove that $\invp AB;$ $=$ $\inv B;\inv A;.$

Consider $(AB)(\inv B;\inv A;)$ $=$ $A(B\inv B;)\inv A;$ $=$ $AI\inv A;$ $=$ $A\inv A;$ $=$ $I.$

[10] Can $\cvt[1 4 3 ]$ be written as a linear combination of $\cvt[1 -2 1 ]$ and $\cvt[1 0 2 ]?$

No.

Suppose $a\cvt[1 -2 1 ] + b \cvt[1 0 2 ]$ $=$ $\cvt[1 4 3 ]$.

Then

$a$	$+$	$b$	=	1
$\tm 2a$	$+$	0	=	4

and

$a$

$+$

$2b$

=

3

From the first set of equations, $a$ $=$ $\tm2$ and $b$ $=$ $3$. However, $\tm2 + 2(3) = 4 \ne 3$. Hence, $\cvt[1 4 3 ]$ cannot be written as a linear combination of $\cvt[1 -2 1 ]$ and $\cvt[1 0 2 ].$

[10] Suppose that $\vv_{{}_{n}}\notin$span$(\{\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}}\})$ and $\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}}$ are linearly independent. Prove that $\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}},\vv_{{}_{n}}$ are linearly independent.

Attempting a contradiction, suppose that there are $a_{{}_{1}},\ldots,a_{{}_{n-1}},a_{{}_{n}}\in\R$ such that $a_{{}_{1}},\ldots,a_{{}_{n-1}},a_{{}_{n}}$ are not all 0 and $a_{{}_{1}}\vv_{{}_{1}} + \cdots + a_{{}_{n-1}}\vv_{{}_{n-1}} + a_{{}_{n}}\vv_{{}_{n}}$ $=$ $\vzero$. Then $a_{{}_{1}}\vv_{{}_{1}} + \cdots + a_{{}_{n-1}}\vv_{{}_{n-1}}$ $=$ $\tm a_{{}_{n}}\vv_{{}_{n}}$. Since $\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}}$ are linearly independent and $a_{{}_{1}},\ldots,a_{{}_{n-1}},a_{{}_{n}}$ are not all $0,$ $a_{{}_{n}}\ne0$. So $\vv_{{}_{n}}$ $=$ $\tm\fric<1,a_{{}_{n}}>\left(a_{{}_{1}}\vv_{{}_{1}} + \cdots + a_{{}_{n-1}}\vv_{{}_{n-1}}\right)$. This contradicts the fact that $\vv_{{}_{n}}\notin$span$(\{\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}}\})$. Hence, $\vv_{{}_{1}},\ldots,\vv_{{}_{n-1}},\vv_{{}_{n}}$ are linearly independent.

[10] Let $A$ $=$ $\twomat[1 0 0 1 ],$ $B$ $=$ $\twomat[0 0 0 -1 ],$ and $C$ $=$ $\twomat[0 1 0 1 ]$. Find span$\left(\{A, B, C\}\right).$

$A + B$ $=$ $\twomat[1 0 0 0 ]$

$B + C$ $=$ $\twomat[0 1 0 0 ]$

$\tm B$ $=$ $\twomat[0 0 0 1 ]$

span$\left(\{A, B, C\}\right)$ $=$ $\left\{\twomat[a b 0 c ]:a,b,c\in\R\right\}$

[10] Let $A$ be any matrix. Explain why $\tranp{\tran A;};$ $=$ $A.$

To form $\tran A;$ interchange the columns and rows of $A$. Interchanging the columns and rows of $\tran A;$ (the definition of $\tranp{\tran A;};$) produces $A.$

[10] Find elementary matrices $E_{{}_{1}}$ and $E_{{}_{2}}$ such that $E_{{}_{2}}E_{{}_{1}}I$ $=$ $\threematb[0 2 0 1 0 0 0 0 1].$

[2](There are two acceptable answers.)

$E_{{}_{1}}$ $=$ $\threematb[0 1 0 1 0 0 0 0 1]$

$E_{{}_{2}}$ $=$ $\threematb[2 0 0 0 1 0 0 0 1]$

$E_{{}_{1}}$ $=$ $\threematb[1 0 0 0 2 0 0 0 1]$

$E_{{}_{2}}$ $=$ $\threematb[0 1 0 1 0 0 0 0 1]$

[2]( For each of the following, find the inverse or explain why it does not exist.)

[10] $\twomat[1 2 -1 3 ]$

$\fric<1,5>\twomat[3 -2 1 1 ]$

[10] $\threematb[1 -2 4 0 5 1 3 4 14]$

Not invertible.

$3R_{{}_{1}} + 2R_{{}_{2}} = R_{{}_{3}}$

$\threematb[1 -2 4 0 5 1 3 4 14]$ $\rref$ $\threematb[1 0 {\tiny\(\fric<22,5>\)} 0 1 {\tiny\(\fric<1,5>\)} 0 0 0]$

[5] Prove that the columns of a square matrix are linearly independent if and only if the rows are also linearly independent.

Let $A$ be an $n\times n$ matrix. Then the following are equivalent.

The columns of $A$ are linearly independent.

The equation $A\vx$ $=$ $\vzero$ has a unique solution.

$A$ is invertible.

$A$ is row equivalent to $I.$

rank$(A)$ $=$ $n.$

The rows of $A$ are linearly independent.

Total Points:

(Exam 3)

[15] Find the $LU$ factorization of $A$ $=$ $\twomat[-2 4 3 -5 ].$

[2]()

$\twomat[-2 4 3 -5 ]$

$\twomat[-2 4 0 1 ] \hspace{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - \tm\fric<3,2>R_{{}_{1}}\) \end{tabular}$

$L$ $=$ $\twomat[1 0 -{\tiny\(\fric<3,2>\)} 1 ]$

$\twomat[-2 4 3 -5 ]$ $=$ $\twomat[1 0 -{\tiny\(\fric<3,2>\)} 1 ] \twomat[-2 4 0 1 ]$

[15] Use the $LU$ factorization from above to solve $A\vx$ $=$ $\cv[6 -7 ].$

[2]()

$L\vc$ $=$ $\cv[6 -7 ]$

$\twomat[1 0 -{\tiny\(\fric<3,2>\)} 1 ] \cv[\(c_{{}_{1}}\) \(c_{{}_{2}}\) ]$ $=$ $\cv[6 -7 ]$

$\vc$ $=$ $\cv[6 2 ]$

$U\vx$ $=$ $\cv[6 2 ]$

$\twomat[-2 4 0 1 ] \cv[\(x_{{}_{1}}\) \(x_{{}_{2}}\) ]$ $=$ $\cv[6 2 ]$

$\vx$ $=$ $\cv[1 2 ]$

[15] Find the $\tran P;LU$ factorization of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 3\\ 1 & 0 & 1\\ 2 & 1 & 2 \end{tabular}\hspace{-4pt}\right].$

[2]()

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 3\\ 1 & 0 & 1\\ 2 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 1 & -1... ...gin{tabular}{l} \tiny\(R_{{}_{2}}\)\\ \tiny\(R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & -1... ...y\(R_{{}_{2}} - R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - 2R_{{}_{1}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...gin{tabular}{l} \\ \tiny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - \tm R_{{}_{2}}\) \end{tabular}$

$P$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 0 ... ...pt}\\ 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$PA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & -1 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...y\(R_{{}_{2}} - 2R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - \tm R_{{}_{2}}\) \end{tabular}$

$L$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 2 & 1 & 0\\ 1 & -1 & 1 \end{tabular}\hspace{-4pt}\right]$

$\tran P;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

$U$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{tabular}\hspace{-4pt}\right]$

$A$ $=$ $\tran P;LU$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 3\\ 1 & 0 & 1\\ 2 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$

$=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 2 & 1 & 0\\ 1 & -1 & 1 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 2 \end{tabular}\hspace{-4pt}\right]$

[15] Let $A$ ben an $n\times n$ matrix. Prove that null$(A)$ is a subspace of $\Rss n .$

Clearly, $\vzero\in$null$(A)$. If $\vx,\vy\in N(A)$ and $a\in\R$, then $A(\vx + \vy) = A\vx + A\vy = \vzero + \vzero = \vzero$ and $A(a\vx) = aA\vx = \vzero.$

[15] Find the row space, column space, and null space of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 0 & 2\\ 0 & 2 & 1 & 0\\ 0 & 0 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$.

Since the rows are linearly independent the row space is the span of the rows. Note that the first three columns are pivot columns. Since three linearly independent vectors span $\Rss 3 ,$ the column space is $\Rss3$. Since rank$(A)$ $=$ $3,$ nullity$(A)$ $=$ $1$. Note that $\cvf[-{\tiny\(\fric<5,2>\)} {\tiny\(\fric<1,2>\)} -1 1 ]$ is in the null space. So the null space of $A$ is span$\left(\cvf[-{\tiny\(\fric<5,2>\)} {\tiny\(\fric<1,2>\)} -1 1 ]\right).$

[15] Find the eigenvalues and corresponding eigenspaces of $A$ $=$ $\twomat[5 -2 4 -1 ].$

[2]()

$\twomat\vert{\(5 - \lambda\)} -2 4 {\(\tm1 - \lambda\)} \vert$

$=$ $\lambda^{2} - 4\lambda + 3$

$\lambda^{2} - 4\lambda + 3$ $=$ 0

$(\lambda - 3)(\lambda - 1)$ $=$ 0

$\lambda$ $=$ $1, 3$

$A - 1I$ $=$ $\twomat[4 -2 4 -2 ]$ $\to$ $\twomat[2 -1 0 0 ]$

$E_{{}_{1}}$ $=$ span$\left(\cv[1 -2 ]\right)$

$A - 3I$ $=$ $\twomat[2 -2 4 -4 ]$ $\to$ $\twomat[1 -1 0 0 ]$

$E_{{}_{3}}$ $=$ span$\left(\cv[1 1 ]\right)$

[15] Consider the linear transformation $T:\Rss 3 \to \Rss 2$ defined by $T\left(\cvt[\(x\) \(y\) \(z\) ]\right)$ $=$ $\cv[\(x+y\) \(z\) ]$. You may assume without proof that $T$ is a linear transformation. Find the transformation matrix $[T].$

$[T]$ $=$ $\twothreemat[1 1 0 0 0 1 ]$

Due: May 2

Suppose that $T:\Rss n \to \Rss m$ is a linear transformation. Then

[1] Prove that $\vzero\in$ker$(T).$

[1] Prove that ker$(T)$ is a subspace of $\Rss n .$

[1] Prove that $T$ is one-to-one if and only if ker$(T)$ $=$ $\{\vzero\}.$

(Final)

Final Exam Math 3720 Spring 2012

Name:

[30] Give the equation of the line (in any form) that contains the points $(1, 0, \tm2)$ and $(0, 1, \tm4).$

[20] Give the equation of the plane (in any form) that contains the points $(0, 0, 0), (2, 0, 2),$ and $(4, 2, 4).$

[30] Suppose that $f$ is a quadratic function such that $f(1)$ $=$ $2,$ $f(\tm1)$ $=$ $6,$ and Find $f(x)$. Hint: We know that $f(x)$ $=$ $a\x2 + bx + c$. Solve for $a,$ $b,$ and $c.$

[20] Solve the following system over $\Z_{{}_{5}}.$

$x$	$+$	$2y$	=	4
$2x$	$+$	$y$	=	1

[30] Find the $3\times3$ matrix $E$ such that left multiplication by $E$ is equivalent to the row operation $(2R_{{}_{2}} + R_{{}_{3}})\to R_{{}_{3}}.$

[40] Find the $\tran P;LU$ factorization of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & -2 & 4\\ 1 & 1 & 1\\ 2 & 4 & 2 \end{tabular}\hspace{-4pt}\right].$

[20] Let $l$ be a line in $\Rss3$ and $\vW$ be the subspace of all vectors in $\Rss3$ parallel to $l$. What is the dimension of $\vW?$

[20] Define $T: \Rss3 \to \Rss 3$ by $T\left(\cvt[\(x\) \(y\) \(z\) ]\right)$ $=$ $\cvt[\(x\) \(1\) \(1\) ]$. Explain why $T$ is not a linear transformation.

[60] Find the row space, column space, and null space of the following matrix.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 3\\ 5 & 2 & 1\\ 0 & 1 & -2\\ -1 & -1 & 1 \end{tabular}\hspace{-4pt}\right]$

[30] Find the eigenvalues of the following matrix and then determine whether or not it is diagonalizable. If it is, you need not find the diagonalization.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 1\\ 1 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

(Spring 2015)

(Quizzes)

[1] In the diagram below, draw the vector $2\vv - \vw.$

[1] Give an example of a right triangle with two of its vertices at $(1, 1, 0)$ and $(0, 1, 1).$

$\vv$ $=$ $\langle1, 0, \tm1\rangle$

$\vn$ $=$ $\langle0,1,0\rangle$ (any vector orthogonal to $\vv$ can be used)

If $(0,1, 1)$ is the terminal point of $\vn,$ then $(0, 0, 1)$ is the initial point.

So the triangle with vertices $(1, 1, 0),$ $(0, 1, 1),$ and $(0, 0, 1)$ is a right triangle.

[1] Does the plane $2x - y + z = 0$ intersect the line with parametric equations given below? If so, where? If not, why not?

$x$ $=$ $1 + t$

$y$ $=$ $1 + t$

$z$ $=$ $3 - t$

No.

$2x - y + z = 0$

$2(1 + t) - (1 + t) + (3 - t) = 0$

$2 + 2t - 1 - t + 3 - t = 0$

$4$ $=$ 0

No solution.

(Page 65: 43) [2] Solve the following.

$\tan x$	$-$	$2\sin y$			=	$2$
$\tan x$	$-$	$\sin y$	$+$	$\cos z$	=	$2$
		$\sin y$	$-$	$\cos z$	=	$\tm 1$

[1] Solve the following system.

$2x$	$-$	$y$	$+$	$z$	$=$	3
$x$	$+$	$y$	$+$	$z$	$=$	$\tm$2
$x$	$+$	$y$	$-$	$z$	$=$	0

[2]()

$R_{{}_{1}} + R_{{}_{3}}$

$3x$ $=$ $3$

$x$ $=$ $1$

$R_{{}_{2}} - R_{{}_{3}}$

$2z$ $=$ $\tm2$

$z$ $=$ $\tm 1$

$x + y - z$ $=$ 0

$1 + y + 1$ $=$ 0

$y$ $=$ $\tm2$

[2] Prove that the following vectors are linearly independent.

$\cvt[1 0 0 ],$ $\cvt[1 0 1 ],$ $\cvt[1 1 1 ]$

Since the second component of the third vector is 1 and the second component of the other two vectors is 0, the third vector cannot be written as a linear combination of the other two.

[2] Write the matrix $\twomat[1 3 3 2 ]$ as a linear combination of the matrices

$\twomat[1 0 0 1 ],$ $\twomat[1 0 0 0 ],$ and $\twomat[0 1 1 0 ].$

$2\twomat[1 0 0 1 ] - \twomat[1 0 0 0 ] + 3\twomat[0 1 1 0 ]$ $=$ $\twomat[1 3 3 2 ]$

[2] Prove that if $A$ is an invertible matrix, then $\tran A;$ is invertible and $\invp{\tran A;}; = \tranp{\inv A;};.$

Consider $\tranp{\inv A;};\tran A;$ $=$ $\tranp{A\inv A;};$ $=$ $\tran I;$ $=$ $I.$

[2] Suppose that $A$ and $B$ are $3\times3$ matrices. Further, suppose that matrix $B$ is formed by performing the following elementary row operations on matrix $A$:

Interchange rows one and three $(R_{{}_{1}} \leftrightarrow R_{{}_{3}});$

Replace row three with the sum of row three and 2 times row two $((R_{{}_{3}} + 2R_{{}_{2}})\to R_{{}_{3}}).$

Find the matrix $E$ such that $EA$ $=$ $B.$

$E$ $=$ $\threematb[1 0 0 0 1 0 0 2 1] \threematb[0 0 1 0 1 0 1 0 0]$ $=$ $\threematb[0 0 1 0 1 0 1 2 0]$

[3] Give the row space, column space, and null space of the following matrix.

$A$ $=$ $\threematb[1 0 1 0 1 0 1 1 1]$

$\threematb[1 0 1 0 1 0 1 1 1]$ $\rref$ $\threematb[1 0 1 0 1 0 0 0 0]$

row$(A)$ $=$ span$\left(\left\{% \cvt[1 0 1 ], \cvt[0 1 0 ] \right\}\right)$

col$(A)$ $=$ span$\left(\left\{% \cvt[1 0 1 ], \cvt[0 1 1 ] \right\}\right)$

null$(A)$ $=$ span$\left(\left\{% \cvt[\tm1 0 1 ] \right\}\right)$

Suppose that $T:\Rss n \to \Rss m$ a linear transformation. The kernel of $T$ is the set $\ker(T)$ $=$ $\{\vx\in\Rss n :T(\vx) = \vzero\in\Rss m \}.$

[3] Show that $\ker(T)$ $\leq$ $\Rss n .$

Suppose that $\vv,\vw\in\ker(T)$. Then $T(\vv + \vw)$ $=$ $T(\vv) + T(\vw)$ $=$ $\vzero + \vzero$ $=$ $\vzero$ which means that $(\vv + \vw)\in\ker(T)$. Now suppose $\vv\in\ker(T)$ and $a\in\R$. Then $T(a\vv)$ $=$ $aT(\vv)$ $=$ $a\vzero$ $=$ $\vzero$ and so $a\vv\in\ker(T).$

[3] Find the eigenvalues and corresponding eigenspaces of $A$ $=$ $\twomat[5 -2 4 -1 ].$

[2]()

$\twomat\vert{\(5 - \lambda\)} -2 4 {\(\tm1 - \lambda\)} \vert$

$=$ $\lambda^{2} - 4\lambda + 3$

$\lambda^{2} - 4\lambda + 3$ $=$ 0

$(\lambda - 3)(\lambda - 1)$ $=$ 0

$\lambda$ $=$ $1, 3$

$A - 1I$ $=$ $\twomat[4 -2 4 -2 ]$ $\to$ $\twomat[2 -1 0 0 ]$

$E_{{}_{1}}$ $=$ span$\left(\cv[1 -2 ]\right)$

$A - 3I$ $=$ $\twomat[2 -2 4 -4 ]$ $\to$ $\twomat[1 -1 0 0 ]$

$E_{{}_{3}}$ $=$ span$\left(\cv[1 1 ]\right)$

Two $n\times n$ matrices $A$ and $B$ are similar if there is an invertible matrix $P$ such that $\inv P;AP$ $=$ $B.$

[2] Suppose that matrix $A$ is similar to matrix $B$. Prove that $A$ and $B$ have the same eigenvalues.

Suppose that $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $\vx$.

Then

$\inv P;AP$ $=$ $B$

$\inv P; A$ $=$ $B\inv P;$

$\inv P; A\vx$ $=$ $B\inv P;\vx$

$\inv P;(\lambda\vx)$ $=$ $B\inv P;\vx$

$\lambda\inv P; \vx$ $=$ $B\inv P;\vx.$

Hence, $\lambda$ is an eigenvalue of $B$ with corresponding eigenvector $\inv P;\vx$.

To see that eigenvalues of $B$ are eigenvalues of $A,$ suppose that $Q$ is invertible such that $\inv Q;BQ$ $=$ $A$ and mimic the argument above.

[3] Suppose that matrix $A$ is similar to matrix $B$. Prove that $A$ and $B$ have the same characteristic polynomial. Hint: Creatively use the fact that the determinant of the product is the product of the determinants.

Suppose that $P$ is invertible such that $\inv P;AP$ $=$ $B$ and consider

$\vert A - \lambda I\vert$

$=$ $\vert I(A - \lambda I)\vert$

$=$ $\vert I\vert\cdot\vert A - \lambda I\vert$

$=$ $\vert P\inv P;\vert\cdot\vert A - \lambda I\vert$

$=$ $\vert P\vert\cdot\vert\inv P;\vert\cdot\vert A - \lambda I\vert$

$=$ $\vert P\vert\cdot \vert A - \lambda I\vert\cdot\vert\inv P;\vert$

$=$ $\vert P(A - \lambda I)\inv P;\vert$

$=$ $\vert PA\inv P; - P\lambda I\inv P;\vert$

$=$ $\vert PA\inv P; - \lambda PI\inv P;\vert$

$=$ $\vert B - \lambda I\vert.$

[3] For the following matrix, find the eigenvalues, a basis for each eigenspace, the algebraic multiplicity of each eigenvalue, and the geometric multiplicity of each eigenvalue.

$A$ $=$ $\threematb[1 1 2 0 3 -1 0 1 1]$

$\threematd\vert\(1-\lambda\) 1 2 0 \(3-\lambda\) -1 0 1 \(1-\lambda\) \vert$ $=$ $(1 - \lambda)[(3 - \lambda)(1 - \lambda) + 1]$ $=$ $(1 - \lambda)(\lambda^{2} - 4\lambda + 4)$ $=$ $(1 - \lambda)(\lambda - 2)^{2}$

Eigenvalues: 1, 2

$A - I$ $=$ $\threematb[0 1 2 0 2 -1 0 1 0]$ $\rref$ $\threematb[0 1 0 0 0 1 0 0 0]$

$E_{{}_{1}}$ $=$ span$\left(\left\{\cvt[1 0 0 ]\right\}\right)$

The algebraic multiplicity of 1 is 1.

The geometric multiplicity of 1 is 1.

$A - 2I$ $=$ $\threematb[-1 1 2 0 1 -1 0 1 -1]$ $\rref$ $\threematb[1 0 -3 0 1 -1 0 0 0]$

$E_{{}_{2}}$ $=$ span$\left(\left\{\cvt[3 1 1 ]\right\}\right)$

The algebraic multiplicity of 2 is 2.

The geometric multiplicity of 2 is 1.

(Page 384: 11) [3] Let $W$ $=$ span$\left(\left\{\cvt[2 1 -2 ],\cvt[4 0 1 ]\right\}\right)$. Find $\per W;.$

[3] Suppose that $\vW$ is vector space and $\vU$ and $\vV$ are subspaces. Prove that $\vU\cap\vV$ is a subspace of $\vW.$

(Exam 1)

Exam 1 Math 3720 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[10] Is the triangle with vertices $(1, 2, 3),$ $(2, 1, 5),$ and $(5, 4, 2)$ a right triangle?

Yes.

Consider the vectors $\vu$ $=$ $\langle1,\tm1,2\rangle,$ $\vv$ $=$ $\langle4,2,\tm1\rangle,$ and $\vw$ $=$ $\langle3,3,\tm3\rangle$ whose initial and terminal points are the vertices of the triangle. Since $\vu\cdot\vv$ $=$ $0,$ $\vu$ and $\vv$ are orthogonal.

Let $\vv$ $=$ $\langle1, \tm2, 1\rangle$ and $\vw$ $=$ $\langle2, \tm1, 0\rangle.$

[10] Calculate $\cos\theta$ where $\theta$ is the angle between the vectors.

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<4,\sqrt{6}\sqrt{5}>$ $=$ $\fric<4,\sqrt{30}>$

[10] Find proj$_{{}_{\vv}}\vw.$

proj$_{{}_{\vv}}\vw$ $=$ $\fricp<\vv\cdot\vw,\Vert\vv\Vert^{2}> \vv$ $=$ $\fric<2,3>\langle1,\tm2,1\rangle$

[10] Give the equation of the plane with normal vector $\vn$ $=$ $\langle1, \tm2, 4\rangle$ that contains the point $(2,\tm3,1).$

$\vx\cdot\vn$ $=$ $\vn\cdot\vp$

$x - 2y + 4z$ $=$ $12$

[2]( [10] Give the parametric equations of the plane that contains the points $(2, 1, 1),$ $(1, 2, 4),$ and $(1, 3, 5).$)

$\vv$ $=$ $\langle\tm1,1,3\rangle$

$\vw$ $=$ $\langle0,1,1\rangle$

$x$ $=$ $2 - r$

$y$ $=$ $1 + r + s$

$z$ $=$ $1 + 3r + s$

Solve each of the following.

[10]

$2x$	$+$	$y$	=	$1$
$x$	$-$	$y$	=	$5$

Adding the equations yields $3x$ $=$ $6$. So $x$ $=$ $2$ and $y$ $=$ $\tm3.$

[10]

$x$	$+$	$y$	$+$	$z$	=	$5$
$2x$	$+$	$y$	$-$	$z$	=	$\tm4$
$3x$			$+$	$2z$	=	$5$

Multiplying the first row by $\tm 1$ and adding all three equations yields $4x$ $=$ $\tm4$. So $x$ $=$ $\tm 1$. Substituting in the third equation gives us $z$ $=$ $4$. Substituting in either the first or second equation produces $y$ $=$ $2$.

[10]

$\e x;$	$+$	$\e y;$	=	$3$
$3\e x;$	$-$	$\e y;$	=	$1$

Adding the equations yields $4\e x;$ $=$ $4$. So $\e x;$ $=$ $1$ which means $x$ $=$ 0. Since $\e x;$ $=$ $1,$ $\e y;$ $=$ $2$ which means $y$ $=$ $\ln 2.$

[10] Explain why a homogeneous system must have at least one solution.

The zero vector is always a solution.

[10] Solve the following system over $\Z_{{}_{5}}.$

$2x$	$+$	$y$	=	$3$
$x$	$+$	$2y$	=	0

Multiplying the first row by $3$ and adding the equations yields $2x$ $=$ $4$ which means that $x$ $=$ $2$. So we have the following in $\Z_{{}_{5}}.$

$2 + 2y$ $=$ 0

$2 + 3 + 2y$ $=$ $3$

$2y$ $=$ $3$

$3\cdot 2y$ $=$ $3\cdot 3$

$y$ $=$ $4$

(Exam 2)

Exam 2 Math 3720 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Consider the set of vectors $S$ $=$ $\left\{\cvt[1 0 1 ],\cvt[1 2 0 ],\cvt[0 1 1 ]\right\}.$

[10] Is $S$ linearly independent or linearly dependent?

Linearly independent.

Suppose there exist $a,b\in\R$ such that $a\cvt[1 2 0 ] + b\cvt[0 1 1 ]$ $=$ $\cvt[1 0 1 ]$. From the first and third components, we conclude that $a$ $=$ $1$ and $b$ $=$ $1$. However, since $2 + 1$ $\ne$ $0,$ $a\cvt[1 2 0 ] + b\cvt[0 1 1 ]$ $\ne$ $\cvt[1 0 1 ]$. Therefore, $S$ linearly independent.

[10] Is $\cvt[1 5 0 ]$ in the span of $S?$

Yes. Since $S$ consists of three linearly independent vectors, $S$ spans $\Rss 3 .$

Also, note that $\tm\cvt[1 0 1 ] + 2\cvt[1 2 0 ] + \cvt[0 1 1 ]$ $=$ $\cvt[1 5 0 ].$

[10] Let $A$ be any matrix. Prove that $A\tran A;$ is square.

Suppose that $A$ is an $m\times n$ matrix. Then $\tran A;$ is an $n\times m$ matrix. So $A\tran A;$ is an $m\times m$ matrix.

[10] Let $A$ be any matrix. Prove that $A\tran A;$ is symmetric.

Note that $\tranp{A\tran A;};$ $=$ $\tranp{\tran A;};\tran A;$ $=$ $A\tran A;.$

[10] Prove that if $A$ and $B$ are same size invertible matrices then $AB$ is invertible.

Note that $(AB)(\inv B;\inv A;)$ $=$ $A(B\inv B;)\inv A;$ $=$ $A(I)\inv A;$ $=$ $A\inv A;$ $=$ $I.$

Let $A$ $=$ $\threematb[1 0 2 1 4 3 0 3 1].$

[10] Find $\tran A;.$

$\tran A;.$ $=$ $\threematb[1 1 0 0 4 3 2 3 1].$

[10] Find $\inv A;.$

[3]()

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1 & 0... ...& 4 & 3 & 0 & 1 & 0\\ 0 & 3 & 1 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1 & 0... ...t} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1 & 0... ...t} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1 & 0... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 3R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & -5 & ... ...} \begin{tabular}{l} \tiny\(R_{{}_{1}} - 2R_{{}_{3}}\)\\ \\ \\ \end{tabular}$

$\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -5 & 6 & -8\\ -1 & 1 & -1\\ 3 & -3 & 4 \end{tabular}\hspace{-4pt}\right]$

[10] Give the $LU$ factorization of $A.$

[3]()

$\threematb[1 0 2 1 4 3 0 3 1]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 4 ... ...R_{{}_{2}} - R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - 0R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 4 ... ...tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - \fric<3,4>R_{{}_{2}}\)\\ \end{tabular}$

$U$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 4 & 1\\ 0 & 0 & \(\fric<1,4>\) \end{tabular}\hspace{-4pt}\right]$

$L$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 1 & 1 & 0\\ 0 & \(\fric<3,4>\) & 1 \end{tabular}\hspace{-4pt}\right]$

$\threematb[1 0 2 1 4 3 0 3 1]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 1 & 1 & 0\\ 0 & \(\fric<3,4>\) & 1 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 4 & 1\\ 0 & 0 & \(\fric<1,4>\) \end{tabular}\hspace{-4pt}\right]$

[10] Solve $A\vx$ $=$ $\cvt[4 11 6 ].$

[3]()

$A\vx$ $=$ $\cvt[4 11 6 ]$

$\inv A;A\vx$ $=$ $\inv A;\cvt[4 11 6 ]$

$\vx$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -5 & 6 & -8\\ -1 & 1 & -1\\ 3 & -3 & 4 \end{tabular}\hspace{-4pt}\right] \cvt[4 11 6 ]$

$\vx$ $=$ $\cvt[-2 1 3 ]$

$A\vx$ $=$ $\cvt[4 11 6 ]$

$LU\vx$ $=$ $\cvt[4 11 6 ]$

$L\vc$ $=$ $\cvt[4 11 6 ]$

$\vc$ $=$ $\cvt[4 7 \(\fric<3,4>\) ]$

$U\vx$ $=$ $\vc$

$\vx$ $=$ $\cvt[-2 1 3 ]$

[10] Show that $\Rss 2$ $=$ span$\left(\cv[1 1 ],\cv[0 1 ]\right).$

Note that for any $\cv[\(a\) \(b\) ]\in\Rss2 ,$ $\cv[\(a\) \(b\) ]$ $=$ $a\cv[1 1 ] + (b - a)\cv[0 1 ].$

[10] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1 & 1... ...& 5 & 5 & 5 & 5 & 5\\ 6 & 6 & 6 & 6 & 6 & 6 \end{tabular}\hspace{-4pt}\right]$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 4 & 4 & 4 & 4 & 4... ...& 2 & 2 & 2 & 2 & 2\\ 6 & 6 & 6 & 6 & 6 & 6 \end{tabular}\hspace{-4pt}\right]$. Find a matrix $E$ such that $EA$ $=$ $B.$

The matrix $E$ is the permutation matrix $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 0 & 0 & 0 & 1 & 0... ... 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

(Exam 3)

Exam 3 Math 3720 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[30] Find the row space, column space, and null space of $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & 1\\ 3 & 4 & 4 \end{tabular}\hspace{-4pt}\right].$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 0 & 1 & 1\\ 3 & 4 & 4 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

row$(A)$ $=$ span$\left(\cvt[1 0 0 ], \cvt[0 1 1 ]\right)$ $=$ span$\left(\cvt[1 1 1 ], \cvt[0 1 1 ]\right)$

col$(A)$ $=$ span$\left(\cvt[1 0 3 ], \cvt[1 1 4 ]\right)$

null$(A)$ $=$ span$\left(\cvt[0 1 -1 ]\right)$

Suppose that $A$ is an $m\times n$ matrix and $m < n$. Then rank$(A) +$   nullity$(A) = n$ and rank$(A) \leq m$. Therefore, nullity$(A) \geq n - m \geq 1$. Since the dimension of the null space is at least 1, the null space is nontrivial.

Define $T:\Rss2 \to \Rss2$ by $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $\cv[\(x+y\) 0 ].$

[10] Show that $T$ is a linear transformation.

Suppose that $\cv[\(r\) \(s\) ],\cv[\(u\) \(v\) ]\in\Rss2$ and $a\in\R$.

Then $T\left(\cv[\(r\) \(s\) ] + \cv[\(u\) \(v\) ]\right)$ $=$ $T\left(\cv[{\(r + u\)} {\(s + v\)} ]\right)$ $=$ $\cv[{\(r + u + s + v\)} 0 ]$ $=$ $\cv[{\(r + s\)} 0 ] + \cv[{\(u + v\)} 0 ]$ $=$ $T\left(\cv[\(r\) \(s\) ]\right) + T\left(\cv[\(u\) \(v\) ]\right)$ and $T\left(a\cv[\(r\) \(s\) ]\right)$ $=$ $T\left(\cv[\(ar\) \(as\) ]\right)$ $=$ $\cv[{\(ar + as\)} 0 ]$ $=$ $a\cv[{\(r + s\)} 0 ]$ $=$ $aT\left(\cv[\(r\) \(s\) ]\right).$

[10] Find $[T]$ (the standard matrix of $T$).

Since $\twomat[1 1 0 0 ]\cv[\(x\) \(y\) ]$ $=$ $\cv[\(x+y\) 0 ],$ $[T]$ $=$ $\twomat[1 1 0 0 ]$. Also, note that $T\left(\cv[1 0 ]\right)$ $=$ $\cv[1 0 ]$ and $T\left(\cv[0 1 ]\right)$ $=$ $\cv[1 0 ]$ are the columns of $[T].$

[3]( For each of the following, find the eigenvalues and corresponding eigenspaces.)

[10]

$\twomat[1 2 2 1 ]$

$\twomat\vert\(1-\lambda\) 2 2 \(1-\lambda\) \vert$

$=$ $(1-\lambda)^{2} - 4$

$=$ $\lambda^{2} - 2\lambda - 3$

$=$ $(\lambda - 3)(\lambda + 1)$

Eigenvalues: $\tm1, 3$

$\twomat[2 2 2 2 ]$ $\rref$ $\twomat[1 1 0 0 ]$

$E_{{}_{\tm1}}$ $=$ span$\left(\cv[1 -1 ]\right)$

$E_{{}_{3}}$ $=$ span$\left(\cv[1 1 ]\right)$

[2]()

[10]

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Eigenvalues: $0, 1$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$E_{{}_{0}}$ $=$ span$\left(\cvt[1 -1 1 ]\right)$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 2 & 1\\ 0 & 0 & 1\\ 0 & 0 & -1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$E_{{}_{1}}$ $=$ span$\left(\cvt[1 0 0 ]\right)$

Calculate the following determinants.

[10] $\left\vert\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $1\twomat\vert 1 1 2 3 \vert - 0\twomat\vert 1 1 3 \vert + 2\twomat\vert 1 1 2 \vert$ $=$ $1 + 2(\tm1)$ $=$ $\tm 1$

[10]

$\left\vert\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4\... ... 4\\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12 \end{tabular}\hspace{-4pt}\right\vert$ $=$ 0

Note that since row 1 and row 2 are the same, the matrix is not invertible. Therefore, the determinant is $0.$

[10]

$\left\vert\hspace{-4pt} \begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0... ... 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $1$

Note that the matrix is formed by applying two row changes to the identity. Therefore, the determinant is $(\tm1)(\tm1)$ $=$ $1.$

(Final)

Final Exam Math 3720 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[10] Let $\theta$ be the angle between $\vv$ $=$ $\cvt[1 0 1 ]$ and $\vw$ $=$ $\cvt[2 1 0 ]$. Find $\cos\theta$.

[10] Solve the following system.

$x$	$+$	$y$	$+$	$z$	=	$1$
$2x$	$-$	$y$	$+$	$z$	=	$6$
$x$	$+$	$2y$	$-$	$z$	=	$\tm5$

[10] Is the following set of vectors linearly independent or linearly dependent?

$\left\{\cvt[1 2 -1 ], \cvt[0 3 2 ], \cvt[2 1 -4 ]\right\}$

[10] Show that $\Rss3$ $=$ span$\left(\cvt[1 0 0 ], \cvt[1 1 0 ], \cvt[1 1 1 ]\right).$

[10] Give a proof or reasonable explanation of the fact that for any matrix $A,$ $A\tran A;$ is square.

[10] Let $A$ $=$ $\threematb[1 2 1 0 1 0 1 0 1]$ and $B$ $=$ $\threematb[2 4 2 1 0 1 0 1 0]$. Find a matrix $E$ such that $EA$ $=$ $B.$

[10] Give the $\tran P;LU$ factorization of $A$ $=$ $\threematb[4 3 3 0 -1 -4 2 1 0].$

Let $\vW$ $=$ span$\left(\cvf[1 1 0 0 ],\cvf[1 0 1 1 ]\right)$ and $A$ $=$ $\twofourmatb[1 1 0 0 1 0 1 1].$

[40] Find row$(A),$ null$(A),$ col$(A),$ and null$\left(\tran A;\right).$

[10] Use an answer from the previous part to find $\per \vW;.$

Let $A$ $=$ $\twomat[-1 -2 3 4 ].$

[10] Find the eigenvalues and corresponding eigenspaces.

[10] Give matrices $D$ and $P$ such that $D$ is a diagonal matrix, $P$ is an invertible matrix, and $\inv P;AP$ $=$ $D.$

[2]( For each of the following, determine whether the given function from $\Rss 2$ to $\Rss 2$ is a linear transformation or not.)

[10] $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $\vzero$

[10] $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $\cv[\(x\) \({\y3 }\) ]$

[10] Let $A$ by any $2\times 2$ matrix and $M_{{}_{A}}$ $=$ $\{B\in M_{{}_{2\times2}}:AB = BA\}$. Show that $M_{{}_{A}}$ is a subspace of $M_{{}_{2\times2}}.$

(Fall 2017)

(Quiz)

[1] Solve the following.

[2]()

$2x$	$+$	$y$	=	1
$\tm x$	$+$	$2y$	=	12

$2x$	$+$	$y$	=	1
$\tm2x$	$+$	$4y$	=	24

$5y$ $=$ $25$

$y$ $=$ $5$

$x$ $=$ $\tm2$

$(\tm2,5)$

[2] Find the distance between the point $(5,\tm9)$ and the line $\tm3x + 2y$ $=$ $6.$

Find the equation of the line through the point $(5,\tm9)$ and perpendicular to the line

The slope of the line $\tm3x + 2y$ $=$ $6$ is $\fric<3,2>.$

The slope of a line perpendicular to $\tm3x + 2y$ $=$ $6$ is $\tm\fric<2,3>.$

$y + 9$ $=$ $\tm\fric<2,3>(x - 5)$

$2x + 3y$ $=$ $\tm17$

Find the point of intersection of the two lines.

[2]()

$2x$	$+$	$3y$	=	$\tm17$
$\tm3x$	$+$	$2y$	=	6

$6x$	$+$	$9y$	=	-51
$\tm6x$	$+$	$4y$	=	12

$13y$ $=$ $\tm39$

$y$ $=$ $\tm3$

$x$ $=$ $\tm4$

The distance between the points $(5,\tm9)$ and $(\tm4,\tm3)$ is $\sqrt{117}.$

[2]( Find all solutions to the following systems of linear equations.)

[1]

$x$	$+$	$y$	$+$	$z$	=	2
$2x$	$-$	$y$	$$	$$	=	3
$x$	$+$	$2y$	$-$	$z$	=	-3

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 2\\ 2 & -1 & 0 & 3\\ 1 & 2 & -1 & -3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 2\\ 0 & -3 & -2 & -1\\ 0 & 1 & -2 & -5 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - 2R_{{}_{1}}\) \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 3 & 7\\ 0 & 1 & -2 & -5\\ 0 & 0 & -8 & -16 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \tiny\(R_{{}_{1}} - R_{{}_{3}}\) \\ \tiny\(R_{{}_{3}}\) \\ \tiny\(R_{{}_{2}} + 3R_{{}_{3}}\) \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 3 & 7\\ 0 & 1 & -2 & -5\\ 0 & 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \tiny\(\tm\fric<1,2>R_{{}_{3}}\) \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \tiny\(R_{{}_{1}} - 3R_{{}_{3}}\) \\ \tiny\(R_{{}_{2}} + 2R_{{}_{3}}\) \\ \\ \end{tabular}$

Solution: $\left(1,\tm1,2\right)$

[1]

$x$	$-$	$y$	$+$	$z$	=	0
$x$	$+$	$3y$	$-$	$z$	=	0
$x$	$-$	$9y$	$+$	$5z$	=	0

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -1 & 1 & 0\\ 1 & 3 & -1 & 0\\ 1 & -9 & 5 & 0 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -1 & 1 & 0\\ 0 & 4 & -2 & 0\\ 0 & -8 & 4 & 0 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\) \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -1 & 1 & 0\\ ... ...\tiny\(\tm\fric<1,2>\)} & 0\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \tiny\(\fric<1,4>R_{{}_{2}}\) \\ \tiny\(R_{{}_{3}} + 2R_{{}_{2}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & {\tiny\(\fr... ...\tiny\(\tm\fric<1,2>\)} & 0\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \\ \end{tabular}$

Free variable: $z$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 4 & 7\\ 0 & 1 & 8 \end{tabular}\hspace{-4pt}\right],$ $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 6\\ 2 & -1\\ -3 & 0 \end{tabular}\hspace{-4pt}\right],$ $C$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & -1 & 4 & 2 & 1\\ 0 & 1 & 8 & 7 & 0\\ -1 & -5 & 6 & 17 & -6 \end{tabular}\hspace{-4pt}\right],$ and $D$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ -1 & 1 & -4 & -2 & ... ...0 & -1 & -8 & -7 & 0\\ 1 & 5 & -6 & -17 & 6 \end{tabular}\hspace{-4pt}\right]$.

Note that $D$ $=$ $\tm C$. Calculate each of the following.

[1] $AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -12 & 2\\ -22 & -1 \end{tabular}\hspace{-4pt}\right]$

[1] $BA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 10 & 55\\ 2 & 7 & 6\\ -3 & -12 & -21 \end{tabular}\hspace{-4pt}\right]$

[1] $AC + AD$ $=$ $A(C + D)$ $=$ $A0_{{}_{3\times5}}$ $=$ $0_{{}_{2\times5}}$

[2] Express the following matrix as a product of elementary matrices.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 1\\ 2 & 0 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

The following elementary row operations transform the identity matrix to the one above.

Add row three to row two.

Multiply row 1 by 2.

Interchange rows 1 and 2.

So $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 1\\ 2 & 0 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 1 & 0 ... ...t}\\ 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

[2]( Calculate the determinant of each of the following matrices.)

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 3\\ -1 & 0 & 4\\ 2 & 1 & 8 \end{tabular}\hspace{-4pt}\right]$

Use the Cofactor Formula.

$\left\vert\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 3\\ -1 & 0 & 4\\ 2 & 1 & 8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $25$

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 31 & -1\\ 0 & 1 & 24\\ 0 & 0 & 8 \end{tabular}\hspace{-4pt}\right]$

Note that this matrix is upper triangular.

$\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 31 & -1\\ 0 & 1 & 24\\ 0 & 0 & 8 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $8$

[3] Let $\vV$ $=$ ${\bf M}_{{}_{2}}$ and $S$ $=$ $\left\{\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(a\) & 0\\ 0 & \(a\) \end{tabular}\hspace{-4pt}\right]:a\in\R\right\}$. Also, let $+:\vV\times\vV\to\vV$ be matrix addition and $\cdot:S\times\vV\to\vV$ be matrix multiplication. Show that $\vV$ is a vector space over $S.$

Verify the eight axioms.

Matrix addition is commutative.

Matrix addition is associative.

Let $\vzero$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 0 & 0 \end{tabular}\hspace{-4pt}\right].$

The additive inverse of $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(x\) & \(y\)\\ \(z\) & \(w\) \end{tabular}\hspace{-4pt}\right]$ is $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(\tm x\) & \(\tm y\)\\ \(\tm z\) & \(\tm w\) \end{tabular}\hspace{-4pt}\right]$ $=$ $\tm\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(x\) & \(y\)\\ \(z\) & \(w\) \end{tabular}\hspace{-4pt}\right].$

Left matrix multiplication distributes over matrix addition.

Right matrix multiplication distributes over matrix addition.

Matrix multiplication is associative.

Note that $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{... ...ce{-12pt}\\ \(x\) & \(y\)\\ \(z\) & \(w\) \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(x\) & \(y\)\\ \(z\) & \(w\) \end{tabular}\hspace{-4pt}\right].$

[2] Do the following vectors form a basis for $\Rss3 ?$

$\cvt[1 1 1 ],$ $\cvt[-1 1 1 ],$ $\cvt[1 -1 1 ]$

Yes.

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 1\\ 1 & 1 & -1\\ 1 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 0 & 0\\ 2 & 2 & 0\\ 1 & 1 & 1\\ \end{tabular}\hspace{-4pt}\right]$ $\begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \tiny\(R_{{}_{2}} + R_{{}_{3}}\)\\ \\ \end{tabular}$ which is formed by performing type III

row operations to $A.$ Note that $B$ is lower triangular and $\det B$ $=$ $4$. So $B,$ and hence $A,$ is invertible. Since $A$ is invertible, the equation $A\vx$ $=$ $\vzero$ has only $\vzero$ as a solution. Therefore, the columns of $A$ (the given vectors) are linearly independent. Three linearly independent vectors form a basis for $\Rss3$.

[2] Let $\vV$ be the vectors in $\Rss3$ that are in the plane $x - 3y - z$ $=$ 0. Find a basis for $\vV.$

A plane has dimension $2$. So we must find two linearly independent vectors in the plane.

One such pair is $\cvt[1 0 1 ]$ and $\cvt[0 1 -3 ].$

[3] For the following matrix, find a basis for the row space, a basis for the column space, and a basis for the null space.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 2\\ -1 & 0 & 2\\ 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 2\\ -1 & 0 & 2\\ 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right].$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 2\\ -1 & 0 & 2\\ 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 0 & 2\\ 0 & 1 & 2\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

row$(A)$ $=$ $\vspan\{[1$ 0 $\tm2], [0$ $1$ $2]\}$

col$(A)$ $=$ $\vspan\left(\left\{\cvt[1 -1 0 ],\cvt[2 0 1 ]\right\}\right)$

nul$(A)$ $=$ $\vspan\left(\left\{\cvt[2 -2 1 ]\right\}\right)$

(Exam 1)

Exam 1 Math 3720 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Solve each of the following completely.

[10]

$x$	$+$	$y$	=	3
$2x$	$+$	$3y$	=	5

Note that $\tm2$Eq$_{{}_{1}} +$   Eq$_{{}_{2}}$ yields $y$ $=$ $\tm 1$. So $(4,\tm1)$ is the solution.

[10]

$x$	$+$	$y$	$+$	$2z$	=	1
$\tm2x$	$-$	$y$	$$	$$	=	-4
$x$	$+$	$y$	$-$	$z$	=	4

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 2 & 1\\ -2 & -1 & 0 & -4\\ 1 & 1 & -1 & 4 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & -1 \end{tabular}\hspace{-4pt}\right]$

Solution: $(1,2,\tm1)$

[10]

$\tm2x$	$-$	$2y$	$+$	$z$	=	0
$3x$	$+$	$2y$	$-$	$3z$	=	-2

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -2 & -2 & 1 & 0\\ 3 & 2 & -3 & -2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & -2 & -2\\ ... ...1 & \raisebox{2pt}{{\Tiny\(\fric<3,2>\)}} & 2 \end{tabular}\hspace{-4pt}\right]$

Free Variable: $z$

$y$ $=$ $2 - \fric<3,2>z$

$x$ $=$ $\tm2 + 2z$

Solutions: $\left(\tm2 + 2z,2 - \fric<3,2>z,z\right)$

[10]

$2x$	$+$	$y$	$+$	$z$	=	1
$x$	$+$	$2y$	$+$	$z$	=	3
$3x$	$+$	$3y$	$+$	$2z$	=	5

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 1 & 1\\ 1 & 2 & 1 & 3\\ 3 & 3 & 2 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2... ...{{\Tiny\(\fric<1,3>\)}} & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

No solution.

Let $A$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & -1\\ 0 & 1 & 6 \end{tabular}\hspace{-4pt}\right],$ $B$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & -3\\ 2 & 1 & 6 \end{tabular}\hspace{-4pt}\right],$ $C$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ -1 & 2 & 4\\ 1 & 1 & 3 \end{tabular}\hspace{-4pt}\right],$ and $\vb$ $=$ $\cvt[1 -2 4 ]$.

Calculate each of the following.

[2]()

[10] $3A + B$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 3 & -5 & -6\\ 2 & 4 & 24 \end{tabular}\hspace{-4pt}\right]$

[10] $AC$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 2 & -5 & -10\\ 5 & 8 & 22 \end{tabular}\hspace{-4pt}\right]$

[10] $\inv C;$ $=$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ -2 & -1 & 2\\ -7 & -2 & 5\\ 3 & 1 & -2 \end{tabular}\hspace{-4pt}\right]$

[10] Solve $C\vx$ $=$ $\vb.$

$C\vx$ $=$ $\vb$

$\inv C;C\vx$ $=$ $\inv C;\vb$

$\vx$ $=$ $\inv C;\vb$

$\vx$ $=$ $\cvt[8 17 -7 ]$

[10] Explain why $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$ has no inverse.

Note that for any $2\times 2$ matrix $\left[\hspace{-4pt}\raisebox{-1.75pt}{\begin{tabular}{rr} \(a\) & \(b\)\\ \(c\) & \(d\) \end{tabular}}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & 0 \end{... ...{tabular}{rr} \(a\) & \(b\)\\ \(c\) & \(d\) \end{tabular}}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\raisebox{-1,5pt}{\begin{tabular}{rr} \(a + 2c\) & \(b + 2d\)\\ 0 & 0 \end{tabular}}\hspace{-4pt}\right]$ $\ne$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ \end{tabular}\hspace{-4pt}\right].$

[10] Prove only one of the following.

If $A$ is symmetric and invertible, then $\inv A;$ is symmetric.

Note that $\tranp {\inv A;};$ $=$ $\invp{\tran A;};$ $=$ $\inv A;.$

If $A$ and $B$ are invertible matrices such that $AB$ $=$ $BA,$ then $\inv A;\inv B;$ $=$ $\inv B;\inv A;.$

Note that $\inv A;\inv B;$ $=$ $\invp BA;$ $=$ $\invp AB;$ $=$ $\inv B;\inv A;.$

Total Points:

(Exam 2)

Exam 2 Math 3720 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

For each of the following, express the matrix as a product of elementary matrices and calculate the determinant.

[20] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 4 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 ... ...pt}\\ 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 4 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm8$

Since this is a diagonal matrix, the deteminant is the product of the diagnonal entries.

[20] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 0 ... ...pt}\\ 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm 1$

Note that this matrix is formed by applying two type III row operations and one type I row operation to the identity matrix which has determinant $1$. The type III row operations do not affect the determinant and the type I row operation changed its sign.

Given that $A$ is a $3\times3$ matrix with $\det A$ $=$ $3,$ calculate the determinant of each of the following.

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right] A$

$\tm3$

A type I row operation changes the sign of the determinant.

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right] A$

$3$

A type III row operation does not affect the determinant.

[10] $\inv A;$

$\fric<1,3>$

$\det\left(\inv A;\right)$ $=$ $\fric<1,\det A>$

[10] Answer the following as true or false (write the entire word). If the statement is true, then prove it. If the statement is false, then give a counter example.

If $A$ and $B$ are square matrices with the same dimension, then $\det(A + B)$ $=$ $\det A + \det B.$

False.

Let $A$ $=$ $B$ $=$ $I$. Then $\det(A + B)$ $=$ $4$ and $\det A + \det B$ $=$ $2.$

[10] Let $\vV$ $=$ $M_{{}_{22}},$ $A$ be a $2\times 2$ matrix, and $\vW$ $=$ $\{B\in M_{{}_{22}}: AB = 0\}$. Show that $\vW\leq\vV.$

Suppose that $C,D\in\vW$ and $\alpha\in\R$. Then $A(C + D)$ $=$ $AC + AD$ $=$ $0 + 0$ $=$ 0 and $A(\alpha C)$ $=$ $\alpha(AC)$ $=$ $\alpha0$ $=$ 0. So $\vW$ is closed under vector addition and scalar multiplication.

[10] Show that $X$ $=$ $\left\{\cvt[\(a\) \(b\) 1 ]: a,b \in\R\right\}$ is not a subspace of $\Rss 3 .$

Note that $\cvt[1 0 1 ],\cvt[0 1 1 ]\in X$ but $\cvt[1 0 1 ] + \cvt[0 1 1 ]$ $=$ $\cvt[1 1 2 ]\notin X.$

For each of the following, determine whether the vectors are linearly independent or linearly dependent. Also, determine whether or not the vectors span $\Rss 3 .$

[20] $\cvt[1 2 0 ],$ $\cvt[3 4 0 ],$ $\cvt[2 2 0 ]$

Dependent.

Note that $\cvt[1 2 0 ] + \cvt[2 2 0 ] - \cvt[3 4 0 ]$ $=$ $\cvt[0 0 0 ]$.

Since any vector in the span of these three vectors must have 0 as its third component, these vectors do not span $\Rss 3 .$

[20] $\cvt[1 0 0 ],$ $\cvt[1 1 0 ],$ $\cvt[0 1 1 ]$

These vectors are linarly independent and do span $\Rss 3 .$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ be the matrix whose columns are the three vectors.

Since $A$ is an upper triangular matrix, the determinant of $A$ is the product of its diagonal entries which is 1. So $A$ is invertible. So the equation $A\vx$ $=$ $\vzero$ has a unique solution in $\Rss3$ which implies the columns of $A$ are linearly independent. Also, for any $\vb\in\Rss 3 ,$ $A\vx$ $=$ $\vb$ has a unique solution which implies that the columns of of $A$ span $\Rss 3 .$

Total Points:

(Exam 3)

Exam 3 Math 3720 Fall 2017

For each of the following subspaces of $\Rss 3 ,$ give a basis.

[2]()

[10] The plane $2x + y - 3z$ $=$ $0.$

$\left\{\cvt[1 -2 0 ],\cvt[0 3 1 ]\right\}$

[10] The line $x$ $=$ $\fric<y,3>$ $=$ $\fric<z,4>.$

$\left\{\cvt[1 3 4 ]\right\}$

[10] Is the following set a basis for $P_{{}_{2}}?$

$\{2x, x + 1, \x2 - 1\}$

Yes.

Note that $1$ $=$ $(x + 1) - \fric<1,2>(2x),$ $x$ $=$ $\fric<1,2>(2x),$ and $\x2$ $=$ $(\x2 - 1) + \fric<1,2>(2x)$. So $P_{{}_{2}}$ $=$ span$\left(\{1,x,\x2 \}\right) \subseteq$span$\left(\{2x,x + 1,\x2 -1\}\right)$ which means that span$\left(\{2x,x + 1,\x2 -1\}\right)$ $=$ $P_{{}_{2}}.$

[2]( Give the row space, column space, and null space of the following matrices.)

[10] $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & 0 & 2\\ 0 & 2 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & 0 & 2\\ 0 & 2 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

So $A$ is an invertible $3\times3$ matrix. Hence, row$(A)$ $=$ col$(A)$ $=$ $\Rss3$ and null$(A)$ $=$ $\left\{\cvt[0 0 0 ]\right\}.$

[2]()

[10] $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 0... ...1 & 1 & 4\\ 1 & 0 & 2 & 5\\ 3 & 2 & 4 & 7 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 0... ...1 & 1 & 4\\ 1 & 0 & 2 & 5\\ 3 & 2 & 4 & 7 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 5\\ 0... ...& -1 & -4\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

row$(A)$ $=$ $\vspan\{[1$ 0 $2$ $5], [0$ $1$ $\tm 1$ $\tm4]\}$

col$(A)$ $=$ $\vspan\left(\cvf[1 0 1 3 ],\cvf[1 -1 0 2 ]\right)$

null$(A)$ $=$ $\vspan\left(\cvf[-2 1 1 0 ],\cvf[-5 4 0 1 ]\right)$

[10] How many rows of zeros are in the reduced row echelon form of a $4\times 6$ matrix with nullity $3?$

One. Since rank$+$   nullity $=$ $6,$ rank $=$ $3$. So the matrix has 3 pivots which means that the reduced row echelon form of the matrix has 3 rows which contain nonzero entries and one row of zeros.

Let $\beta$ $=$ $\left\{\cvt[0 1 1 ],\cvt[1 0 1 ],\cvt[1 1 0 ]\right\}$ and $\mu$ $=$ $\left\{\cvt[1 0 0 ],\cvt[0 1 0 ],\cvt[0 0 1 ]\right\}.$

[10] Show that $\beta$ is a basis for $\Rss 3 .$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$. Verify that $A$ is invertible and $\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ {\tiny\(\tm\fric<1,2>... ...iny\(\fric<1,2>\)} & {\tiny\(\tm\fric<1,2>\)} \end{tabular}\hspace{-4pt}\right]$. Since $A$ is invertible, its columns are linearly independent vectors in $\Rss3$. Hence, the columns of $A$ form a basis for $\Rss3$.

[10] Give the transition matrix from $\beta$ to $\mu.$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

[10] Give the transition matrix from $\mu$ to $\beta.$

$\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ {\tiny\(\tm\fric<1,2>... ...iny\(\fric<1,2>\)} & {\tiny\(\tm\fric<1,2>\)} \end{tabular}\hspace{-4pt}\right]$

[10] Write $\cvt[4 -1 1 ]_{{}_{\mu}}$ in terms of $\beta.$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ {\tiny\(\tm\fric<1,2>... ...2>\)} & {\tiny\(\tm\fric<1,2>\)} \end{tabular}\hspace{-4pt}\right]\cvt[4 -1 1 ]$ $=$ $\cvt[-2 3 1 ]$

Verify that $\tm2\cvt[0 1 1 ] + 3\cvt[1 0 1 ] + \cvt[1 1 0 ]$ $=$ $\cvt[4 -1 1 ].$

(Final)

Final Exam Math 3720 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

For each of the following, find all solutions to the equation $A\vx$ $=$ $\vb.$

[10] $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 2 & 1\\ -1 & 1 \end{tabular}\hspace{-4pt}\right],$ $\vb$ $=$ $\cv[3 -3 ]$

[10] $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 2 & 1 & 2\\ -1 & 2 & 1 \end{tabular}\hspace{-4pt}\right],$ $\vb$ $=$ $\cvt[3 8 1 ]$

[10] Multiply the following matrices. Hint: Use block multiplication. Partition each matrix into four $2\times 2$ matrices.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 0... ...0 & 0 & 1\\ 1 & 2 & 1 & 2\\ 3 & 4 & 3 & 4 \end{tabular}\hspace{-4pt}\right]$

[10] Does there exist a quadratic function $f(x)$ $=$ $a\x2 + bx + c$ such that $f(1)$ $=$ $5,$ $f(\tm1)$ $=$ $7,$ and $f'(2)$ $=$ $7$? Hint: Create a system of equations with unknown variables $a,$ $b,$ and $c.$

[10] Calculate the determinant of the following matrix. Is it invertible?

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 1 & 6\\ 0 & 0 & -3 \end{tabular}\hspace{-4pt}\right]$

[10] Give an example of two square matrices $A$ and $B$ such that $\vert A + B\vert$ $\ne$ $\vert A\vert + \vert B\vert.$

[10] For the following matrix, find a basis for the row space, a basis for the column space, and a basis for the null space.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 1 & 1\\ 2... ... & 0 & -1\\ 4 & 1 & 2 & 1\\ 3 & 1 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

[10] Give a basis for the plane $2x + 3y - z$ $=$ 0 in $\Rss 3 .$

[10] Choose one of the following.

Prove that the inverse of an invertible matrix is unique.

Prove that the product of two invertible matrices is invertible.

[10] Choose one of the following.

Prove that if $\lambda$ is an eigenvalue of an invertible matrix $A,$ then $\fric<1,\lambda>$ is an eigenvalue of $\inv A;.$

Prove that if $\lambda$ is an eigenvalue of a matrix $A,$ then $\lambda^{{}^{n}}$ is an eigenvalue of $A^{{}^{n}}.$

[10] Answer the following as true or false (write the entire word). If the statement is true, then prove it. If the statement is false, then give a counter example.

A type I elementary matrix is an orthogonal matrix.

For each of the following, determine whether or not the given function is a linear transformation. If it is, give its range and kernel.

[10] $T: \Rss3 \to \Rss 3$ defined by $T\left(\cvt[\(x\) \(y\) \(z\) ]\right)$ $=$ $\cvt[\(x\) \(y\) 2 ]$

[10] $T:\Rss2 \to \R$ defined by $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $x + y$

Let $A$ $=$ $\twomat[-8 10 -5 7 ].$

[10] Find the eigenvalues and corresponding eigenspaces of $A.$

[10] Is $A$ diagonalizable? If so, find an invertible matrix $P$ such that $\inv P;AP$ is a diagonal matrix.

(Summer 2018)

(Quizzes)

[05-16-18]

Let $\vv$ $=$ $\rvf[1 \tm1 0 2 ]$ and $\vw$ $=$ $\rvf[3 \tm2 1 4 ].$

[3]()

[1] Calculate $2\vv - 3\vw.$

$2\vv - 3\vw$ $=$ $\rvf[\tm7 4 \tm3 \tm8 ]$

[1] Calculate $\vv\cdot\vw.$

$\vv\cdot\vw$ $=$ $3 + 2 + 0 + 8$ $=$ $13$

[1] Calculate $\Vert\vv\Vert.$

$\Vert\vv\Vert$ $=$ $\sqrt{1 + 1 + 0 + 4}$ $=$ $\sqrt{6}$

[2] If possible, find $\alpha$ and $\beta$ such that $\alpha\vv + \beta\vw$ $=$ $\rvf[1 0 1 0 ]$. If this is not possible, explain why it is not.

Suppose that $\alpha\vv + \beta\vw$ $=$ $\rvf[1 0 1 0 ]$. Then $\rvf[\(\alpha\) \(\tm\alpha\) 0 \(2\alpha\) ] + \rvf[\(3\beta\) \(\tm2\beta\) \(\beta\) \(4\beta\) ]$ $=$ $\rvf[1 0 1 0 ]$ which implies

$\alpha$	$+$	$3\beta$	=	1
$\tm\alpha$	$-$	$2\beta$	=	0
0	$+$	$\beta$	=	1
$2\alpha$	$+$	$4\beta$	=	0

From the third equation, we see that $\beta$ $=$ $1$. Substituting in any of the other three equations, yields $\alpha$ $=$ $\tm2$. So $\tm2\vv + \vw$ $=$ $\rvf[1 0 1 0 ].$

[05-18-18]

Let $\vv$ $=$ $\rvt[1 \tm3 2 ],$ $\vw$ $=$ $\rvt[2 0 5 ],$ and $\theta$ be the angle between $\vv$ and $\vw$.

[2]( [2] Calculate each of the following.)

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<12,\sqrt{14}\sqrt{29}>$

$\sin\theta$ $=$ $\sqrt{1 - \cos^{2}\theta}$ $=$ $\sqrt{1 - \fric<144,14\cdot29>}$ $=$ $\sqrt{\fric<131,203>}$

[1] Find two vectors $\va$ and $\vb$ with the following three properties:

$\va + \vb = \vw;$

$\va$ and $\vv$ are parallel;

$\vb$ and $\vv$ are orthogonal.

Let $\va$ $=$ proj$_{{}_{\vv}}\vw$ $=$ $\fric<6,7>\rvt[1 \tm3 2, ]$ $=$ $\rvt[\(\fric<6,7>\) \(\tm\fric<18,7>\) \(\fric<12,7>\) ]$ and $\vb$ $=$ $\vw - \va$ $=$ $\vw -$   proj$_{{}_{\vv}}\vw$ $=$ $\rvt[\(\fric<8,7>\) \(\fric<18,7>\) \(\fric<23,7>\) ].$

Note that $\va + \vb = \vw,$ $\va$ and $\vv$ are parallel since $\va$ $=$ $\fric<6,7>\vv,$ and $\vb$ and $\vv$ are orthogonal since $\vb\cdot\vv$ $=$ $0.$

[1] Suppose that $\vv$ and $\vw$ are nonzero vectors in $\Rss3$ such that the angle between them is 0. Explain why $\fric<\Vert\vw\Vert,\Vert\vv\Vert>\vv$ $=$ $\vw.$

Since the angle between $\vv$ and $\vw$ is $0,$ $\vv$ and $\vw$ have the same direction. Also, $\fric<\Vert\vw\Vert,\Vert\vv\Vert>\vv$ and $\vv$ have the same direction since they are scalar multiples of each other. Also, note that $\left\Vert\fric<\Vert\vw\Vert,\Vert\vv\Vert>\vv\right\Vert$ $=$ $\Vert\vw\Vert$. So $\fric<\Vert\vw\Vert,\Vert\vv\Vert>\vv$ is the vector in the direction of $\vw$ with magnitude $\Vert\vw\Vert$ which means that $\fric<\Vert\vw\Vert,\Vert\vv\Vert>\vv$ is $\vw.$

[05-23-18]

[3] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 5\\ 0 & 2 & 1 \end{tabular}\hspace{-4pt}\right]$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 2 & 3\\ -1 & 2 \end{tabular}\hspace{-4pt}\right]$. Perform the indicated operation if possible. If it is not possible, explain why it is not.

$A + B$

This is not possible since $A$ and $B$ have different dimensions.

$AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 5\\ 0 & 2... ...\vspace{-12pt}\\ 1 & -1\\ 2 & 3\\ -1 & 2 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -6 & 6\\ 3 & 8 \end{tabular}\hspace{-4pt}\right]$

$BA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 2 & 3\\ -... ...} \vspace{-12pt}\\ 1 & -1 & 5\\ 0 & 2 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -3 & 4\\ 2 & 4 & 13\\ -1 & 5 & -3 \end{tabular}\hspace{-4pt}\right]$

[1] Explain why $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 1\\ 0 & 2 \end{tabular}\hspace{-4pt}\right]$ is not invertible.

Note that for any $2\times 2$ matrix $\left[\hspace{-4pt}\raisebox{-1.75pt}{\begin{tabular}{rr} \(a\) & \(b\)\\ \(c\) & \(d\) \end{tabular}}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 1\\ 0 & 2 \end{... ...{tabular}{rr} \(a\) & \(b\)\\ \(c\) & \(d\) \end{tabular}}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\raisebox{-1,5pt}{\begin{tabular}{rr} \(c\) & \(d\)\\ \(2c\) & \(2d\) \end{tabular}}\hspace{-4pt}\right]$ $\ne$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{tabular}\hspace{-4pt}\right]$.

Also, note that $\left[\hspace{-4pt}\raisebox{-1.75pt}{\begin{tabular}{rr} \(a\) & \(b\)\\ \(c\... ...bular}{rr} \vspace{-12pt}\\ 0 & 1\\ 0 & 2 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\raisebox{-1,5pt}{\begin{tabular}{rr} 0 & \(a + 2b\)\\ 0 & \(c + 2d\) \end{tabular}}\hspace{-4pt}\right]$ $\ne$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{tabular}\hspace{-4pt}\right]$.

[1] Suppose that $A$ and $B$ are matrices such that $\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 0 & 1 \end{tabular}\hspace{-4pt}\right]$ and $AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 8 & 3\\ -1 & 5 & 4 \end{tabular}\hspace{-4pt}\right]$. Find $B.$

$B$ $=$ $\inv A;AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 0 & 1 \end... ...r} \vspace{-12pt}\\ 1 & 8 & 3\\ -1 & 5 & 4 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 3 & -1\\ -1 & 5 & 4 \end{tabular}\hspace{-4pt}\right]$

[05-25-18]

[2]( Use Gauss-Jordan elimination to transform the given matrix into reduced row echelon form.)

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 0 & 2\\ -1 & 1 & 8 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ -1 & 1 ... ...ace{-8pt} \begin{tabular}{l} \tiny\(\fric<1,2>R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + R_{{}_{1}}\)\\ \end{tabular}$

[1] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \raisebox{2pt}{\tiny\(\fric<1,2>\)} & 1 \\ -1 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ -1 & 3 \end{... ...ght] \hspace{-8pt} \begin{tabular}{l} \tiny\(2R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & 5 \end{... ...{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & 1 \end{... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,5>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} - 2R_{{}_{2}}\)\\ \\ \end{tabular}$

[3] For the given matrix $A,$ calculate $\inv A;,$ $\tran A;,$ and $\invp{\tran A;};.$

[2]()

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 2 & -1 & -2\\ 0 & 1 & 5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ...-1 & -2 & 0 & 1 & 0\\ 0 & 1 & 5 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ...} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - 2R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ...t} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} + R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ...} \\ \tiny\(\tm R_{{}_{2}}\)\\ \tiny\(\fric<1,5>R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & \ra... ...t} \begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{3}}\)\\ \\ \\ \end{tabular}$

$\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{\tiny... ...1,5>\)} & \raisebox{2pt}{\tiny\(\fric<1,5>\)} \end{tabular}\hspace{-4pt}\right]$

$\tran A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 0 & -1 & 1\\ -1 & -2 & 5 \end{tabular}\hspace{-4pt}\right]$

$\invp{\tran A;};$ $=$ $\tranp {\inv A;};$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{\tiny\... ...\)} & 0 & \raisebox{2pt}{\tiny\(\fric<1,5>\)} \end{tabular}\hspace{-4pt}\right]$

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 0 & -1 & 1\\ -1 & -2 & 5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 0 & 1 & ... ...-1 & 1 & 0 & 1 & 0\\ -1 & -2 & 5 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 0 & 1 & ... ...\ \tiny\(\tm R_{{}_{2}}\)\\ \tiny\(R_{{}_{3}} + R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 2 & 1 & ... ..._{{}_{1}} - 2R_{{}_{2}}\)\\ \\ \tiny\(\fric<1,5>R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & \rai... ...}_{1}} - 2R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} + R_{{}_{3}}\)\\ \\ \end{tabular}$

$\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{\tiny\... ...\)} & 0 & \raisebox{2pt}{\tiny\(\fric<1,5>\)} \end{tabular}\hspace{-4pt}\right]$

$\tran A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 2 & -1 & -2\\ 0 & 1 & 5 \end{tabular}\hspace{-4pt}\right]$

$\invp{\tran A;};$ $=$ $\tranp {\inv A;};$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{\tiny... ...1,5>\)} & \raisebox{2pt}{\tiny\(\fric<1,5>\)} \end{tabular}\hspace{-4pt}\right]$

Calculate the following determinants.

[1] $\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -2\\ -1 & 4 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $4 - 2$ $=$ $2$

[1] $\left\vert\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 2 & -3 & 10\\ 1 & 0 & 1\\ -4 & 10 & 12 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -3 & 10\\ 10 ... ...rr} \vspace{-12pt}\\ 2 & -3\\ -4 & 10 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm136 - 8$ $=$ $\tm128$

[1] $\left\vert\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\\ 0 & 3 & 3 & 3\... ...0 & 1\\ 4 & 4 & 4 & 4\\ 0 & 0 & 2 & 2 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $(\tm1)\left\vert\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\\ 0 & 3 & ... ...4 & 4\\ 0 & 0 & 0 & 1\\ 0 & 0 & 2 & 2 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $(\tm1)(\tm1)\left\vert\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\\ 0 ... ...4 & 4\\ 0 & 0 & 2 & 2\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $(\tm1)(\tm1)(\tm1)\left\vert\hspace{-4pt} \begin{tabular}{rrrr} \vspace{-12pt}\... ...3 & 3\\ 0 & 0 & 2 & 2\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right\vert$ $\tm24$

[1] Prove that the additive identity (zero vector) in a vector space is unique.

See problem 53.

[1] Let $A$ be an $m\times n$ matrix. Prove that $N(A)\leq\Rss n .$

See problem 60.

Transform the matrix to reduced row echelon form. For each free column, find a nonzero vector in the null space.

[1] $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -2 & 0 & 0\\ 0 & 0 & 3 & -5\\ 0 & 0 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

Free column: 2

$\cvf[2 1 0 0 ]$

[1] $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 2\\ 0 & 1 & 1\\ 1 & 2 & 3\\ 1 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Free column: 3

$\cvt[-1 -1 1 ]$

[1] $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 0 & 1 & 0 & -3 \\ 0 & 1 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 6 \end{tabular}\hspace{-4pt}\right]$

Free columns: 3, 5

$\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -1\\ -2\\ 1\\ 0\\ 0 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ 3\\ -1\\ 0\\ -6\\ 1 \end{tabular}\hspace{-4pt}\right]$

[2]( Determine whether the given vectors are linearly independent or linearly dependent.)

[1] $\cv[1 -1 ],$ $\cv[0 7 ]$

Independent.

Using column vectors:

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ -1 & 7 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmattwo$

Pivots: 2

Using row vectors:

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 0 & 7 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmattwo$

No zero rows.

[1] $\cv[1 4 ],$ $\cv[\raisebox{5pt}{\(\tm\pi\)} \(\sqrt{2}\) ],$ $\cv[27 -6 ]$

Dependent.

Three vectors in $\Rss 2 .$

[1] $\cvt[-6 1 8 ],$ $\cvt[4 1 2 ]$

Independent.

Niether vector is a multiple of the other.

Using column vectors:

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -6 & 4\\ 1 & 1\\ 8 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Pivots: 2

Using row vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ -6 & 1 & 8\\ 4 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2... ...& 1 & \raisebox{2pt}{\tiny\(\fric<22,5>\)}\\ \end{tabular}\hspace{-4pt}\right]$

No zero rows.

[1] $\cvt[-3 -4 4 ],$ $\cvt[2 3 -5 ],$ $\cvt[6 5 13 ]$

Dependent.

Using column vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ -3 & 2 & 6\\ -4 & 3 & 5\\ 4 & -5 & 13 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -8\\ 0 & 1 & -9\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Pivots: 2

Using row vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ -3 & -4 & 4\\ 2 & 3 & -5\\ 6 & 5 & 13 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 8\\ 0 & 1 & -7\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Row of zeros.

[2] Find a vector $\vv\in\Rss3$ such that $\left\{\cvt[1 1 1 ],\cvt[1 1 0 ],\vv\right\}$ is a basis for $\Rss 3 .$

Choose any vector not in the span of $\left\{\cvt[1 1 1 ],\cvt[1 1 0 ]\right\}$.

One such example is $\cvt[1 0 0 ]$.

[2] Find an orthonormal set with the same span as $\left\{\cvt[1 1 1 ],\cvt[1 0 2 ]\right\}.$

Use the Gram-Schmidt orthogonalization process.

Let $\vv_{{}_{1}}$ $=$ $\cvt[1 1 1 ]$ and $\vv_{{}_{2}}$ $=$ $\cvt[1 0 2 ]$.

Also, !et $\vw_{{}_{1}}$ $=$ $\vv_{{}_{1}}$ $=$ $\cvt[1 1 1 ]$ and $\vw_{{}_{2}}$ $=$ $\vv_{{}_{2}} - \fricp<\vw_{{}_{1}}\cdot\vv_{{}_{2}},\Vert\vw_{{}_{1}}\Vert^{2}> \vw_{{}_{1}}$ $=$ $\cvt[1 0 2 ] - \fric<3,3> \cvt[1 1 1 ]$ $=$ $\cvt[0 -1 1 ]$.

Finally, let $\vu_{{}_{1}}$ $=$ $\fric<\vw_{{}_{1}},\Vert\vw_{{}_{1}}\Vert>$ $=$ $\fric<1,\sqrt{3}>\cvt[1 1 1 ]$ and $\vu_{{}_{2}}$ $=$ $\fric<\vu_{{}_{2}},\Vert\vu_{{}_{2}}\Vert>$ $=$ $\fric<1,\sqrt{2}>\cvt[0 -1 1 ].$

[4] Find the four fundamental subspaces of the following matrix.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 1 & 1 & 2\\ 1 & 1 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 1 & 1 & 2\\ 1 & 1 & 3 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$R(A)$ $=$ span$\left(\left\{\cvt[1 1 0 ],\cvt[0 0 1 ]\right\}\right)$

$N(A)$ $=$ span$\left(\left\{\cvt[-1 1 0 ]\right\}\right)$

$C(A)$ $=$ span$\left(\left\{\cvt[1 1 1 ],\cvt[1 2 3 ]\right\}\right)$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 2 & 3 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 0 & 1 & 2\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$N\left(\tran A;\right)$ $=$ span$\left(\left\{\cvt[1 -2 1 ]\right\}\right)$

[2]( Solve completely.)

[1]

$x$	$+$	$y$	=	-1
$3x$	$-$	$y$	=	-15

Add the two equations.

$4x$ $=$ $\tm16$

$x$ $=$ $\tm4$

Substitute.

$y$ $=$ $3$

[1]

$2x$	$-$	$y$	$+$	$z$	=	2
$\tm x$	$+$	$y$	$+$	$2z$	=	7
$3x$	$-$	$y$	$+$	$4z$	=	10

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & -1 & 1 & 2\\ -1 & 1 & 2 & 7\\ 3 & -1 & 4 & 10 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 3 & 0\\ 0 & 1 & 5 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

No solution.

[1]

$x$	$+$	$2y$	$-$	$z$	=	2
$x$	$-$	$y$	$+$	$2z$	=	5
$x$	$+$	$y$			=	3

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & -1 & 2\\ 1 & -1 & 2 & 5\\ 1 & 1 & 0 & 3 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 1 & 4\\ 0 & 1 & -1 & -1\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Particular solution: $\cvt[4 -1 0 ]$

Special solution: $\cvt[-1 1 1 ]$

Complete solution: $\left\{ \cvt[4 -1 0 ] + z\cvt[-1 1 1 ]:z\in\R\right\}$

Total Points:

(Exam 1)

Exam 1 Math 3720 Summer 2018

Let $\vv_{{}_{1}}$ $=$ $\rvt[2 -1 1 ],$ $\vv_{{}_{2}}$ $=$ $\rvt[2 1 -1 ],$ $\vw_{{}_{1}}$ $=$ $\rvt[3 0 8 ],$ and $\vw_{{}_{2}}$ $=$ $\rvt[8 0 3 ]$. Also, let $\theta_{{}_{11}}$ be the angle between $\vv_{{}_{1}}$ and $\vw_{{}_{1}},$ $\theta_{{}_{12}}$ be the angle between $\vv_{{}_{1}}$ and $\vw_{{}_{2}},$ $\theta_{{}_{21}}$ be the angle between $\vv_{{}_{2}}$ and $\vw_{{}_{1}},$ and $\theta_{{}_{22}}$ be the angle between $\vv_{{}_{2}}$ and $\vw_{{}_{2}}$. Compute/answer the following.

[2]()

[10]

$\vv_{{}_{1}}\cdot\vw_{{}_{1}}$ $=$ $6 + 0 + 8$ $=$ $14$

$\vv_{{}_{1}}\cdot\vw_{{}_{2}}$ $=$ $16 + 0 + 3$ $=$ $19$

$\vv_{{}_{2}}\cdot\vw_{{}_{1}}$ $=$ $6 + 0 - 8$ $=$ $\tm2$

$\vv_{{}_{2}}\cdot\vw_{{}_{2}}$ $=$ $16 + 0 - 3$ $=$ $13$

[10]

$\cos\theta_{{}_{11}}$ $=$ $\fric<\vv_{{}_{1}}\cdot\vw_{{}_{1}},\Vert\vv_{{}_{1}}\Vert\Vert\vw_{{}_{1}}\Vert>$ $=$ $\fric<14,\sqrt{6}\sqrt{73}>$ $=$ $\fric<14,\sqrt{438}>$

$\cos\theta_{{}_{12}}$ $=$ $\fric<\vv_{{}_{1}}\cdot\vw_{{}_{2}},\Vert\vv_{{}_{1}}\Vert\Vert\vw_{{}_{2}}\Vert>$ $=$ $\fric<19,\sqrt{6}\sqrt{73}>$ $=$ $\fric<19,\sqrt{438}>$

$\cos\theta_{{}_{21}}$ $=$ $\fric<\vv_{{}_{2}}\cdot\vw_{{}_{1}},\Vert\vv_{{}_{2}}\Vert\Vert\vw_{{}_{1}}\Vert>$ $=$ $\fric<\tm2,\sqrt{6}\sqrt{73}>$ $=$ $\tm\fric<2,\sqrt{438}>$

$\cos\theta_{{}_{22}}$ $=$ $\fric<\vv_{{}_{2}}\cdot\vw_{{}_{2}},\Vert\vv_{{}_{2}}\Vert\Vert\vw_{{}_{2}}\Vert>$ $=$ $\fric<13,\sqrt{6}\sqrt{73}>$ $=$ $\fric<13,\sqrt{438}>$

[10]

comp$_{{}_{\vv_{{}_{1}}}}\vw_{{}_{1}}$ $=$ $\fric<\vv_{{}_{1}}\cdot\vw_{{}_{1}},\Vert\vv_{{}_{1}}\Vert>$ $=$ $\fric<14,\sqrt{6}>$

comp$_{{}_{\vv_{{}_{1}}}}\vw_{{}_{2}}$ $=$ $\fric<\vv_{{}_{1}}\cdot\vw_{{}_{2}},\Vert\vv_{{}_{1}}\Vert>$ $=$ $\fric<19,\sqrt{6}>$

comp$_{{}_{\vv_{{}_{2}}}}\vw_{{}_{1}}$ $=$ $\fric<\vv_{{}_{2}}\cdot\vw_{{}_{1}},\Vert\vv_{{}_{2}}\Vert>$ $=$ $\fric<\tm2,\sqrt{6}>$ $=$ $\tm\fric<2,\sqrt{6}>$

comp$_{{}_{\vv_{{}_{2}}}}\vw_{{}_{2}}$ $=$ $\fric<\vv_{{}_{2}}\cdot\vw_{{}_{2}},\Vert\vv_{{}_{2}}\Vert>$ $=$ $\fric<13,\sqrt{6}>$

[10]

[10]

Find a unit vector parallel to $\vw_{{}_{1}}.$

$\fric<1,\Vert\vw_{{}_{1}}\Vert>\vw_{{}_{1}}$ $=$ $\fric<1,\sqrt{73}>\rvt[3 0 8 ]$

Find a unit vector parallel to $\vw_{{}_{2}}.$

$\fric<1,\Vert\vw_{{}_{2}}\Vert>\vw_{{}_{2}}$ $=$ $\fric<1,\sqrt{73}>\rvt[8 0 3 ]$

[10]

Find a vector orthogonal to $\vw_{{}_{1}}.$

$\rvt[\tm8 0 3 ]$

Find a vector orthogonal to $\vw_{{}_{2}}.$

$\rvt[\tm3 0 8 ]$

[2]()

[10]

Suppose that $\vv_{{}_{1}} + \vw_{{}_{1}} + \vx$ $=$ $\vzero$.
Give $\vx$ in component form.

$\vv_{{}_{1}} + \vw_{{}_{1}} + \vx$ $=$ $\vzero$

$\rvt[5 -1 9 ] + \vx$ $=$ $\vzero$

$\vx$ $=$ $\rvt[-5 1 -9 ]$

Suppose that $\vv_{{}_{1}} + \vw_{{}_{2}} + \vx$ $=$ $\vzero$.
Give $\vx$ in component form.

$\vv_{{}_{1}} + \vw_{{}_{2}} + \vx$ $=$ $\vzero$

$\rvt[10 -1 4 ] + \vx$ $=$ $\vzero$

$\vx$ $=$ $\rvt[-10 1 -4 ]$

Suppose that $\vv_{{}_{2}} + \vw_{{}_{1}} + \vx$ $=$ $\vzero$.
Give $\vx$ in component form.

$\vv_{{}_{2}} + \vw_{{}_{1}} + \vx$ $=$ $\vzero$

$\rvt[5 1 7 ] + \vx$ $=$ $\vzero$

$\vx$ $=$ $\rvt[-5 -1 -7 ]$

Suppose that $\vv_{{}_{2}} + \vw_{{}_{2}} + \vx$ $=$ $\vzero$.
Give $\vx$ in component form.

$\vv_{{}_{2}} + \vw_{{}_{2}} + \vx$ $=$ $\vzero$

$\rvt[10 1 2 ] + \vx$ $=$ $\vzero$

$\vx$ $=$ $\rvt[-10 -1 -2 ]$

[10] Suppose that $\vu,\vv,\vw\in\Rss 3$ such that $\vu\perp\vv$ and $\vu\perp\vw$. Prove that $\vu\perp(\vv + \vw).$

Since $\vu\perp\vv$ and $\vu\perp\vw,$ $\vu\cdot\vv$ $=$ 0 and $\vu\cdot\vw$ $=$ 0. Then $\vu\cdot(\vv + \vw)$ $=$ $\vu\cdot\vv + \vu\cdot\vw$ $=$ $0 + 0$ $=$ 0. Therefore, $\vu\perp(\vv + \vw).$

[10] Suppose that $\vu,$ $\vv,$ and $\vw$ are nonzero vectors in $\Rss3$ such that $\vu$ is parallel to $\vv$ and $\vu$ is parallel to $\vw$. Prove that $\vv$ and $\vw$ are parallel.

Since $\vu$ is parallel to $\vv,$ there is $\alpha\in\R$ such that $\alpha\ne 0$ and $\vu$ $=$ $\alpha\vv$. Likewise, there is $\beta\in\R$ such that $\beta\ne 0$ and $\vu$ $=$ $\beta\vw$. Then $\alpha\vv$ $=$ $\beta\vw$ which means that $\vv$ $=$ $\fric<\beta,\alpha>\vw$. Hence, $\vv$ is parallel to $\vw.$

Let $A_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 2 & 1\\ -3 & 0\\ 1 & 2 \end{tabular}\hspace{-4pt}\right],$ $A_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & -3\\ 2 & 1 \end{tabular}\hspace{-4pt}\right],$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$. Calculate each of the following.

[10]

[10]

$BA_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ... \vspace{-12pt}\\ 2 & 1\\ -3 & 0\\ 1 & 2 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 3 & 3\\ -3 & 0 \end{tabular}\hspace{-4pt}\right]$

$BA_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 ... ...\vspace{-12pt}\\ 1 & 2\\ 0 & -3\\ 2 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 3 & 3\\ 0 & -3 \end{tabular}\hspace{-4pt}\right]$

[10]

$\tran A_{{}_{1}};\tran B;$ $=$ $\tranp BA_{{}_{1}};$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 3 & -3\\ 3 & 0 \end{tabular}\hspace{-4pt}\right]$

$\tran A_{{}_{2}};\tran B;$ $=$ $\tranp BA_{{}_{2}};$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 3 & 0\\ 3 & -3 \end{tabular}\hspace{-4pt}\right]$

[3]( Give the reduced row echelon form of each of the following.)

[10] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -3\\ -2 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -3\\ 0 & -2 \en... ...-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -3\\ 0 & 1 \end... ...{-8pt} \begin{tabular}{l} \\ \tiny\(\tm\fric<1,2>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} + 3R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 3\\ -2 & -4 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 3\\ 0 & 2 \end{... ...-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 3\\ 0 & 1 \end{... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,2>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} - 3R_{{}_{2}}\)\\ \\ \end{tabular}$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & -5 & 3\\ 1 & 1 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 4\\ 2 & -5... ... \begin{tabular}{l} \tiny\(R_{{}_{2}}\)\\ \tiny\(R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 4\\ 0 & -7... ...-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 4\\ 0 & 1 ... ...{-8pt} \begin{tabular}{l} \\ \tiny\(\tm\fric<1,7>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2p... ...{-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} - R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 2 & 5 & 3\\ 1 & -1 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 4\\ 2 & 5... ... \begin{tabular}{l} \tiny\(R_{{}_{2}}\)\\ \tiny\(R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 4\\ 0 & 7... ...-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 4\\ 0 & 1... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,7>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2p... ...{-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \end{tabular}$

[10] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 2 & 7\\ -1 & 5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 7\\ 0... ...R_{{}_{2}} - 2R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} + R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ 0... ...-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,7>R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ 0... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 5R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 1\\ 2 & 7\\ 1 & -5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -5\\ 0 & 1\\ ... ...ny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{1}}\)\\ \tiny\(R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -5\\ 0 & 1\\ ... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1\\ 0... ...{1}} + 5R_{{}_{2}}\)\\ \\ \tiny\(R_{{}_{3}} - 17R_{{}_{2}}\)\\ \end{tabular}$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 4 & 8\\ -3 & 1 & -2\\ 5 & -3 & 2 \end{tabular}\hspace{-4pt}\right],$ $A_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -3 & 1 & -2\\ 1 & 4 & 8\\ 5 & -3 & 2 \end{tabular}\hspace{-4pt}\right],$ $A_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 4 & 8\\ 5 & -3 & 2\\ -3 & 1 & -2 \end{tabular}\hspace{-4pt}\right],$ and $A_{{}_{3}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & 5 & 6\\ -3 & 1 & -2\\ 5 & -3 & 2 \end{tabular}\hspace{-4pt}\right].$

[10] Find $B$ such that $BA$ $=$ $A_{{}_{1}}.$

$B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[10] Find $B$ such that $BA$ $=$ $A_{{}_{2}}.$

$B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

[10] Find $B$ such that $BA$ $=$ $A_{{}_{3}}.$

$B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[10] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$. If possible, find a nonzero matrix $B$ such that $AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$. If this is not possible, explain why it is not.

$B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 0 & 0\\ 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[10] Prove that if $A$ and $B$ are symmetric matrices of the same dimension, then $A + B$ is symmetric.

Note that $\tranp A + B;$ $=$ $\tran A; + \tran B;$ $=$ $A + B.$

[10] Prove that for any matrix $A,$ $\tran A;A$ is symmetric.

Note that $\tranp{\tran A;A};$ $=$ $\tran A;\tranp{\tran A;};$ $=$ $\tran A;A.$

Prove that for any matrix $A,$ $A\tran A;$ is symmetric.

Note that $\tranp{A\tran A;};$ $=$ $\tranp{\tran A;};\tran A;$ $=$ $A\tran A;.$

[3]( For each of the following, compute the inverse if it exists. If the inverse does not exist, explain why it does not.)

[10] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 5 & -1\\ 5 & 3 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 5 & -1 & 1 & 0\\ 5 & 3 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 5 & -1 & 1 & 0\\ ... ...{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 5 & -1 & 1 & 0\\ ... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,4>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 5 & 0 & \raisebox{... ...{-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & \raisebox{... ...ace{-8pt} \begin{tabular}{l} \tiny\(\fric<1,5>R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 3 & -1\\ 3 & 5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 3 & -1 & 1 & 0\\ 3 & 5 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 3 & -1 & 1 & 0\\ ... ...{-8pt} \begin{tabular}{l} \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 3 & -1 & 1 & 0\\ ... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,6>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 3 & 0 & \raisebox{... ...{-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & \raisebox{... ...ace{-8pt} \begin{tabular}{l} \tiny\(\fric<1,3>R_{{}_{1}}\)\\ \\ \end{tabular}$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 1 & -3 & 0\\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 1 & 1 & ... ... -3 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 1 & 1 & ... ...l} \\ \tiny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 1 & 1 & ... ..._{1}} - 2R_{{}_{2}}\)\\ \\ \tiny\(R_{{}_{3}} + 5R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ...tiny\(R_{{}_{1}} + R_{{}_{3}}\)\\ \\ \tiny\(\tm R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 3\\ 0 & 0 & 1\\ 1 & -1 & -5 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0\\ 0 & -2 & 1\\ 1 & 2 & 0 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & 1 & 0 & 1 & ... ... -2 & 1 & 0 & 1 & 0\\ 1 & 2 & 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & 1 & 0 & 1 & ... ..._{{}_{2}} + 2R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - 2R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & -2 &... ...ny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{1}}\)\\ \tiny\(R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & 0 & 1\\ 1 & 0 & 0\\ 2 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & -5 & -1\\ 1 & 2 & 1\\ 3 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & -5 & -1 & 1 ... ...& 2 & 1 & 0 & 1 & 0\\ 3 & 1 & 2 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & -5 & -1 & 1 ... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 3R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & -5 & -1 & 1 ... ...t} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\)\\ \end{tabular}$

Not invertible.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -3 & 4\\ 1 & 0 & -1\\ 4 & -3 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & -3 & 4 & 1 &... ...0 & -1 & 0 & 1 & 0\\ 4 & -3 & 1 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & -3 & 5 & 1 &... ...}_{1}} - R_{{}_{2}}\)\\ \\ \tiny\(R_{{}_{3}} - 4R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & -3 & 5 & 1 &... ...t} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\)\\ \end{tabular}$

Not invertible.

[10] Suppose that $A$ is an invertible matrix and $\vx\vy\in\Rss 3$ such that $A\vx$ $=$ $\cvt[3 -19 -13 ]$ and $A\vy$ $=$ $\cvt[-2 21 17 ]$. Given that $\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & -2 & 3\\ 0 & -3 & 4 \end{tabular}\hspace{-4pt}\right],$ find $\vx$ and $\vy.$

[2]()

$A\vx$ $=$ $\cvt[3 -19 -13 ]$

$\inv A;A\vx$ $=$ $\inv A;\cvt[3 -19 -13 ]$

$\vx$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & -2 & 3\\ 0 & -3 & 4 \end{tabular}\hspace{-4pt}\right]\cvt[3 -19 -13 ]$

$\vx$ $=$ $\cvt[-10 5 5 ]$

$A\vy$ $=$ $\cvt[-2 21 17 ]$

$\inv A;A\vy$ $=$ $\inv A;\cvt[-2 21 17 ]$

$\vy$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & -2 & 3\\ 0 & -3 & 4 \end{tabular}\hspace{-4pt}\right]\cvt[-2 21 17 ]$

$\vy$ $=$ $\cvt[15 5 5 ]$

[10] Suppose that $A$ and $B$ are invertible matrices of the same dimension. Prove that $\invp AB;$ $=$ $\inv A;\inv B;$ if and only if $AB$ $=$ $BA.$

First, suppose that $AB$ $=$ $BA$. Then $\invp AB;$ $=$ $\invp BA;$ $=$ $\inv A;\inv B;$. Now suppose that $\invp AB;$ $=$ $\inv A;\inv B;$. Then $AB$ $=$ $\invb{\invp AB;};$ $=$ $\invp {\inv A;\inv B;};$ $=$ $\invp {\inv B;}; \invp {\inv A;};$ $=$ $BA.$

(Final)

Final Exam Math 3720 Summer 2018

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero. The use of calculators, phones, electronic devices, or outside sources will result in a score of 0 on the exam.

[2]( Calculate the following determinants.)

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -4 & 1\\ 5 & 8 \end{tabular}\hspace{-4pt}\right\vert$

[10] $\left\vert\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 4 & 3 & 0\\ 1 & 0 & 0\\ 10 & -5 & 6 \end{tabular}\hspace{-4pt}\right\vert$

Suppose that $A$ is a $4\times4$ matrix, $\vert A\vert$ $=$ $\tm3,$ $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 0... ... & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ and $F$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$. Compute the determinant of each of the following.

[3]()

[10] $EA$

[10] $FA$

[10] $2A$ (Be careful)

[10] Suppose that $\vv,\vw\in\Rss3$. Prove that $\vw -$   proj$_{{}_{\vv}}\vw$ is orthogonal to $\vv.$

[10] Let $\vW$ $=$ $\left\{\cv[\(x\) \(\tm x\) ]:x\in\R\right\}$. Prove that $\vW\leq\Rss2 .$

[2]( For each of the following, determine whether or not the given set is a basis for $\Rss 3 .$)

[10] $\left\{\cvt[1 -1 0 ],\cvt[1 1 1 ],\cvt[0 1 1 ]\right\}$

[10] $\left\{\cvt[1 2 -1 ],\cvt[-1 3 2 ],\cvt[5 0 -7 ]\right\}$

[10] Suppose that $A$ is an $m\times n$ matrix and $n < m$ ($A$ has more rows than columns). Prove that the rows of $A$ are linearly dependent.

[10] Find an orthogonal basis for $\Rss3$ that includes the vectors $\cvt[1 0 1 ]$ and $\cvt[1 1 -1 ]$. Note that $\cvt[1 0 1 ]$ and $\cvt[1 1 -1 ]$ are orthogonal to each other.

[10] Note that $\left\{\cvt[1 -1 2 ],\cvt[1 1 0 ],\cvt[-1 1 1 ]\right\}$ is an orthogonal set of vectors. Express $\cvt[-2 8 3 ]$ as a linear combination of the vectors in the given set.

[10] Suppose that $S = \left\{\vv_{{}_{1}},\vv_{{}_{2}},\ldots,\vv_{{}_{n}} \right\}\subseteq\Rss m ,$ $\vw\in\Rss m ,$ and $\vw$ is orthogonal to $S$. Prove that $\vw$ is orthogonal to span$(S)$.

[10] Let $\vW$ $=$ span$\left(\left\{\cvf[-1 2 0 1 ],\cvf[1 -3 1 0 ]\right\}\right)\leq\Rss4$. Find a basis for $\per \vW;.$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & -3 ... ... 0\\ 0 & 0 & 1 & 0 & \(\pi\) & 17 & -5 & 17 \end{tabular}\hspace{-4pt}\right]$. Find each of the following.

[2]()

[10] The row space of $A.$

[10] The nullity (not the null space) of $A.$

[2]()

[10] The column space of $A.$

[10] The null space of $\tran A;.$

Solve completely.

[10]

$\tm3x$	$+$	$4y$	=	11
$6x$	$+$	$5y$	=	43

[10]

$2x$	$-$	$y$	$+$	$z$	=	-1
$x$	$+$	$y$	$+$	$z$	=	4

[10]

$\tm3x$	$+$	$4y$	$+$	$7z$	=	4
$x$	$+$	$3y$	$-$	$6z$	=	-5
$2x$	$-$	$7y$	$-$	$z$	=	2

(Summer 2019)

(Quizzes)

[05-24-19]

Let $\vv$ $=$ $\rvf[2 0 \tm3 4 ]$ and $\vw$ $=$ $\rvf[1 0 2 \tm1 ].$

[3]()

[1] Calculate $4\vv - 2\vw.$

$4\vv - 2\vw$ $=$ $\rvf[6 0 \tm16 18 ]$

[1] Calculate $\vv\cdot\vw.$

$\vv\cdot\vw$ $=$ $2 + 0 - 6 - 4$ $=$ $\tm8$

[1] Calculate $\Vert\vw\Vert.$

$\Vert\vw\Vert$ $=$ $\sqrt{1 + 0 + 4 + 1}$ $=$ $\sqrt{6}$

[2] If possible, find $\alpha$ and $\beta$ such that $\alpha\vv + \beta\vw$ $=$ $\rvf[7 0 0 5 ]$. If this is not possible, explain why it is not.

Suppose that $\alpha\vv + \beta\vw$ $=$ $\rvf[7 0 0 5 ]$. Then $\rvf[\(2\alpha\) 0 \(\tm3\alpha\) \(4\alpha\) ] + \rvf[\(\beta\) 0 \(2\beta\) \(\tm\beta\) ]$ $=$ $\rvf[7 0 0 5 ]$ which implies

$2\alpha$	$+$	$\beta$	=	7
0	$+$	0	=	0
$\tm3\alpha$	$-$	$2\beta$	=	0
$4\alpha$	$-$	$\beta$	=	5

Adding the first equation to the fourth, we see that $6\alpha$ $=$ $12$ which of course means that $\alpha$ $=$ $2$. Substituting in any of the three non-trivial equations yields $\beta$ $=$ $3$. So $2\vv + 3\vw$ $=$ $\rvf[7 0 0 5 ].$

[2] Suppose that $\vv,\vw\in\Rss n$. Prove that $\vv\cdot\vw$ $=$ $\vw\cdot\vv.$

See the proof of Theorem 22.

[05-29-19]

Let $\vv$ $=$ $\rvt[-1 0 8 ],$ $\vw$ $=$ $\rvt[2 1 6 ],$ and $\theta$ be the angle between $\vv$ and $\vw$. Find each of the following.

[2]()

[1] $\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<46,\sqrt{65}\sqrt{41}>$

[1] $\comp(\vv,\vw)$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert>$ $=$ $\fric<46,\sqrt{65}>$

[2]()

[1] $\proj(\vv,\vw)$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert^{2}>\vv$ $=$ $\fric<46,65>\rvt[-1 0 8 ]$

[1] A unit vector parallel to $\vv.$

$\fric<\vv,\Vert\vv\Vert>$ $=$ $\fric<1,\sqrt{65}>\rvt[-1 0 8 ]$

Suppose that $\vv,\vw\in\Rss3$ such that $\Vert\vv\Vert$ $=$ $2$ and $\Vert\vw\Vert$ $=$ $5$.

[1] Is it possible that $\vv\cdot\vw$ $=$ $12$? Why or why not?

No. By the Cauchy-Bunyakovsky-Schwarz Inequality, $\vert\vv\cdot\vw\vert$ $\leq$ $\Vert\vv\Vert\Vert\vw\Vert$ $=$ $10.$

[1] Is it possible that $\Vert\vv + \vw\Vert$ $=$ $9$? Why or why not?

No. By the Triangle Inequality, $\Vert\vv + \vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\vw\Vert$ $=$ $7.$

[1] Let $A$ $=$ $\twomat[1 2 3 4 ]$ and $B$ $=$ $\twomat[-1 0 1 2 ]$. Calculate $A + 2B$.

$A + 2B$ $=$ $\twomat[-1 2 5 8 ]$

[05-31-19]

[2] Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 8\\ -2 & 1 & -5 \end{tabular}\hspace{-4pt}\right]$ and $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -5 & 0\\ 1 & 6\\ 3 & -1 \end{tabular}\hspace{-4pt}\right]$. Calculate the following products.

[2]()

$AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 19 & -8\\ -4 & 11 \end{tabular}\hspace{-4pt}\right]$

$BA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -5 & 0 & -40\\ -11 & 6 & -22\\ 5 & -1 & 29 \end{tabular}\hspace{-4pt}\right]$

[1] Multiply. Hint: Use block multiplication.

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & 0 & 0 & 1 & ... ...& 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1 & ... ...& 5 & 5 & 5 & 5 & 5\\ 6 & 6 & 6 & 6 & 6 & 6 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 4 & 4 & 4 & 4 & ... ...& 2 & 2 & 2 & 2 & 2\\ 3 & 3 & 3 & 3 & 3 & 3 \end{tabular}\hspace{-4pt}\right]$

Also, note that the first matrix is the identity matrix with the rows permuted. Rows 1 and 4 are interchanged as are rows 2 and 5 as well as rows 3 and 6. So the result of multiplying this matrix on the left is the same as performing the corresponding row operations on the matrix on the right.

[06-03-19]

[2]( For each of the following, determine whether the given matrices are inverses, one-sided inverses, or neither.)

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 1 & 2\\ 0 & 1 & -3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 1 & 5\\ 0 & 1 & 3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 1 & 2\\ 0 & 1 ... ...pt}\\ -1 & 1 & 5\\ 0 & 1 & 3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\idmatthree$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 1 & 5\\ 0 & 1 ... ...t}\\ -1 & 1 & 2\\ 0 & 1 & -3\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\idmatthree$

Inverses.

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 1 & -1 & 1 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -1 & 2\\ 1 & -1\\ 2 & -2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 0\\ 1 & -1... ...vspace{-12pt}\\ -1 & 2\\ 1 & -1\\ 2 & -2 \end{tabular}\hspace{-4pt}\right]$ $=$ $\idmattwo$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -1 & 2\\ 1 & -1\\ ... ...} \vspace{-12pt}\\ 1 & 2 & 0\\ 1 & -1 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -4 & 2\\ 0 & 3 & -1\\ 0 & 6 & -2 \end{tabular}\hspace{-4pt}\right]$

One-sided inverses.

[1] Suppose that $A,$ $B,$ and $X$ are matrices such that $\inv A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ $B$ $=$ $\cvt[2 0 1 ],$ and $AX$ $=$ $B$. Find $X.$

Since $AX$ $=$ $B,$ $X$ $=$ $\inv A;B$ $=$ $\cvt[2 -1 1 ].$

[06-05-19]

[2]( For each of the following, find the inverse or show that it does not exist.)

[1] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -2\\ 0 & 4 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & -2 & 1 & 0\\ 0 & 4 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & \raise... ...gin{tabular}{l} \tiny\(R_{{}_{1}} + \fric<1,2>R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & \raise... ...ace{-8pt} \begin{tabular}{l} \\ \tiny\(\fric<1,4>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & \raisebox{2pt}{\T... ...}\\ 0 & \raisebox{2pt}{\Tiny\(\fric<1,4>\)} \end{tabular}\hspace{-4pt}\right]$

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 3 & 4 & 4\\ 5 & 4 & 2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ...& 4 & 4 & 0 & 1 & 0\\ 5 & 4 & 2 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & -1 & 1 &... ..._{{}_{2}} - 3R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} - 5R_{{}_{1}}\)\\ \end{tabular}$

Not invertible. Note rows 2 and 3.

[1] Prove that a matrix with a column of zeros has no left inverse.

See problem 44 from the homework.

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & 0 & 8\\ -1 & 3 & -2 \end{tabular}\hspace{-4pt}\right]$. For each of the following, find an elementary matrix $E$ such that $EA$ $=$ $B.$

[3]()

[1] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 3 & -2\\ 2 & 0 & 8\\ 1 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

[1] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 1 & 0 & 4\\ -1 & 3 & -2 \end{tabular}\hspace{-4pt}\right]$

$E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & \r... ...x{2pt}{\Tiny\(\fric<1,2>\)} & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[1] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & 0 & 8\\ 1 & 3 & 6 \end{tabular}\hspace{-4pt}\right]$

$E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$

[06-12-19]

(DefinitionDefinitionOfVectorSpace, 0)A vector space over a field $S$ (usually $\R$ or $C$) is a set $\vV$ along with two operations $+$ (vector addition) and $\cdot$ (scalar multiplication). Addition is a function from $\vV\times\vV\to\vV$ and scalar multiplication is a function from $S\times\vV\to\vV$. Also, the following conditions must be satisfied.

(SubdefinitionVectorSpaceAdditionIsCommutative, 0)For all $\vv,\vw\in\vV,$ $\vv + \vw$ $=$ $\vw + \vv.$

(SubdefinitionVectorSpaceAdditionIsAssociative, 0)For all $\vv,\vw,\vz\in\vV,$ $(\vv + \vw) + \vz$ $=$ $\vv + (\vw + \vz).$

(SubdefinitionVectorSpaceExistenceOfZeroVector, 0)There exists $\vzero\in\vV$ such that $\vv + \vzero$ $=$ $\vv.$

(SubdefinitionVectorSpaceExistenceOfAdditiveInverse, 0)For all $\vv\in\vV$ there exists $\tm\vv\in\vV$ such that $\vv + \tm\vv$ $=$ $\vzero.$

(SubdefinitionVectorSpaceScalarMultiplicationDistributesVectors, 0)For all $\vv,\vw,\in\vV$ and $\alpha\in S,$ $\alpha(\vv + \vw)$ $=$ $\alpha\vv + \alpha\vw.$

(SubdefinitionVectorSpaceScalarMultiplicationDistributesScalars, 0)For all $\vv\in\vV$ and $\alpha,\beta\in S,$ $(\alpha + \beta)\vv$ $=$ $\alpha\vv + \beta\vv.$

(SubdefinitionVectorSpaceScalarMultiplicationAssociative, 0)For all $\vv\in\vV$ and $\alpha,\beta\in S,$ $(\alpha\beta)\vv$ $=$ $\alpha(\beta\vv).$

(SubdefinitionVectorSpaceScalarUnit, 0)For all $\vv\in\vV,$ $1\vv$ $=$ $\vv.$

Prove each of the following.

[1] The additive identity in a vector space is unique.

See problem 55 from the homework.

[1] The additive inverse of each element in a vector space is unique.

See problem 56 from the homework.

[1] For each vector $\vv$ in a vector space, $(\tm1)\vv$ $=$ $\tm\vv.$

See the proof of Theorem 161 from the class notes.

[06-14-19]

[1] Suppose that $A$ is an $m\times n$ matrix. Prove that $N(A)\leq\Rss n .$

See problem 62 from the homework.

[1] Transform the following matrix to reduced row echelon form. For each free column, find a nonzero vector in the null space.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 0 & 1\\ 2 & 4 & 1 & -1\\ 1 & 2 & 1 & -2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 0 & 1\\ 2 & 4 & 1 & -1\\ 1 & 2 & 1 & -2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 0 & 1\\ 0 & 0 & 1 & -3\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Free columns: 2, 4

$\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -2\\ 1\\ 0\\ 0 \end{tabular}\hspace{-4pt}\right],$ $\left[\hspace{-4pt}\begin{tabular}{r} \vspace{-12pt}\\ -1\\ 0\\ 3\\ 1 \end{tabular}\hspace{-4pt}\right]$

[06-19-19]

[2]( [4] Determine whether the given vectors are linearly independent or linearly dependent.)

$\cvt[1 2 1 ],$ $\cvt[0 -1 3 ],$ $\cvt[1 0 5 ]$

Independent.

Using column vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 2 & -1 & 0\\ 1 & 3 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmatthree$

Pivots: 3

Using row vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & 1\\ 0 & -1 & 3\\ 1 & 0 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmatthree$

No zero rows.

$\cvt[7 5 -4 ],$ $\cvt[1 -1 -2 ],$ $\cvt[2 4 1 ]$

Dependent.

Using column vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 7 & 1 & 2\\ 5 & -1 & 4\\ -4 & -2 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2p... ...ox{2pt}{\Tiny\(\tm\fric<3,2>\)}\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Pivots: 2

Using row vectors:

$\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 7 & 5 & -4\\ 1 & -1 & -2\\ 2 & 4 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2p... ...sebox{2pt}{\Tiny\(\fric<5,6>\)}\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Row of zeros.

$\cvt[1 0 0 ],$ $\cvt[1 1 0 ],$ $\cvt[1 1 1 ],$ $\cvt[1 -1 1 ]$

Dependent. Four vectors in $\Rss 3 .$

[2]()

$\cvf[1 0 0 0 ],$ $\cvf[1 1 0 0 ],$ $\cvf[1 1 1 0 ],$ $\cvf[1 1 1 1 ]$

Independent.

Using column vectors:

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmatfour$

Pivots: 4

Using row vectors:

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\idmatfour$

No zero rows.

[6-21-19]

[1] Prove that $\beta$ $=$ $\left\{ \twomat[1 0 0 0 ], \twomat[1 1 0 0 ], \twomat[1 1 1 0 ], \twomat[1 1 1 1 ] \right\}$ is a basis for $\Mss2\times2 .$

Since dim$\Mss2\times2$ $=$ $4,$ we need only show that $\beta$ is a spanning set. Note that

$\twomat[0 1 0 0 ]$ $=$ $\twomat[1 1 0 0 ] - \twomat[1 0 0 0 ],$ $\twomat[0 0 1 0 ]$ $=$ $\twomat[1 1 1 0 ] - \twomat[1 1 0 0 ],$ and $\twomat[0 0 0 1 ]$ $=$ $\twomat[1 1 1 1 ] - \twomat[1 1 1 0 ]$.

Therefore, $\Mss2\times2$ $=$ span$\left(\left\{ \twomat[1 0 0 0 ], \twomat[0 1 0 0 ], \twomat[0 0 1 0 ], \twomat[0 0 0 1 ] \right\}\right)$ $\subseteq$ span$\left(\beta\right),$ as desired.

[1] Suppose that $\vv_{{}_{1}},\vv_{{}_{2}},\ldots,\vv_{{}_{n}}$ are linearly independent vectors in a vector space $\vV$ and
$\vv\in$span$(\{\vv_{{}_{1}},\vv_{{}_{2}},\ldots,\vv_{{}_{n}}\})$. Prove that there are unique scalars $a_{{}_{1}},a_{{}_{2}},\ldots,a_{{}_{n}}$ such that
$\vv$ $=$ $a_{{}_{1}}\vv_{{}_{1}} + a_{{}_{2}}\vv_{{}_{2}} + \cdots + a_{{}_{n}}\vv_{{}_{n}}.$

See Problem 75.

[1] Suppose that $S\subseteq\Rss n$ and $\vv\in\Rss n$. Prove that $\vv$ is orthogonal to $S$ if and only if $\vv$ is orthogonal to span$(S).$

See Problem 78.

[2]( [1] Find an orthogonal set of vectors with the same span as the set below.)

$\left\{\cvt[1 0 2 ], \cvt[2 1 0 ]\right\}$

Let $\vv_{{}_{1}}$ $=$ $\cvt[1 0 2 ]$ and $\vv_{{}_{2}}$ $=$ $\cvt[2 1 0 ]$.

Use the Gram-Schmidt orthogonalization
process.

Let $\vw_{{}_{1}}$ $=$ $\vv_{{}_{1}}$ $=$ $\cvt[1 0 2 ]$ and

$\vw_{{}_{2}}$ $=$ $\vv_{{}_{2}} - \fric<\vv_{{}_{2}}\cdot\vw_{{}_{1}},\Vert\vw_{{}_{1}}\Vert^{2}>\vw_{{}_{1}}$ $=$ $\cvt[2 1 0 ] - \fric<2,5>\cvt[1 0 2 ].$

$\left\{\cvt[1 0 2 ], \cvt[2 1 0 ] - \fric<2,5>\cvt[1 0 2 ]\right\}$

[06-24-19]

[4] Find the four fundamental subspaces of the following matrix.

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & -3 & 1\\ ... ... & 1 & 0\\ 0 & 1 & 1 & 1\\ 1 & 3 & -2 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & -3 & 1\\ ... ... & 1 & 0\\ 0 & 1 & 1 & 1\\ 1 & 3 & -2 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & -5 & 0\\ ... ...1 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$R(A)$ $=$ span$\left(\left\{\cvf[1 0 -5 0 ],\cvf[0 1 1 0 ],\cvf[0 0 0 1 ]\right\}\right)$

$N(A)$ $=$ span$\left(\left\{\cvf[5 -1 1 0 ]\right\}\right)$

$C(A)$ $=$ span$\left(\left\{\cvf[1 0 0 1 ],\cvf[2 1 1 3 ],\cvf[1 0 1 1 ]\right\}\right)$

$\tran A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 2... ... & 1 & 3\\ -3 & 1 & 1 & -2\\ 1 & 0 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 2... ... & 1 & 3\\ -3 & 1 & 1 & -2\\ 1 & 0 & 1 & 1 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$N(\tran A;)$ $=$ span$\left(\left\{\cvf[-1 -1 0 1 ]\right\}\right)$

[06-26-19]

[2]([3] Find all solutions (if any) to the following systems of linear equations.)

$x$	$+$	$y$	$+$	$z$	=	-1
$x$	$$	$$	$+$	$z$	=	-2
$\tm2x$	$+$	$y$	$+$	$z$	=	5

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & -1\\ 1 & 0 & 1 & -2\\ -2 & 1 & 1 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 0 & 0 & -2\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

Solution: $\cvt[-2 1 0 ]$

$x$	$+$	$y$	$+$	$z$	=	1
$2x$	$-$	$y$	$+$	$z$	=	0
$x$	$-$	$2y$	$$	$$	=	2

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 2 & -1 & 1 & 0\\ 1 & -2 & 0 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 0 & \raisebox{... ...{{\Tiny\(\fric<1,3>\)}} & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

No solution.

$x$	$+$	$y$	$+$	$z$	=	0
$x$	$-$	$y$	$-$	$z$	=	6
$3x$	$+$	$y$	$+$	$z$	=	6

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & 0\\ 1 & -1 & -1 & 6\\ 3 & 1 & 1 & 6 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 0 & 0 & 3\\ 0 & 1 & 1 & -3\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Particular solution: $\cvt[3 -3 0 ]$

Special solution: $\cvt[0 -1 1 ]$

Complete solution: $\left\{ \cvt[3 -3 0 ] + z\cvt[0 -1 1 ]:z\in\R\right\}$

(Midterm)

Midterm Exam Math 3720 Summer 2019

Let $\vv$ $=$ $\rvt[-2 1 3 ],$ $\vw$ $=$ $\rvt[4 -1 -7 ],$ and $\theta$ be the angle between $\vv$ and $\vw.$

Calculate each of the following.

[3]()

[10] $3\vv + 2\vw$

$\rvt[2 1 -5 ]$

[10] $\vv\cdot\vw$

$\tm8 - 1 - 21$ $=$ $\tm30$

[10] $\Vert\vv\Vert$

$\sqrt{4 + 1 + 9}$ $=$ $\sqrt{14}$

$\Vert\vw\Vert$

$\sqrt{16 + 1 + 49}$ $=$ $\sqrt{66}$

[10] $\cos\theta$

$\fric<\tm30,\sqrt{14}\sqrt{66}>$ $=$ $\tm\fric<15,\sqrt{231}>$

[10] $\sin\theta$

$\sqrt{1 - \cos^{2}\theta}$

$=$ $\sqrt{1 - \fric<225,231>}$

$=$ $\sqrt{\fric<2,77>}$

[10] $\comp(\vv,\vw)$

$\fric<\tm30,\sqrt{14}>$

[10] $\proj(\vv,\vw)$

$\tm\fric<30,14>\rvt[2 1 3 ]$

$=$ $\tm\fric<15,7>\rvt[2 1 3 ]$

[10] If possible find $\alpha,\beta\in\R$ such that $\alpha\vv + \beta\vw$ $=$ $\rvt[0 1 -1 ]$. If this is not possible, explain why it is not.

Suppose that $\alpha\vv + \beta\vw$ $=$ $\rvt[0 1 -1 ]$. Then $\rvt[\(\tm2\alpha\) \(\alpha\) \(3\alpha\) ] + \rvt[\(4\beta\) \(\tm\beta\) \(\tm7\beta\) ]$ $=$ $\rvt[0 1 -1 ]$ which implies

$\tm2\alpha$	$+$	$4\beta$	=	0
$\alpha$	$-$	$\beta$	=	1
$3\alpha$	$-$	$7\beta$	=	-3

From the second equation, we see that $\alpha$ $=$ $\beta + 1$. Substituting in any of the other two equations, yields $\beta$ $=$ $1$ which then implies that $\alpha$ $=$ $2$. So $2\vv + \vw$ $=$ $\rvt[0 1 -1 ].$

[10] Find a unit vector orthogonal (not parallel) to $\vv.$

First note that $\rvt[1 2 0 ]$ is orthogonal to $\vv$. Then $\fric<1,\sqrt{5}>\rvt[1 2 0 ]$ is a unit vector which is orthogonal to $\vv$.

Let $A$ $=$ $\twofourmatb[1 0 -1 3 2 -1 6 1 ],$ $B$ $=$ $\twofourmatb[2 1 -3 5 0 1 -4 1 ],$ and $C$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -2\\ 1 & 1 & 2\\ -2 & 1 & 5\\ -1 & 2 & 0 \end{tabular}\hspace{-4pt}\right]$. For each of the following, perform the indicated operation if possible. If it is not possible explain why it is not.

[3]()

[10] $A - B$

$\twofourmatb[-1 -1 2 -2 2 -2 10 0 ]$

[10] $AB$

Not possible. Note that matrix $A$ has four columns and matrix $B$ has two rows.

[10] $AC$

$\twothreemat[0 5 -7 -12 7 24 ]$

[10] Prove that matrix addition is associative.

Suppose that $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{ccccc} \(a_{{}_{11}}\) & \(a_{{}_{12}}\) & \... ...\ldots\) & \(a_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}\hspace{-4pt}\right],$ $B$ $=$ $\left[\hspace{-4pt}\raisebox{-1pt}{\begin{tabular}{ccccc} \(b_{{}_{11}}\) & \(b... ...ldots\) & \(b_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right],$ and $C$ $=$ $\left[\hspace{-4pt}\raisebox{-1pt}{\begin{tabular}{ccccc} \(c_{{}_{11}}\) & \(c... ...\ldots\) & \(c_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$.

Then

$(A + B) + C$

$=$ $\left(\left[\hspace{-4pt}\begin{tabular}{ccccc} \(a_{{}_{11}}\) & \(a_{{}_{12}}... ...\ldots\) & \(c_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$

$=$ $\left[\hspace{-4pt}\raisebox{-1pt}{\begin{tabular}{ccccc} \(a_{{}_{11}} + b_{{}... ...\ldots\) & \(c_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$

$=$ $\left[\hspace{-4pt}\raisebox{-1pt}{\begin{tabular}{ccccc} \(\left(a_{{}_{11}} +... ...n}}\right) + c_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$

$=$ $\left[\hspace{-4pt}\raisebox{-1pt}{\begin{tabular}{ccccc} \(a_{{}_{11}} + \left... ...n}} + c_{{}_{mn}}\right)\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$

$=$ $\left[\hspace{-4pt}\begin{tabular}{ccccc} \(a_{{}_{11}}\) & \(a_{{}_{12}}\) & \... ..._{{}_{mn}} + c_{{}_{mn}}\)\\ \vspace{-12pt} \end{tabular}}\hspace{-4pt}\right]$

$=$

$A + (B + C).$

Let $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 2 & 2 & 2 & 2\\ 3 & 3 & 3 & 3\\ 4 & 4 & 4 & 4 \end{tabular}\hspace{-4pt}\right]$. For each of the following, find an elementary matrix $E$ such that $EA$ $=$ $B.$

[10] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 3 & 3 & 3 & 3\\ 2 & 2 & 2 & 2\\ 1 & 1 & 1 & 1\\ 4 & 4 & 4 & 4 \end{tabular}\hspace{-4pt}\right]$

Since $B$ is formed by switching rows 1 and 3 of $A,$ $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 0 & 0 & 1 & 0\\ 0... ... & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

[10] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 1 & 1\\ 2... ... 2 & 2\\ -1 & -1 & -1 & -1\\ 4 & 4 & 4 & 4 \end{tabular}\hspace{-4pt}\right]$

Since $B$ is formed by subtracting row 4 of $A$ from row 3 of $A,$ $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 0... ...& 0 & 0\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

Since $B$ is formed by multiplying row 3 of $A$ by $\tm\fric<1,3>,$ $E$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0\\ 0... ...iny\(\tm\fric<1,3>\)}} & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

[10] Give the transpose of the following matrix.

[2]()

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 4 & 5 & 0\\ -3 & 2 & 8 & 3\\ 1 & 7 & -2 & 7 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -3 & 1\\ 4 & 2 & 7\\ 5 & 8 & -2\\ 0 & 3 & 7 \end{tabular}\hspace{-4pt}\right]$

[10] Prove that for any matrix $A,$ $A\tran A;$ is symmetric.

See problem 50 from the notes.

[2]( For each of the following, find the inverse or show that it does not exist.)

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{\Tiny... ...ic<1,2>\)} & 0\\ 0 & -1 & 0\\ -1 & 0 & -1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ \raisebox{2pt}{\... ...1 & 0 & 0 & 1 & 0\\ -1 & 0 & -1 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 1 & 0 & -2 &... ...}_{1}}\)\\ \tiny\(\tm R_{{}_{2}}\)\\ \tiny\(\tm R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & -2 &... ...t} \begin{tabular}{l} \tiny\(R_{{}_{1}} - R_{{}_{2}}\)\\ \\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & -2 &... ...t} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\)\\ \end{tabular}$

Inverse: $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -2 & 1 & 0\\ 0 & -1 & 0\\ 2 & -1 & -1 \end{tabular}\hspace{-4pt}\right]$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ -1 & 1 & 3\\ 2 & -1 & 0\\ 1 & 1 & 9 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ -1 & 1 & 3 & 1 & ... ... -1 & 0 & 0 & 1 & 0\\ 1 & 1 & 9 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & -1 & -3 & -1... ...R_{{}_{2}} + 2R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} + R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & -1 & -3 & -1... ...} \begin{tabular}{l} \\ \\ \tiny\(R_{{}_{3}} - 2R_{{}_{2}}\)\\ \end{tabular}$

Not invertible.

[2]()

[10] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -2 & 5\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Not invertible.

[10] $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 2 & 0 & 0 & 0 & 0... ...& 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

Inverse: $\left[\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ \raisebox{2pt}{\T... ...& 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[10] Suppose that $A$ is an invertible matrix and $B$ is a singular matrix. Is $AB$ invertible or singular? Justify your answer.

Singular. See problem 38 from the notes.

[2]( Calculate each determinant below.)

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -1\\ 0 & 6 \end{tabular}\hspace{-4pt}\right\vert$

$6 - 0$ $=$ $6$

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ -3 & 1 & 6\\ 1 & 1 & -1 \end{tabular}\hspace{-4pt}\right\vert$

$1\cdot\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 6\\ 1... ...r}{rr} \vspace{-12pt}\\ -3 & 1\\ 1 & 1 \end{tabular}\hspace{-4pt}\right\vert$

$=$ $1(\tm1 - 6) + 2(\tm3 - 1)$

$=$ $\tm15$

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4\\ ... ... & 4\\ 0 & 0 & 3 & 4\\ 0 & 0 & 0 & 4\\ \end{tabular}\hspace{-4pt}\right\vert$ $=$ $24$

The determinant of an upper triangular matrix, is the product of the diagonal entries.

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 4 & -8 & ... ... 3 & 7 & 8 & 6\\ 0 & 1 & 7 & -5 & 1 & 2 \end{tabular}\hspace{-4pt}\right\vert$ $=$ 0

Note that rows 3 and 6 are identical.

(Final)

Final Exam Math 3720 Summer 2019

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Engaging in any of the following will result in a 0 on the exam.
Using a calculator, phone, smart watch, or any other electronic device.
Using an outside source.
Attempting to block my view.

[10] Prove that the additive identity in a vector space is unique.

[10] Let $\vW$ $=$ $\left\{\cvf[\(a\) \(b\) 0 1 ]:a,b\in\R\right\}$. Is $\vW\leq\Rss4 ?$

[10] Do the following vectors span $\Rss3 ?$

$\cvt[1 0 1 ],\cvt[1 1 1 ], \cvt[-1 1 1 ]$

[2]( Find the four fundamental subspaces of each of the following.)

[40] $\twothreemat[1 2 5 -2 -3 -9 ]$

[40] $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 3 & 0 & 2\\ 2... ... & 1 & 8\\ -1 & -3 & 1 & 2\\ 1 & 3 & 1 & 6 \end{tabular}\hspace{-4pt}\right]$

[10] Suppose that a $4\times4$ matrix is in reduced row echelon form, the first three columns are pivot columns, and the fourth column is a free column. Explain why the fourth column is in the span of the first three.

[10] Find a single nonzero vector in $\Rss3$ that is orthogonal to both of the vectors below.

$\cvt[2 -1 6 ],\cvt[3 0 -1 ]$

[10] Note that $\left\{\cvt[2 -1 1 ],\cvt[1 0 -2 ],\cvt[2 5 1 ]\right\}$ is an orthogonal set of vectors. Express $\cvt[5 3 5 ]$ as a linear combination of the vectors in the set.

[10] Suppose that $\vW\leq\Rss n$. Prove that $\vW\cap\vW^{{}^{\perp}}$ $=$ $\{\vzero\}.$

[10] Suppose that $\vW\leq\Rss n ,$ $S\subseteq\vW,$ $T\subseteq\vW^{{}^{\perp}},$ and both $S$ and $T$ are linearly independent. Prove that $S\cup T$ is linearly independent. Hint: Use the previous part.

[2]( Find all solutions (if any) to the following systems of linear equations.)

[10]

$x$	$+$	$y$	=	3
$2x$	$-$	$y$	=	12
$x$	$-$	$y$	=	7

[10]

$x$	$+$	$y$	$+$	$z$	=	2
$\tm x$	$+$	$2y$	$+$	$z$	=	6
$3x$	$-$	$4y$	$+$	$2z$	=	2

[2]()

[10]

$x$	$+$	$y$	$+$	$z$	=	9
$2x$	$+$	$3y$	$+$	$z$	=	8
$x$	$+$	$2y$	$$	$$	=	0

[10]

$x$	$+$	$y$	$+$	$z$	=	6
$3x$	$+$	$2y$	$+$	$z$	=	10

(Spring 2020)

(Quizzes)

[01-22-20]

[2] Let $\vv$ $=$ $\cvf[-1 0 4 3 ]$ and $\vw$ $=$ $\cvf[2 1 8 \tm1 ].$

$\vw - 2\vv$ $=$ $\cvf[4 1 0 \tm7 ]$ $\vv\cdot\vw$ $=$ $27$ $\Vert\vv\Vert$ $=$ $\sqrt{26}$ $\Vert\vw\Vert$ $=$ $\sqrt{70}$

[1] Prove that for any vector $\vv\in\Rss n ,$ $\Vert\tm\vv\Vert$ $=$ $\Vert\vv\Vert.$

Suppose that $\vv$ $=$ $\cvv\in\Rss n$. Then $\tm\vv$ $=$ $\cvn \tm v;$ and $\Vert\tm\vv\Vert$ $=$ $\sqrt{\left(\tm v_{{}_{1}}\right)^{2} + \left(\tm v_{{}_{2}}\right)^{2} + \cdots + \left(\tm v_{{}_{n}}\right)^{2}}$ $=$ $\sqrt{v_{{}_{1}}^{2} + v_{{}_{2}}^{2} + \cdots + v_{{}_{n}}^{2}}$ $=$ $\Vert\vv\Vert.$

[01-27-20]

[1] Find the angle between the following vectors.

$\cvt[1 0 -1 ],$ $\cvt[0 1 1 ]$

Let $\theta$ be the angle between the vectors. By the Cosine Formula, $\cos\theta$ $=$ $\tm\fric<1,2>$. Therefore, $\theta$ $=$ $\fric<2\pi,3>$.

[1] Let $\vv$ $=$ $\cvt[6 0 8 ]$. Find a vector $\vw$ that is parallel to $\vv$ and has magnitude $7.$

Since $\Vert\vv\Vert$ $=$ $10,$ both $\fric<7,10>\vv$ and $\tm\fric<7,10>\vv$ are parallel to $\vv$ and have magnitude $7.$

[1] Let $\vv$ $=$ $\cvt[1 -2 2 ]$. If possible, find a unit vector $\vu$ such that $\vu + \vv$ $=$ $\cvt[4 3 5 ]$. If this is not possible, explain why it is not.

This is not possible.

Solution 1:

Suppose that $\vu\in\Rss3$ is a unit vector. By the Triangle Inequality, $\Vert\vu + \vv\Vert$ $\leq$ $\Vert\vu\Vert + \Vert\vv\Vert$ $=$ $4$.

Since $\left\Vert\cvt[4 3 5 ]\right\Vert$ $=$ $\sqrt{50}$ $>$ $4,$ $\vu + \vv$ $\ne$ $\cvt[4 3 5 ]$.

Solution 2:

Suppose that $\vw\in\Rss3$ and $\vv + \vw$ $=$ $\cvt[4 3 5 ]$. Then $\vw$ $=$ $\cvt[3 5 3 ]$ and $\left\Vert\cvt[3 5 3 ]\right\Vert$ $=$ $\sqrt{43}.$

[01-31-20]

[1] Set $\vv$ $=$ $\rvt[1 -3 1 ]$ and $\vw$ $=$ $\rvt[4 0 -1 ]$. Find $\proj(\vv,\vw).$

$\proj(\vv,\vw)$ $=$ $\left(\fric<\vv\cdot\vw,\Vert\vv\Vert^{2}>\right)\vv$ $=$ $\fric<3,11>\rvt[1 -3 1 ]$

[1] Suppose that $A$ and $B$ are $m\times n$ matrices. Prove that $A + B$ $=$ $B + A.$

See notes.

For each pair of matrices, determine whether or not they are inverses of each other.

[1] $A$ $=$ $\twomat[1 0 3 -1 ],$ $B$ $=$ $\twomat[1 0 3 -1 ]$

Yes. Note that $AB$ $=$ $I.$

[1] $A$ $=$ $\twothreemat[1 1 0 1 -1 0 ],$ $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \raisebox{2pt}{{\Tiny... ...sebox{2pt}{{\Tiny -\(\fric<1,2>\)}}\\ 0 & 0 \end{tabular}\hspace{-4pt}\right]$

No. Although $AB$ $=$ $\idmattwo,$ $BA$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\ne$ $I.$

[1] Suppose that $A$ and $B$ are matrices such that $\inv A;$ $=$ $\twomat[1 0 -1 2 ],$ and $AB$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & -4 \\ \raisebox{2pt}{{\Tiny\(\fric<1,2>\)}} & -1 \end{tabular}\hspace{-4pt}\right]$. Find $B.$

Since matrix multiplication is associative, $B$ $=$ $IB$ $=$ $(\inv A;A)B$ $=$ $\inv A;(AB)$ $=$ $\twomat[1 -4 0 2 ].$

[02-24-20]

[3] Set $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -2 \\ 0 & ... ... & \raisebox{2pt}{{\Tiny\(\fric<1,3>\)}} & 0 \end{tabular}\hspace{-4pt}\right],$ $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -2 \\ 0 & ... ...t}{{\Tiny\(\fric<1,3>\)}} & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ $C$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right],$ and $D$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

Give matrices $E_{{}_{1}},$ $E_{{}_{2}},$ and $E_{{}_{3}}$ such that $E_{{}_{1}}A$ $=$ $B,$ $E_{{}_{2}}B$ $=$ $C,$ and $E_{{}_{3}}C$ $=$ $D$. Give a left inverse of $A$ as a product of $E_{{}_{1}},$ $E_{{}_{2}},$ and $E_{{}_{3}}.$

Since matrix $A$ is transformed into matrix $B$ by interchanging rows 2 and 3, $E_{{}_{1}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right].$

Since matrix $B$ is transformed into matrix $C$ by multiplying row 2 by 3, $E_{{}_{2}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

Since matrix $C$ is transformed into matrix $D$ by adding 2 times row 3 to row 1, $E_{{}_{3}}$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

Since $E_{{}_{3}}E_{{}_{2}}E_{{}_{1}}A$ $=$ $I,$ $E_{{}_{3}}E_{{}_{2}}E_{{}_{1}}$ is a left inverse of $A.$

[1] Give the transpose of the following matrix.

$A$ $=$ $\twothreemat[-3 2 1 0 5 4 ]$

$\tran A;$ $=$ $\threetwomat[-3 0 2 5 1 4 ]$

[2]( For each of the following, find the inverse or explain why the inverse does not exist.)

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{{\Tiny... ...} & \raisebox{2pt}{{\Tiny\(\fric<1,7>\)}} \\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ \raisebox{2pt}{{\... ...x{2pt}{{\Tiny\(\fric<1,7>\)}} & 0 & 0 & 1 \\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 1 & 0 & 7 & 0... ...ny\(7R_{{}_{1}}\)\\ \tiny\(7R_{{}_{2}}\)\\ \tiny\(7R_{{}_{3}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 2 & 7 & 7... ...)\\ \tiny\(\tm R_{{}_{2}}\)\\ \tiny\(R_{{}_{3}} + 3R_{{}_{2}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 2 & 7 & 7... ...y\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \tiny\(\fric<1,7>R_{{}_{3}}\) \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & 7 & 1... ..._{1}} - 2R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} + 2R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 7 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & 3 & 1 \end{tabular}\hspace{-4pt}\right]$

[1] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & -1 \\ 3 & 1 & 8 \\ 6 & 7 & 5 \end{tabular}\hspace{-4pt}\right]$

Since $R_{{}_{3}}$ $=$ $3R_{{}_{1}} + R_{{}_{2}},$ the matrix is not invertible.

Also, since $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 2 & -1 \\ 3 & 1 & 8 \\ 6 & 7 & 5 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & \raisebox{2p... ...pt}{{\Tiny\(\tm\fric<11,5>\)}}\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right],$ the matrix is not invertible.

[2] Let $\vV$ $=$ ${\bf M}_{{}_{2}}$ and $S$ $=$ $\left\{\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ \(a\) & 0\\ 0 & \(a\) \end{tabular}\hspace{-4pt}\right]:a\in\R\right\}$. Also, let $+:\vV\times\vV\to\vV$ be matrix addition and $\cdot:S\times\vV\to\vV$ be matrix multiplication. Show that $\vV$ is a vector space over $S.$

[2] Discuss the following statement.

The vector spaces ${\bf M}_{{}_{2}}$ (with the usual scalar multiplication, not the one defined above) and $\Rss 4$ are the same.

[2] Give an example of a matrix in reduced row echelon form that has 3 pivot columns and 3 free columns. Find three linearly independent vectors in the null space of the matrix.

[2] Find an orthogonal basis for $\Rss 4$ that contains the vector $\cvf[1 2 3 4 ]$. Express $\cvf[1 1 1 1 ]$ as a linear combination of the basis.

Set $\vw$ $=$ $\cvf[1 1 1 1 ],$ $\vv_{{}_{1}}$ $=$ $\cvf[1 2 3 4 ],$ $\vv_{{}_{2}}$ $=$ $\cvf[2 -1 0 0 ],$ $\vv_{{}_{3}}$ $=$ $\cvf[3 6 -5 0 ],$ and $\vv_{{}_{4}}$ $=$ $\cvf[4 8 12 -14 ]$.

By Theorem 220,

$\vw$ $=$ $\fricp<\vw\cdot\vv_{{}_{1}},\vv_{{}_{1}}\cdot\vv_{{}_{1}}> \vv_{{}_{1}} + \fric... ..._{3}} + \fricp<\vw\cdot\vv_{{}_{4}},\vv_{{}_{4}}\cdot\vv_{{}_{4}}> \vv_{{}_{4}}$ $=$ $\fric<1,3>\cvf[1 2 3 4 ] + \fric<1,5>\cvf[2 -1 0 0 ] + \fric<2,35>\cvf[3 6 -5 0 ] + \fric<1,42>\cvf[3 6 -5 0 ]$

[2] Prove that a type 1 elementary matrix is an orthogonal matrix.

Suppose that $E$ is a type 1 elementary matrix. By Theorem 115, $E$ $=$ $\inv E;$ and by Proposition 136, $E$ $=$ $\tran E;$. Since $\inv E;$ $=$ $\tran E;,$ $E$ is symmetric by Corollary 233.

[2] Set $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 0 & 1 & 0 & 0 \\ ... ...& 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$. Suppose that $\vx,\vy\in\Rss4$ such that $(A\vx)\cdot(A\vy)$ $=$ $3$. Find $\vx\cdot\vy.$

[2] Find the four fundamental subspaces of $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 1 & 0 & 2 \\ ... ...& 4 & 5 \\ 0 & 0 & 2 & 6 \\ 1 & 1 & 2 & 2 \end{tabular}\hspace{-4pt}\right]$.

Note that $A$ is an orthogonal matrix. Then by Corollary 240, $\vx\cdot\vy$ $=$ $(A\vx)\cdot(A\vy)$ $=$ $3.$

Total Points:

(Exam 1)

Exam 1 Math 3720 Spring 2020

Name:

Set $\vv$ $=$ $\rvt[1 -5 7 ],$ $\vw$ $=$ $\rvt[-2 1 3 ],$ and set $\theta$ to be the angle between $\vv$ and $\vw.$

[3]()

$\vv\cdot\vw$ $=$ $14$

$\Vert\vv\Vert$ $=$ $5\sqrt{3}$

$\cos\theta$ $=$ $\fric<14,5\sqrt{3}\sqrt{14}>$ $=$ $\fric<\sqrt{42},15>$

$\sin\theta$ $=$ $\sqrt{1 - \cos^{2}\theta}$ $=$ $\fric<\sqrt{183},15>$

Find a unit vector in the direction of $\vv.$

$\fric<1,5\sqrt{3}>\rvt[1 -5 7 ]$

$\comp(\vv,\vw)$ $=$ $\fric<14,5\sqrt{3}>$

$\proj(\vv,\vw)$ $=$ $\fric<14,75>\rvt[1 -5 7 ]$

Find $\vw_{{}_{1}},\vw_{{}_{2}}\in\Rss3$ such that $\vw_{{}_{1}}$ is parallel to $\vv,$ $\vw_{{}_{2}}$ is orthogonal to $\vv,$ and $\vw_{{}_{1}} + \vw_{{}_{2}}$ $=$ $\vw.$

$\vw_{{}_{1}}$ $=$ $\proj(\vv,\vw)$ $=$ $\fric<14,75>\rvt[1 -5 7 ]$

$\vw_{{}_{2}}$ $=$ $\vw - \vw_{{}_{1}}$ $=$ $\rvt[-2 1 3 ] - \fric<14,75>\rvt[1 -5 7 ]$

Calculate $d(\vv,\vw)$ where $d$ is the Euclidean metric on $\Rss 3 .$

$d(\vv,\vw)$ $=$ $\Vert\vv - \vw\Vert$ $=$ $\Vert\rvt[3 -6 4 ]\Vert$ $=$ $\sqrt{61}$

Suppose that $\vv\in\Rss n$ and $c\in\R$. Prove that $\Vert c\vv\Vert$ $=$ $\vert c\vert\Vert\vv\Vert.$

See notes.

Set $A$ $=$ $\threetwomat[1 -1 2 2 0 -5 ],$ $B$ $=$ $\threetwomat[3 2 0 1 -3 4 ],$ $C$ $=$ $\twomat[1 -2 -4 0 ],$ $D$ $=$ $\threematb[-3 0 4 1 2 1 -1 0 -2],$ $E$ $=$ $\rvt[0 -3 1 ],$ and $F$ $=$ $\cvt[2 -8 5 ].$

Perform the indicated operation if possible. If not possible, explain why it is not.

[3]()

$2A - B$ $=$ $\threetwomat[-1 -4 4 3 3 -14 ]$

[10] $B + C$

Not possible. The matrices have differenct dimensions.

[10] $BC$ $=$ $\threetwomat[-5 -6 -4 0 -19 6 ]$

[10] $AD$

Not possible. Invalid dimensions.

[10] $DA$ $=$ $\threetwomat[-3 -17 5 -2 -1 11 ]$

[10] $EF$ $=$ $\left[29\right]$

[10] $FE$ $=$ $\threematb[0 -6 2 0 24 -8 0 -15 5]$

[10] Multiply the following matrices. Hint: You may want to use block multiplication.

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & 0\\ ... ...\hdashline 3 & 1 & -1 & 0\\ 3 & 1 & 0 & -1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 4 & 3 & -1 & 0\\ ... ... 0 & -1\\ 3 & 1 & -1 & 0\\ 3 & 1 & 0 & -1 \end{tabular}\hspace{-4pt}\right]$

(Exam 2)

Exam 2 Math 3720 Spring 2020

[10] Give the transpose of the following matrix.

$\twothreemat[3 -1 4 -2 7 8 ]$ $\threetwomat[3 -2 -1 7 4 8 ]$

Calculate the following determinants.

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -2 & 1 \\ \rais... ...\raisebox{2pt}{{\Tiny\(\tm\fric<1,2>\)}} \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm2\cdot\tm\fric<1,2> - \fric<3,2>\cdot 1$ $=$ $\tm\fric<1,2>$

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 2 \\ 0 & 1 & 8 \\ -1 & 2 & 6 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\tm0\left\vert\hspace{-4pt}\begin{tabular}{rr} -1 & 2 \\ 2 & 6 \end{tabular}\... ...pt}\begin{tabular}{rr} 1 & -1 \\ -1 & 2 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $1\cdot8 - 8\cdot1$ $=$ 0

[10] $\left\vert\hspace{-4pt}\begin{tabular}{rrrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & ... ...& 0 & \raisebox{2pt}{\Tiny\(\fric<1,7>\)} \end{tabular}\hspace{-4pt}\right\vert$ $=$ $\left(1\right) \left(\tm\fric<1,2>\right) \left(\fric<1,3>\right) \left(\tm\fri... ...ght) \left(\fric<1,5>\right) \left(\tm\fric<1,6>\right) \left(\fric<1,7>\right)$ $=$ $\tm\fric<1,7!>$ $=$ $\tm\fric<1,5040>$

[2]( For each of the following, find the inverse or explain why the inverse does not exist.)

[10] $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ -2 & 1 \\ \raisebox... ...} & \raisebox{2pt}{{\Tiny\(\tm\fric<1,2>\)}} \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ -2 & 1 & 1 & 0\\ ... ...}} & \raisebox{2pt}{{\Tiny\(\tm\fric<1,2>\)}} & 0 & 1\\ \end{tabular}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ -2 & 1 & 1 & 0\\ ... ...ght] \hspace{-8pt} \begin{tabular}{l} \\ \tiny\(2R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & 2\\ ... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} + 2R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & 2\\ ... ...-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - 3R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & 1 & 2\\ ... ...t] \hspace{-8pt} \begin{tabular}{l} \\ \tm\tiny\(R_{{}_{2}}\)\\ \end{tabular}$

Inverse: $\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 3 & 4 \end{tabular}\hspace{-4pt}\right]$

[10] $\twothreemat[3 -1 4 -2 7 8 ]$

The matrix is not invertible since it is not square.

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 0 \\ 1 & 0 & 3 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & -2 & 0 & 1 &... ... 0 & 3 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 3 & 0 & ... ...gin{tabular}{l} \tiny\(R_{{}_{2}}\)\\ \tiny\(R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ...}_{1}} - 3R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ...} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} + 3R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ...t} \begin{tabular}{l} \\ \tiny\(\tm\fric<1,2>R_{{}_{2}}\)\\ \\ \end{tabular}$

Inverse: $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & -3 \\ \raise... ...x{2pt}{{\Tiny\(\tm\fric<3,2>\)}}\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 2 \\ 0 & 1 & 8 \\ -1 & 2 & 6 \end{tabular}\hspace{-4pt}\right]$

The matrix is not invertible since $R_{{}_{2}} - R_{{}_{1}}$ $=$ $R_{{}_{3}}.$

Also, recall that $\left\vert\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 2 \\ 0 & 1 & 8 \\ -1 & 2 & 6 \end{tabular}\hspace{-4pt}\right\vert$ $=$ $0.$

[10] $\left[\hspace{-4pt}\begin{tabular}{rrrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ...& 0 & 0 & \raisebox{2pt}{\Tiny\(\fric<1,7>\)} \end{tabular}\hspace{-4pt}\right]$

Inverse: $\left[\hspace{-4pt}\begin{tabular}{rrrrrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 0 & ... ... 0 & 0 & -6 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 7 \end{tabular}\hspace{-4pt}\right]$

[10] Express the following as a product of elementary matrices.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 0 \\ 1 & 0 & 3 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

First, transform the identy matrix into the given matrix using elementary row operations.

[5]()

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\ 1 & -2 ... ...} \begin{tabular}{l} \\ \tiny\(R_{{}_{1}} - 2R_{{}_{3}}\)\\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 3 \\ 1 & -2 ... ...} \begin{tabular}{l} \tiny\(R_{{}_{1}} + 3R_{{}_{3}}\)\\ \\ \\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 0 \\ 1 & 0 ... ...gin{tabular}{l} \tiny\(R_{{}_{2}}\)\\ \tiny\(R_{{}_{1}}\)\\ \\ \end{tabular}$

Note that the identity matrix is transfomed into the given matrix by:

multiplying row 2 by $\tm2$;

adding row 1 to row 2;

adding 3 times row 3 to row 1;

interchanging rows 1 and 2.

Therefore, $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -2 & 0 \\ 1 & 0 & 3 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 1 & 0 \\ 1 & 0... ...\ 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right].$

[10] Choose one of the following. Give a proof.

The product of an invertible matrix and a singular matrix

is always singular.

is always invertible.

may be either invertible or singular depending on the matrices.

The product of an invertible matrix and a singular matrix is always singular. See Problem 38.

[10] Prove that the transpose of an invertible matrix is invertible.

See the proof of Corollary 135.

(Exam 3)

Exam 3 Math 3720 Spring 2020

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero. Do your own work. Do not ask others (except your instructor) for help. Do not research the answers. Write your answers on the space provided.

[10] Set $\vV$ $=$ $\left\{\cvt[\(a\) 0 0 ]:a\in\R\right\}$ and $\vW$ $=$ $\left\{\cvt[0 \(b\) 0 ]:b\in\R\right\}$. Find $\vV\cap\vW,$ $\vV\cup\vW,$ and $\vV + \vW$. Which of the three are subspaces of $\Rss3 ?$

$\vV\cap\vW$ $=$ $\left\{\cvt[0 0 0 ] \right\}$ $\leq$ $\Rss3$ (Theorem 167)

$\vV\cup\vW$ $=$ $\left\{\cvt[\(a\) \(b\) 0 ]:a,b\in\R,\text{ and } a = 0\text{ or }b = 0\right\}$ $\nleq$ $\Rss3$

Note that $\cvt[1 0 0 ]\in\vV\cup\vW$ and $\cvt[0 1 0 ]\in\vV\cup\vW$ but $\cvt[1 0 0 ] + \cvt[0 1 0 ]$ $=$ $\cvt[1 1 0 ]\notin\vV\cup\vW.$

$\vV + \vW$ $=$ $\left\{\cvt[\(a\) \(b\) 0 ]:a,b\in\R\right\}$ $\leq$ $\Rss3$ (Problem 60)

[10] Suppose that $A$ and $B$ are $n\times n$ matrices and $A$ is invertible. Show that $N(AB)$ $=$ $N(B).$

First, suppose that $\vx\in N(B)$. Then $(AB)\vx$ $=$ $A(B\vx)$ $=$ $A\vzero$ $=$ $\vzero$. Therefore, $N(B)$ $\subseteq$ $N(AB).$

Now suppose that $\vx\in N(AB)$. Then we have that

$(AB)\vx$ $=$ $\vzero$

$\inv A;(AB)\vx$ $=$ $\inv A;\vzero$

$(\inv A;A)B\vx$ $=$ $\inv A;\vzero$

$B\vx$ $=$ $\vzero.$

Therefore, $N(AB)$ $\subseteq$ $N(B)$.

Since $N(B)$ $\subseteq$ $N(AB)$ and $N(AB)$ $\subseteq$ $N(B),$ $N(AB)$ $=$ $N(B).$

[10] Find a nonzero matrix whose null space includes the following.

$\cvf[1 0 -1 0 ],$ $\cvf[1 1 1 0 ],$ $\cvf[0 0 -1 1 ]$

$\rvf[1 -2 1 1 ]$

[10] Suppose that $A$ is an $n\times n$ invertible matrix and $\vb\in\Rss n$. Show that $\vb\in$col$(A).$

By Corollary 186, $\vb\in$col$(A)$ if and only if there is $\vx\in\Rss n$ such that $A\vx$ $=$ $\vb$. Set $\vx$ $=$ $\inv A;\vb$ and note that $A\vx$ $=$ $A(\inv A;\vb)$ $=$ $(A\inv A;)\vb$ $=$ $\vb$. Therefore, $\vb\in$col$(A).$

For each of the following, determine whether the vectors are linearly dependent or linearly independent.

[10] $\cvf[1 0 -2 1 ],$ $\cvf[-3 1 4 1 ],$ $\cvf[-1 2 -2 1 ],$ $\cvf[1 -1 0 3 ]$

Set $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -3 & -1 & 1\\ ... ...& 2 & -1\\ -2 & 4 & -2 & 0\\ 1 & 1 & 1 & 3 \end{tabular}\hspace{-4pt}\right]$. Then rref$(A)$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 0 & 3\\ 0... ... & 0 & 1\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$. The columns are linearly dependent by Theorem 195.

[10] $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} 0\\ -3\\ 2\\ 4\\ 1\\ -5 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} -1\\ 3\\ -5\\ 2\\ 0\\ 4 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} 3\\ 1\\ -2\\ 0\\ 1\\ 3 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} -2\\ 0\\ -1\\ 3\\ 0\\ 3 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} 1\\ -2\\ 3\\ -4\\ 5\\ 0 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} 3\\ -2\\ -4\\ 1\\ 0\\ 5 \end{tabular}} \hspace{-4pt}\right],$ $\left[\hspace{-4pt}% \raisebox{-1.5pt}{% \begin{tabular}{r} -3\\ 4\\ 0\\ -1\\ 2\\ 1 \end{tabular}} \hspace{-4pt}\right]$

Seven vectors in $\Rss6$ are linearly dependent by Corollary 194.

[10] Give a basis for ${\bf M}_{{}_{3\times3}}$. Prove that it is in fact a basis.

$\left\{ \left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0 \\... ...& 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]\right\}$

Note that for any $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \(a_{{}_{11}}\) & \(... ... \(a_{{}_{33}}\) \end{tabular}\hspace{-4pt}\right] \in {\bf M}_{{}_{3\times3}},$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \(a_{{}_{11}}\) & \(... ...}_{31}}\) & \(a_{{}_{32}}\) & \(a_{{}_{33}}\) \end{tabular}\hspace{-4pt}\right]$ $=$ $a_{{}_{11}}\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0... ...}\\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$. Also, if $c_{{}_{11}}, c_{{}_{12}}, c_{{}_{13}}, c_{{}_{21}}, c_{{}_{22}}, c_{{}_{23}}, c_{{}_{31}}, c_{{}_{32}}, c_{{}_{33}}\in\R$ such that $c_{{}_{11}}\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0... ...}\\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right],$ then $c_{{}_{11}}$ $=$ $c_{{}_{12}}$ $=$ $c_{{}_{13}}$ $=$ $c_{{}_{21}}$ $=$ $c_{{}_{22}}$ $=$ $c_{{}_{23}}$ $=$ $c_{{}_{31}}$ $=$ $c_{{}_{32}}$ $=$ $c_{{}_{33}}$ $=$ 0. Therefore, the set given above is a basis for ${\bf M}_{{}_{3\times3}}.$

[10] Suppose that $A$ is an $m\times n$ matrix and $m < n$. What can be said about the dimension of $N(A)?$

It is at least $n - m.$

Since $A$ has $m$ rows, it has at most $m$ pivots which means it has at least $n - m$ free columns. Therefore, $n(A)\geq n - m.$

Suppose that $A$ and $B$ are $m\times n$ matrices.

[10] Show that $N(A)\cap N(B)\subseteq N(A + B).$

Suppose that $\vx\in N(A)\cap N(B)$. Then $(A + B)\vx$ $=$ $A\vx + B\vx$ $=$ $\vzero + \vzero$ $=$ $\vzero$. Therefore, $\vx\in N(A + B)$.

[10] Is it necessarily true that $N(A + B)$ $=$ $N(A) + N(B)?$

No.

Set $A$ $=$ $\twomat[1 0 0 0 ]$ and $B$ $=$ $\twomat[0 0 0 1 ]$. Then $N(A)$ $=$ $\left\{\cv[0 \(b\) ]: b\in\R\right\},$ $N(B)$ $=$ $\left\{\cv[\(a\) 0 ]: a\in\R\right\},$ $N(A) + N(B)$ $=$ $\left\{\cv[\(a\) \(b\) ]: a,b\in\R\right\}$ $=$ $\Rss2 ,$ and $N(A + B)$ $=$ $N\left(\idmattwo\right)$ $=$ $\left\{\cv[0 0 ]\right\}.$

(Final Exam)

Final Exam Math 3720 Spring 2020

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero. Do your own work. Do not ask others (except your instructor) for help. Do not research the answers. Write your answers on the space provided. Due: Friday May 8, 2020 at 12:00 p.m.

[10] Find an orthogonal basis for $\Rss 4$ that contains the vectors $\cvf[1 -1 0 2 ],$ $\cvf[1 1 1 0 ],$ and $\cvf[1 1 -2 0 ]$. Express $\cvf[1 2 3 4 ]$ as a linear combination of the basis.

Name:

[10] Give an example of a $3\times 5$ matrix in reduced row echelon form with rank 3. Find the four fundamental subspaces of the matrix.

Name:

[10] Answer the following as true or false (write the entire word). If the statement is true, then prove it. If the statement is false, then give a counterexample.

If $A$ is an orthogonal matrix and $B$ is row equivalent to $A,$ then $B$ is an orthogonal matrix.

Name:

[10] Give an example of two linearly independent vectors $\vv,\vw\in\Rss4$ such that each vector has at least three nonzero components. Set $\vW$ $=$ span$\left(\left\{\vv,\vw\right\}\right)$ and give a basis for $\per \vW;.$

Name:

[2]()

Solve completely.

[10]

$x$	$+$	$y$	$+$	$2z$	=	3
$2x$	$-$	$y$	$-$	$4z$	=	-5
$\tm6x$	$+$	$3y$	$+$	$4z$	=	-5

[10]

$x_{{}_{1}}$	$+$	$x_{{}_{2}}$	$+$	$x_{{}_{3}}$	$+$	$x_{{}_{4}}$	=	1
$\tm x_{{}_{1}}$	$+$	$2x_{{}_{2}}$	$-$	$x_{{}_{3}}$	$+$	$3x_{{}_{4}}$	=	-6
$2x_{{}_{1}}$	$-$	$x_{{}_{2}}$	$+$	$x_{{}_{3}}$	$+$	$2x_{{}_{4}}$	=	-6
$\tm x_{{}_{1}}$	$-$	$x_{{}_{2}}$	$-$	$2x_{{}_{3}}$	$+$	$3x_{{}_{4}}$	=	4

Name:

Let $P_{{}_{2}}$ be the set of polynomials with degree at most 2 and $P_{{}_{3}}$ be the set of polynomials with degree at most 3. Recall that $P_{{}_{2}}$ and $P_{{}_{3}}$ are vector spaces over $\R$. Set $\vW$ $=$ $\{f\in P_{{}_{3}}: f(0) = 0\}$.

[10] Show that $\vW\leq P_{{}_{3}}$.

[10] Define $T: P_{{}_{2}}\to \vW$ by $T(f)$ $=$ $\dint f(x)dx$. Recall from your calculus classes that $T$ is a linear transformation. Give the matrix (as in Note 301) for $T.$

Name:

[10] Find the kernel of $T$. Is $T$ 1 –1? Is $T$ onto?

Name:

[10] Find the eigenvalues of the following matrix.

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & -1 & 2 \\ 3 & 2 & -1 \\ 1 & 4 & -5 \end{tabular}\hspace{-4pt}\right]$

(Spring 2023)

(Quizzes)

Set $\vv$ $=$ $\rvt[1 \tm2 0 ]$ and $\vw$ $=$ $\rvt[\tm1 0 3 ]$ and let $\theta$ be the angle between $\vv$ and $\vw.$

[2] Compute each of the following.

[3]()

$2\vv - \vw$ $=$ $\rvt[3 \tm4 \tm3 ]$

$\vv\cdot\vw$ $=$ $\tm 1$

$\Vert\vv\Vert$ $=$ $\sqrt{5}$

[2]()

$\Vert\vw\Vert$ $=$ $\sqrt{10}$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\tm\fric<1,5\sqrt{2}>$

[1]
Find a vector that is orthogonal to $\vv.$

Any vector whose dot product with $\vv$ is 0. For example, $\rvt[2 1 0 ].$

Give the vector in the same direction as $\vv$ with magnitude 4.

$\fric<4,\Vert\vv\Vert>\vv$ $=$ $\fric<4,\sqrt{5}>\rvt[1 \tm2 0 ]$

[1] Suppose that $\vv = \rvv,\vw = \rvw\in\Rss n$. Prove that $\vv\cdot\vw$ $=$ $\vw\cdot\vv.$

See class notes.

[1] If possible, give two unit vectors in $\Rss 2$ whose sum is $\rv[1 2 ]$. If this is not possible, explain why it is not.

This is not possible. Suppose that $\vv$ and $\vw$ are unit vectors By the triangle inequality, $\Vert\vv + \vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\vw\Vert$ $=$ $2$. Since $\left\Vert\rv[1 2 ]\right\Vert$ $=$ $\sqrt{5}$ $>$ $2,$ $\rv[1 2 ]$ is not the sum of two unit vectors.

[1] Set $\vv$ $=$ $\rvt[1 -1 2 ]$ and $\vw$ $=$ $\rvt[0 1 -1 ]$. Calculate each of the following.

$\Vert\vv\Vert$ $=$ $\sqrt{6}$ $\comp(\vv,\vw)$ $=$ $\fric<\vv\cdot\vw,\Vert v\Vert>$ $=$ $\fric<\tm3,\sqrt{6}>$ $\proj(\vv,\vw)$ $=$ $\fric<\vv\cdot\vw,\Vert v\Vert^{2}>\vv$ $=$ $\tm\fric<1,2>\rvt[1 -1 2 ]$

[1] Set $A$ $=$ $\twofourmatb[1 0 1 4 0 -2 8 -3]$ and $B$ $=$ $\twofourmatb[1 0 0 -1 -1 0 -2 4]$. Compute the following.

$A + 2B$ $=$ $\twofourmatb[2 0 1 3 -1 -2 6 1]$

[1] Suppose that $A$ $=$ $\twothreemat[\(a\) \(b\) \(c\) \(d\) \(e\) \(f\) ]$. For each of the following, give an elementary matrix $E$ such that

$B$ $=$ $\twothreemat[\(d\) \(e\) \(f\) \(a\) \(b\) \(c\) ]$ $B$ $=$ $\twothreemat[\(a\) \(b\) \(c\) \(\tm d\) \(\tm e\) \(\tm f\) ]$ $B$ $=$ $\twothreemat[\(a+2d\) \(b+2e\) \(c+2f\) \(d\) \(e\) \(f\) ]$

[2] For the matrix below, use Gauss-Jordan elimination to transform the matrix into the identity matrix. Find a left inverse.

$\twomat[1 2 1 3 ]$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 2\\ 0 & 1 \end{... ...{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr} \vspace{-12pt}\\ 1 & 0\\ 0 & 1 \end{... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} - 2R_{{}_{2}}\)\\ \\ \end{tabular}$

Set $E_{{}_{1}}$ $=$ $\twomat[1 0 -1 1 ],$ $E_{{}_{2}}$ $=$ $\twomat[1 -2 0 1 ],$ and $E$ $=$ $E_{{}_{2}}E_{{}_{1}}$ $=$ $\twomat[3 -2 -1 1 ]$. Check that $\twomat[3 -2 -1 1 ]\twomat[1 2 1 3 ]$ $=$ $\idmattwo.$

[2] For each of the following, find the inverse or show that it does not exist.

[2]()

$\threematb[0 3 -1 -1 -1 1 1 2 -1 ]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 0 & 3 & -1 & 1 &... ...-1 & 1 & 0 & 1 & 0\\ 1 & 2 & -1 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ -1 & -1 & 1 & 0 &... ...}\)\\ \tiny\(R_{{}_{1}}\)\\ \tiny\(R_{{}_{3}} + R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ -1 & 0 & 1 & 0 & ... ...\)\\ \tiny\(R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} - 3R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ -1 & 0 & 0 & 1 & ... ...R_{{}_{3}}\)\\ \tiny\(R_{{}_{3}}\)\\ \tiny\(\tm R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & -1 &... ...begin{tabular}{l} \tiny\(\tm R_{{}_{1}} + R_{{}_{3}}\)\\ \\ \\ \end{tabular}$

$\threematb[1 2 3 4 5 6 5 7 9 ]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 3 & 1 & ... ...& 5 & 6 & 0 & 1 & 0\\ 5 & 7 & 9 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 3 & 1 & ... ...ular}{l} \\ \\ \tiny\(R_{{}_{3}} - R_{{}_{1}} - R_{{}_{2}}\)\\ \end{tabular}$

Not invertible.

BONUS: [1] Compute the determinants of the matrices above.

$\threematd\vert 3 -1 -1 -1 1 1 2 -1 \vert$ $=$ $\tm3 \twomat\vert-1 1 1 -1 \vert - \twomat\vert-1 -1 1 2 \vert$ $=$ $0 + 1$ $=$ $1$

$\threematd\vert 1 2 3 4 5 6 5 7 9 \vert$ $=$ $\twomat\vert 5 6 7 9 \vert - 2\twomat\vert 4 6 5 9 \vert + 3\twomat\vert 4 5 5 7 \vert$ $=$ $3 - 12 + 9$ $=$ 0

[1] Give the transpose of the following matrix.

$\threetwomat[1 4 3 -5 0 2 ]$

$\twothreemat[1 3 0 4 -5 2 ]$

[1] Prove that for any matrix, $A,$ $A\tran A;$ is symmetric.

See class notes.

[2] Calculate the following determinants.

[2]()

$\twomat\vert 3 4 1 2 \vert$ $=$ $6 - 4$ $=$ $2$

$\threematd\vert 1 0 4 -2 1 0 0 0 -3\vert$ $=$ $\tm3\twomat\vert 1 0 -2 1 \vert$ $=$ $\tm3(1 - 0)$ $=$ $\tm3$

Suppose that $A$ and $B$ are $n\times n$ matrices, $\vert A\vert$ $=$ $2,$ and $\vert B\vert$ $=$ $\tm 1$. Compute each of the following. Be careful with the last one.

[3]()

$\vert AB\vert$ $=$ $\vert A\vert\vert B\vert$ $=$ $\tm2$

$\left\vert\inv A;\right\vert$ $=$ $\fric<1,\vert A\vert>$ $=$ $\fric<1,2>$

$\vert 2A\vert$ $=$ $2^{{}^{n}}\vert A\vert$ $=$ $2^{{}^{n+1}}$

For each of the following, transform the matrix to reduced row echelon form. For each free column, find a nonzero vector in the null space.

[1] $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 0 & 4 & 0 & 0\\ 0 & 1 & 0 & 0 & -1\\ 0 & 0 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

The matrix is in reduced row echelon form. The free columns are 3 and 5. The corresponding vectors in the null space are below.

$\cvfive[-4 0 1 0 0 ]$ $\cvfive[0 1 0 0 1 ]$

[1] $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & 3\\ 0 & 1 & 2\\ 0 & 2 & 4\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt} \begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Free column: 3

$\cvt[-1 -2 1 ]$

[3] Determine whether the given vectors are linearly independent or linearly dependent.

[2]()

$\cv[1 2 ],$ $\cv[3 4 ]$

Independent.

See class notes.

$\cv[1 2 ],$ $\cv[3 4 ],$ $\cv[5 6 ]$

Dependent.

Three vectors in $\Rss 2 .$

[2]()

$\cvt[1 2 3 ],$ $\cvt[4 1 5 ],$ $\cvt[5 -4 1 ]$

Dependent.

$2\cvt[4 1 5 ] -3\cvt[1 2 3 ]$ $=$ $\cvt[5 -4 1 ]$

Using column vectors:

$\threematb[1 4 5 2 1 -4 3 5 1 ]$ $\rref$ $\threematb[1 0 -3 0 1 2 0 0 0 ]$

Pivots: 2

Using row vectors:

$\threematb[1 2 3 4 1 5 5 -4 1 ]$ $\rref$ $\threematb[1 0 1 0 1 1 0 0 0 ]$

Row of zeros.

$\cvt[2 0 1 ],$ $\cvt[1 0 2 ],$ $\cvt[4 1 3 ]$

Independent.

Using column vectors:

$\threematb[2 1 4 0 0 1 1 2 3 ]$ $\rref$ $\idmatthree$

Pivots: 3

Using row vectors:

$\threematb[2 0 1 1 0 2 4 1 3 ]$ $\rref$ $\idmatthree$

No zero rows.

[2] Determine whether or not the given vectors form a basis for $\Rss3$.

$\cvt[1 2 3 ],$ $\cvt[1 2 0 ],$ $\cvt[1 0 0 ]$

Set $A$ $=$ $\threematb[1 1 1 2 2 0 3 0 0 ]$. It is readily seen that $\vert A\vert$ $=$ $\tm6$. Therefore, $A$ is invertible which means that its columns are linearly independent. Three linearly independent vectors in $\Rss3$ form a basis.

$\cvt[1 0 1 ],$ $\cvt[0 1 0 ],$ $\cvt[2 5 2 ]$

Since $\cvt[2 5 2 ]$ $=$ $2\cvt[1 0 1 ] + 5\cvt[0 1 0 ],$ the vectors are linearly dependent and therefore do not form a basis.

Note that $\left\{\vv_{{}_{1}} = \cvt[1 -1 0 ], \vv_{{}_{2}} = \cvt[2 2 1 ], \vv_{{}_{3}} = \cvt[1 1 -4 ]\right\}$ is an orthogonal set of vectors.

[1] Prove or reasonably explain that $\left\{\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}}\right\}$ is a basis for $\Rss3$.

Recall that an orthogonal set of vectors is linearly independent. Three linearly independent vectors in $\Rss3$ form a basis.

[1] Express $\vw = \cvt[1 -3 13 ]$ as a linear combination of $\vv_{{}_{1}}, \vv_{{}_{2}},$ and $\vv_{{}_{3}}.$

$\vw$ $=$ $\fricp<\vw\cdot\vv_{{}_{1}},\Vert\vv_{{}_{1}}\Vert^{2}> \vv_{{}_{1}} + \fricp<\... ...{{}_{2}} + \fricp<\vw\cdot\vv_{{}_{3}},\Vert\vv_{{}_{3}}\Vert^{2}> \vv_{{}_{3}}$ $=$ $\fric<4,2>\vv_{{}_{1}} + \fric<9,9>\vv_{{}_{2}} + \fric<\tm54,18>\vv_{{}_{2}}$ $=$ $2\vv_{{}_{1}} + \vv_{{}_{2}} - 3\vv_{{}_{2}}$

$\cvt[1 -3 13 ]$ $=$ $2\cvt[1 -1 0 ] + \cvt[2 2 1 ] - 3\cvt[1 1 -4 ]$

[2] Find the four fundamental subspaces of the following matrix.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -4 & 0 & 1\\ 0 & 0 & 1 & -2\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

[2]()

$R(A)$ $=$ span$\left(\cvf[1 -4 0 1 ],\cvf[0 0 1 -2 ]\right)$

$N(A)$ $=$ span$\left(\cvf[4 1 0 0 ],\cvf[-1 0 2 1 ]\right)$

$C(A)$ $=$ span$\left(\cvt[1 0 0 ],\cvt[0 1 0 ]\right)$

$\tran A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ -4 & 0 & 0\\ 0 & 1 & 0\\ 1 & -2 & 0 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$N\left(\tran A;\right)$ $=$ span$\left(\cvt[0 0 1 ]\right)$

[1] Set $\vW$ $=$ span$\left(\cvt[1 4 1 ]\right)$. Find a basis for $\per \vW;.$

See class notes.

Alternatively, note that $\cvt[1 0 -1 ]$ and $\cvt[-4 1 0 ]$ are both orthogonal to $\vW$ and are linearly independent of each other. Since $\dim\vW + \dim\per\vW;$ $=$ $3,$ $\left\{\cvt[1 0 -1 ],\cvt[-4 1 0 ]\right\}$ is a basis for $\per \vW;.$

(Exam 1)

Exam 1 Math 3720 Spring 2023 Name:

Set $\vv$ $=$ $\rvt[1 \tm2 0 ]$ and $\vw$ $=$ $\rvt[\tm1 0 3 ]$ and let $\theta$ be the angle between $\vv$ and $\vw$.

[70] Compute each of the following.

[3]()

$2\vv + \vw$ $=$ $\rvt[1 -4 3 ]$

$\vv\cdot\vw$ $=$ $\tm 1$

$\Vert\vv\Vert$ $=$ $\sqrt{5}$

$\Vert\vw\Vert$ $=$ $\sqrt{10}$

$\cos\theta$ $=$ $\tm\fric<1,5\sqrt{2}>$

$\comp(\vv,\vw)$ $=$ $\tm\fric<1,\sqrt{5}>$

$\proj(\vv,\vw)$ $=$ $\tm\fric<1,5>\rvt[1 \tm2 0 ]$

[10] Suppose that $\vv,\vw\in\Rss n$. Prove that $\Vert\vv - \vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\vw\Vert.$

By the Triangle Inequality, $\Vert\vv - \vw\Vert$ $=$ $\Vert\vv + \tm\vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\tm \vw\Vert$ $=$ $\Vert\vv\Vert + \vert\tm1\vert\cdot\Vert\vw\Vert$ $=$ $\Vert\vv\Vert + \Vert\vw\Vert.$

[40] Set $A$ $=$ $\twothreemat[1 0 2 -1 1 3 ],$ $B$ $=$ $\threetwomat[1 2 -3 0 4 -1 ],$ and $C$ $=$ $\twothreemat[5 1 -2 0 -2 5 ]$. For each of the following, either perform the calculation or explain why it is not possible.

[2]()

$2A + C$ $=$ $\twothreemat[7 1 2 -2 0 11 ]$

$AB$ $=$ $\twomat[9 0 8 -5 ]$

$BA$ $=$ $\threematb[-1 2 8 -3 0 -6 5 -1 5 ]$

$AC$ is undefined

[20] Set $A$ $=$ $\threematb[1 2 -4 1 0 2 3 -1 4]$ and $B$ $=$ $\threematb[3 2 0 -3 2 4 1 0 1]$. Compute each of the following.

[2]()

$\vr_{{}_{2}}(AB)$ $=$ $\vr_{{}_{2}}(A)B$ $=$ $\rvt[5 2 2 ]$

$\vc_{{}_{1}}(AB)$ $=$ $A\vc_{{}_{1}}(B)$ $=$ $\cvt[-7 5 16 ]$

[10] Set $A$ $=$ $\threematb[1 0 0 0 0 0 0 0 1]$. Compute $A^{{}^{10}}.$

Since $A^{{}^{2}}$ $=$ $A,$ $A^{{}^{10}}$ $=$ $A.$

[10] Multiply.

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rr:rr}% 1 & 0 & 0 & 0 \... ... & -1 & -2 & -3 \\ 1 & 2 & 3 & -1 & -2 & -3 \end{tabular}}\hspace{-4pt}\right]$ $=$ $\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr:rrr}% 1 & 2 & 3 & 1... ... 0 & 0 & 2 & 4 & 6 \\ 0 & 0 & 0 & 2 & 4 & 6 \end{tabular}}\hspace{-4pt}\right]$

[10] Suppose that $A$ and $B$ are $3\times3$ matrices such that $\inv A;$ $=$ $\threematb[1 1 -1 0 1 1 1 1 1 ]$ and $AB$ $=$ $\threematb[-4 8 -12 4 -5 7 2 5 9 ]$. Find $B.$

$B$ $=$ $(\inv A;A)B$ $=$ $\inv A;(AB)$ $=$ $\threematb[-2 -2 -14 6 0 16 2 8 4 ]$

[20] Suppose that $A$ is a $2\times 2$ matrix and $\inv A;$ $=$ $\twomat[1 -4 0 2 ]$. Find each of the following.

[2]()

$\invp A^{{}^{2}};$ $=$ $\left(\inv A;\right)^{{}^{2}}$ $=$ $\twomat[1 -12 0 4 ]$

$\invp\tm3 A;$ $=$ $\tm\fric<1,3>\inv A;$ $=$ $\twomat[\raisebox{2pt}{\Tiny\(\tm\fric<1,3>\)} \raisebox{2pt}{\Tiny\(\fric<4,3>\)} 0 \raisebox{2pt}{\Tiny\(\tm\fric<2,3>\)} ]$

[10] Show that the following matrix is not invertible.

$\twomat[1 0 3 0 ]$

Suppose that $\twomat[\(a\) \(b\) \(c\) \(d\) ]$ is any $2\times 2$ matrix. Then $\twomat[\(a\) \(b\) \(c\) \(d\) ]\twomat[1 0 3 0 ]$ $=$ $\twomat[\(a+3b\) \(0\) \(c+3d\) \(0\) ]$ which is not the identity matrix.

(Exam 2)

Exam 2 Math 3720 Spring 2023 Name:

For each of the following, find the inverse or show that it does not exist.

[3]()

[10] $\twomat[1 -3 0 2 ]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rr:rr}% 1 & -3 & 1 & 0 \\ 0 & 2 & 0 & 1 \end{tabular}}\hspace{-4pt}\right]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rr:rr}% 1 & -3 & 1 & 0 ... ... & 0 & \raisebox{2pt}{{\tiny\(\fric<1,2>\)}} \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \Tiny\(\fric<1,2>R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rr:rr}% 1 & 0 & 1 & \ra... ... & 0 & \raisebox{2pt}{{\tiny\(\fric<1,2>\)}} \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \Tiny\(R_{{}_{1}} + 3R_{{}_{2}}\)\\ \\ \end{tabular}$

$\twomat[1 \raisebox{2pt}{{\tiny\(\fric<3,2>\)}} 0 \raisebox{2pt}{{\tiny\(\fric<1,2>\)}} ]$

[10] $\twomat[4 -2 2 -1 ]$

Singular

$\twomat\vert 4 -2 2 -1 \vert$ $=$ 0

[10] $\threematb[1 1 1 0 1 1 0 0 1 ]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr:rrr}% 1 & 1 & 1 & 1... ... 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{tabular}}\hspace{-4pt}\right]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr:rrr}% 1 & 0 & 0 & 1... ...0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \Tiny\(R_{{}_{1}} - R_{{}_{2}}\)\\ \Tiny\(R_{{}_{2}} - R_{{}_{3}}\)\\ \\ \end{tabular}$

$\threematb[1 -1 0 0 1 -1 0 0 1 ]$

Set $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8\\ 9 & 10 & 11 & 12 \end{tabular}\hspace{-4pt}\right]$. For each of the following, find a matrix $E$ such that $EA$ $=$ $B.$

[2]()

[10] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ -5 & -6 & -7 & -8\\ 1 & 2 & 3 & 4 \\ 9 & 10 & 11 & 12 \end{tabular}\hspace{-4pt}\right]$

$E$ $=$ $\threematb[0 -1 0 1 0 0 0 0 1 ]$

[10] $B$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8\\ 15 & 18 & 21 & 24 \end{tabular}\hspace{-4pt}\right]$

$E$ $=$ $\threematb[1 0 0 0 1 0 0 3 0 ]$ or $E$ $=$ $\threematb[1 0 0 0 1 0 1 1 1 ]$

[10] Give the transpose of the following.

$\twothreemat[1 2 3 4 5 6 ]$

$\threetwomat[1 4 2 5 3 6 ]$

[10] Give the definition of a symmetric matrix.

See class notes.

[10] Prove that for any square matrix $A,$ $A + \tran A;$ is symmetric.

See class notes.

Calculate the following determinants.

[10] $\twomat\vert 1 -7 4 8 \vert$ $=$ $36$

[10] $\threematd\vert 3 1 4 1 5 9 2 6 5 \vert$ $=$ $3\twomat\vert 5 9 6 5 \vert - 1\twomat\vert 1 9 2 5 \vert + 4\twomat\vert 1 5 2 6 \vert$ $=$ $3(\tm29) - (\tm13) + 4(\tm4)$ $=$ $\tm90$

Bonus [5] Identify the pattern of the entries in the second determinant above.

The entries are the first nine digits of $\pi.$

[10] Explain why the following determinant is 0.

$\left\vert\hspace{-4pt}\begin{tabular}{rrrrrr} \vspace{-12pt}\\ 1 & 2 & 3 & 4... ...& 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right\vert$

Two of the rows are identical.

Suppose that $A$ is a square matrix with $\vert A\vert$ $=$ $\tm2$. Find the following determinants.

[2]()

[10] $\left\vert\inv A;\right\vert$ $=$ $\fric<1,\vert A\vert>$ $=$ $\tm\fric<1,2>$

[10] $\left\vert A\tran A;\right\vert$ $=$ $\left\vert A\right\vert\cdot\left\vert\tran A;\right\vert$ $=$ $\vert A\vert\cdot\vert A\vert$ $=$ $4$

[10] Prove that the additive identity in a vector space is unique.

See class notes.

[10] Suppose that $A$ is an $m\times n$ matrix. Prove that $N(A)\leq\Rss n .$

See class notes.

(Exam 3)

Exam 3 Math 3720 Spring 2023 Name:

[2]()

[10] For each free column, find a nonzero vector in the null space.

[10] $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 0 & 1 & 0 & 2\\ 0 & 1 & -2 & 0 & 1\\ 0 & 0 & 0 & 1 & -6 \end{tabular}\hspace{-4pt}\right]$

Free columns: 3, 5

$\cvfive[-1 2 1 0 0 ],$ $\cvfive[-2 -1 0 6 1 ]$

[10] Find the null space of the following.

$\threetwomat[1 0 0 1 0 0 ]$

$\left\{\cv[0 0 ]\right\}$

Determine whether or not the given set of vectors is a basis for $\Rss 2 .$

[10] $\left\{\cv[1 2 ], \cv[3 4 ]\right\}$

Yes. Since neither vector is a scalar multiple of the other, the vectors are linearly independent. Two linearly independent vectors form a basis for $\Rss 2 .$

[10] $\left\{\cv[\(\sqrt{2}\) 7 ], \cv[1 \(\pi\) ], \cv[\(e\) 4 ]\right\}$

No. Three vectors in $\Rss 2$ are linearly dependent and therefore not a basis.

[10] Determine whether the following set of vectors is linearly independent or linearly dependent.

[3]()

$\left\{\cvt[1 0 3 ],\cvt[0 1 2 ],\cvt[1 0 1 ]\right\}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 1 \\ 0 & 1 & 0 \\ 3 & 2 & 1 \end{tabular}}\hspace{-4pt}\right]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 2 & -2 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \Tiny\(R_{{}_{3}} - 3R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & -1 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \Tiny\(\fric<1,2>R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \\ \Tiny\(R_{{}_{3}} - R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \Tiny\(R_{{}_{1}} + R_{{}_{3}}\)\\ \\ \Tiny\(\tm R_{{}_{3}}\)\\ \end{tabular}$

Linearly independent.

[10]

See class notes.

[10] Note that the following is an orthogonal set of vectors. Express $\cvt[1 2 3 ]$ as a linear combination of the vectors in the set.

[2]()

$\left\{\cvt[1 1 1 ],\cvt[1 -1 0 ],\cvt[1 1 -2 ]\right\}$

$\cvt[1 2 3 ]\cdot\cvt[1 1 1 ]$ $=$ $6$

$\cvt[1 2 3 ]\cdot\cvt[1 -1 0 ]$ $=$ $\tm 1$

$\cvt[1 2 3 ]\cdot\cvt[1 1 -2 ]$ $=$ $\tm3$

$\fric<6,3>\cvt[1 1 1 ] - \fric<1,2>\cvt[1 -1 0 ] - \fric<3,6>\cvt[1 1 -2 ]$ $=$ $\cvt[1 2 3 ]$

$2\cvt[1 1 1 ] - \fric<1,2>\cvt[1 -1 0 ] - \fric<1,2>\cvt[1 1 -2 ]$ $=$ $\cvt[1 2 3 ]$

[10] Suppose that $\left\{\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}}\right\}$ is an orthogonal set of vectors in $\Rss 4$ and $\vw\notin$span$\left(\left\{\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}}\right\}\right)$. Describe how the Gram-Schmidt orthogonalization process is used to define a vector $\vv_{{}_{4}}\in\Rss 4$ such that $\left\{\vv_{{}_{1}},\vv_{{}_{2}},\vv_{{}_{3}},\vv_{{}_{4}}\right\}$ is an orthogonal basis for $\Rss 4 .$

Set $\vv_{{}_{4}}$ $=$ $\vw - \fricp<\vw\cdot\vv_{{}_{1}},\Vert\vv_{{}_{1}}\Vert^{2}> \vv_{{}_{1}} - \... ...}_{2}} - \fricp<\vw\cdot\vv_{{}_{3}},\Vert\vv_{{}_{3}}\Vert^{2}> \vv_{{}_{3}}.$

[10] Suppose that $\vv,\vw\in\Rss n$ and $\vw$ $\ne$ $\vzero$. Prove that $\vv - \fricp<\vw\cdot\vv,\Vert\vw\Vert^{2}> \vw$ is orthogonal to $\vw.$

Note that $\left[\vv - \fricp<\vw\cdot\vv,\Vert\vw\Vert^{2}> \vw\right]\cdot\vw$ $=$ $\vv\cdot\vw - \fricp<\vw\cdot\vv,\Vert\vw\Vert^{2}> \vw\cdot\vw$ $=$ $\vv\cdot\vw - \fricp<\vw\cdot\vv,\Vert\vw\Vert^{2}> \Vert\vw\Vert^{2}$ $=$ $\vv\cdot\vw - \vw\cdot\vv$ $=$ $0.$

[10] Recall that a matrix $A$ is symmetric if $A$ $=$ $\tran A;$. Prove that an orthogonal symmetric matrix is its own inverse.

Suppose that $A$ is both orthogonal and symmetric. Then $A$ $=$ $\tran A;$ $=$ $\inv A;.$

(Final Exam)

Final Exam Math 3720 Spring 2023 Name:

Directions: Show all of your work and justify all of your answers.

[10] Find the four fundamental subspaces of the following matrix.

$\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & 0 & 2 & 0\\ 0 & 1 & -1 & 1\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

[10] Give an example of an orthogonal basis for $\Rss3$ other than the standard unit basis.

[10] Express $\cvt[1 2 3 ]$ as a linear combination of the basis from the previous part.

[10] The dimension of the row space of a $4\times 6$ matrix is 3. What is the dimension of its null space?

[10] Set $\vW$ $=$ span$\left(\cvf[1 0 1 1 ],\cvf[0 1 2 1 ]\right)$. Find a basis for $\vW^{{}^{\perp}}.$

[30] Give the complete solution of each of the following.

[3]()

$x$	$+$	$y$	$+$	$z$	$=$	$\tm2$
$\tm x$	$+$	$2y$	$-$	$z$	$=$	$\tm4$
$2x$	$+$	$3y$	$+$	$4z$	$=$	$\tm4$

$x$	$-$	$2y$	$+$	$z$	$=$	$2$
$\tm x$	$+$	$3y$	$-$	$2z$	$=$	1
$2x$	$-$	$3y$	$+$	$z$	$=$	$6$

$x$	$+$	$y$	$+$	$z$	$=$	$\tm 1$
$\tm2x$	$+$	$y$	$-$	$z$	$=$	4
$x$	$+$	$4y$	$+$	$2z$	$=$	$1$

Define $T:\Rss 2 \to \Rss 3$ by $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $\cvt[\(2x\) \(x+y\) \(y-x\) ].$

[10] Prove that $T$ is a linear transformation.

[10] Find a matrix $A$ such that $T(\vx)$ $=$ $A\vx$ for all $\vx\in\Rss 3 .$

[10] Explain why $T:\Rss2 :\to\Rss2$ defined by $T\left(\cv[\(x\) \(y\) ]\right)$ $=$ $\cv[\(\vert x\vert\) \(y\) ]$ is not a linear transformation.

[20] For the following matrix, find the eigenvalues, a basis for each eigenspace, the algebraic multiplicity of each eigenvalue, and the geometric multiplicity of each eigenvalue.

$\threematb[3 2 -1 0 2 1 0 1 2 ]$

(Summer 2023)

(Quizzes)

[3] Set $\vv$ $=$ $\rvt[\tm1 0 4 ],$ $\vw$ $=$ $\rvt[3 \tm1 2 ],$ and let $\theta$ be the angle between $\vv$ and $\vw.$

Compute each of the following.

[3]()

$2\vv + \vw$ $=$ $\rvt[1 \tm1 10 ]$

$\vv\cdot\vw$ $=$ $5$

$\Vert\vv\Vert$ $=$ $\sqrt{17}$

[2]()

$\Vert\vw\Vert$ $=$ $\sqrt{14}$

$\cos\theta$ $=$ $\fric<5,\sqrt{17}\sqrt{14}>$ $=$ $\fric<5,\sqrt{238}>$

Find a nonzero vector that is orthogonal to $\vv.$

Any vector whose dot product with $\vv$ is 0 is orthogonal to $\vv$.

One example is $\rvt[4 0 1 ].$

[1] Suppose that $\vx\in \Rss 3$ and $\Vert\vx\Vert$ $=$ 0. Prove that $\vx$ $=$ $\vzero.$

See class notes.

BONUS (2) Set $A$ $=$ $\twothreemat[2 -1 5 0 1 3 ],$ $B$ $=$ $\twothreemat[1 0 6 -1 -2 5 ],$ and $C$ $=$ $\threetwomat[1 2 0 1 -1 8 ]$. perform the indicated operation if possible. If it is not possible, explain why it is not.

[2]()

$2A - B$ $=$ $\twothreemat[3 -2 4 1 4 1 ]$

$A + C$

Undefined. The dimensions are different.

[2]()

$AC$ $=$ $\twomat[-3 43 -3 25 ]$

$AB$

Undefined. Matrix $A$ has 3 columns and matrix $B$ has 2 rows.

[1] If possible, give two unit vectors in $\Rss3$ whose sum is $\rvt[2 0 1 ]$. If this is not possible, explain why it is not.

This is not possible. Suppose that $\vu$ and $\vv$ are unit vectors. By the Triangle Inequality, $\Vert\vu + \vv\Vert$ $\leq$ $\Vert\vu\Vert + \Vert\vv\Vert$ $=$ $2$. Since $\left\Vert\rvt[2 0 1 ]\right\Vert$ $=$ $\sqrt{5}$ $>$ $2,$ $\vu + \vv$ $\ne$ $\left\Vert\rvt[2 0 1 ]\right\Vert.$

[1] Set $\vv$ $=$ $\rvt[1 1 1 ]$ and $\vw$ $=$ $\rvt[3 1 4 ]$. Calculate each of the following.

[3]()

$\Vert\vv\Vert$ $=$ $\sqrt{3}$

$\comp(\vv,\vw)$ $=$ $\fric<\vv\cdot\vw,\Vert v\Vert>$ $=$ $\fric<8,\sqrt{3}>$

$\proj(\vv,\vw)$ $=$ $\fricp<\vv\cdot\vw,\Vert v\Vert^{2}> \vv$ $=$ $\fric<8,3>\rvt[1 1 1 ]$

[1] Compute each of the following products.

[2]()

$\threematb[3 0 -1 -1 0 1 -2 -4 5] \threematb[7 1 -1 2 -1 5 0 1 1 ]$ $=$ $\threematb[21 2 -4 -7 0 2 -22 2 -13 ]$

$\threematb[7 1 -1 2 -1 5 0 1 1 ] \threematb[3 0 -1 -1 0 1 -2 -4 5]$ $=$ $\threematb[22 4 -11 -15 -20 21 -3 -4 6 ]$

Suppose that $A,$ $B,$ and $C$ are $2\times 2$ matrices such that $AB$ $=$ $\twomat[1 -1 3 4 ]$ and $A(B + C)$ $=$ $\twomat[2 1 0 7 ].$

[1] Find $AC.$

$A(B + C)$ $=$ $AB + AC$

$\twomat[2 1 0 7 ]$ $=$ $\twomat[1 -1 3 4 ] + AC$

$AC$ $=$ $\twomat[2 1 0 7 ] - \twomat[1 -1 3 4 ]$

$AC$ $=$ $\twomat[1 2 -3 3 ]$

[1] Use block multiplication to compute the following product. Then express your answer as a single $4\times4$ matrix.

$\twomat[\(A\) \(A\) \(0\) \(A\) ] \twomat[\(B\) \(B\) \(C\) \(0\) ]$ $=$ $\twomat[\(AB+AC\) \(AB\) \(AC\) \(0\) ]$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 2 & 1 & 1 & -1\\ ... ...7 & 3 & 4\\ 1 & 2 & 0 & 0\\ -3 & 3 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

[1] Given that $A$ $=$ $\twomat[1 2 0 -1 ]$ is invertible and $\inv A;B$ $=$ $\twomat[-1 8 1 -3 ],$ find $B.$

$B$ $=$ $A\inv A;B$ $=$ $\twomat[1 2 0 -1 ] \twomat[-1 8 1 -3 ]$ $=$ $\twomat[1 2 -1 3 ]$

[2] Given that $\inv A;$ $=$ $\twomat[1 2 1 1 ]$ and $\inv B;$ $=$ $\twomat[1 1 0 1 ],$ calculate each of the following.

$\invp AB;$ $=$ $\inv B;\inv A;$ $=$ $\twomat[1 1 0 1 ] \twomat[1 2 1 1 ]$ $=$ $\twomat[2 3 1 1 ]$

$\invp BA;$ $=$ $\inv A;\inv B;$ $=$ $\twomat[1 2 1 1 ] \twomat[1 1 0 1 ]$ $=$ $\twomat[1 3 1 2 ]$

$\invp A^{{}^{2}};$ $=$ $\left(\inv A;\right)^{{}^{2}}$ $=$ $=$ $\twomat[1 2 1 1 ] \twomat[1 2 1 1 ]$ $=$ $\twomat[3 4 2 3 ]$

$\invp 2A;$ $=$ $\fric<1,2>\inv A;$ $=$ $\twomat[% \raisebox{2pt}{\tiny\(\fric<1,2>\)} 1 \raisebox{2pt}{\tiny\(\fric<1,2>\)} \raisebox{2pt}{\tiny\(\fric<1,2>\)} ]$

[1] Set $A$ $=$ $\twothreemat[1 1 2 1 0 2 ]$. Give the reduced row echelon form of $A$ and find a matrix $B$ such that $A\rref BA.$

[2]()

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 1 & 2 \\ 1 & 0 & 2 \end{tabular}}\hspace{-4pt}\right]$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 1 & 2 \\ 0 & -1 & 0 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \Tiny\(R_{{}_{2}} - R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 2 \\ 0 & -1 & 0 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \Tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \\ \end{tabular}$

$\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 2 \\ 0 & 1 & 0 \end{tabular}}\hspace{-4pt}\right]$ $\begin{tabular}{l} \\ \Tiny\(\tm R_{{}_{2}}\)\\ \end{tabular}$

Set $E_{{}_{1}}$ $=$ $\twomat[1 0 -1 1 ],$ $E_{{}_{2}}$ $=$ $\twomat[1 1 0 1 ],$ $E_{{}_{3}}$ $=$ $\twomat[1 0 0 -1 ],$ and $B$ $=$ $E_{{}_{3}}E_{{}_{2}}E_{{}_{1}}$ $=$ $\twomat[0 1 1 -1 ].$

Check that $BA$ $=$ $\twomat[0 1 1 -1 ] \twothreemat[1 1 2 1 0 2 ]$ $=$ $\left[\hspace{-8pt}% \raisebox{-1.5pt}{ \begin{tabular}{rrr}% 1 & 0 & 2 \\ 0 & 1 & 0 \end{tabular}}\hspace{-4pt}\right].$

[1] Prove that the additive identity in a vector space is unique.

See class notes.

[1] Let $A$ be an $m\times n$ matrix. Prove that $N(A)\leq\Rss n .$

See class notes.

[2] For each free column, find a nonzero vector in the null space.

[2]()

$\twothreemat[1 0 2 0 1 -1 ],$ $\cvt[1 -2 1 ]$

$\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & 3 & 0 & 0 & -2\\ 0 & 0 & 1 & 0 & 4\\ 0 & 0 & 0 & 1 & 7 \end{tabular}\hspace{-4pt}\right],$ $\cvfive[-3 1 0 0 0 ],$ $\cvfive[2 0 -4 -7 1 ]$

[1] Give the span of the set $\left\{\cv[2 0 ], \cv[1 1 ]\right\}.$

$\Rss 2$

Note that $\cv[1 0 ]$ $=$ $\fric<1,2>\cv[2 0 ]$ and $\cv[0 1 ]$ $=$ $\cv[1 1 ] - \fric<1,2>\cv[2 0 ]$. So

$\Rss 2$ $=$ span$\left(\cv[1 0 ],\cv[0 1 ]\right)$ $\subseteq$ span$\left(\cv[2 0 ],\cv[1 1 ]\right)$ $\subseteq$ $\Rss 2$ which means that span$\left(\cv[2 0 ],\cv[1 1 ]\right)$ $=$ $\Rss 2 .$

[1] Define $f:M_{{}_{2\times2}}\to\Rss 4$ by $f\left(\twomat[\(a\) \(b\) \(c\) \(d\) ]\right)$ $=$ $\cvf[\(a\) \(b\) \(c\) \(d\) ]$. Prove that for all $A,B\in M_{{}_{2\times2}},$ $f(A + B)$ $=$ $f(A) + f(B).$

Suppose that $\twomat[\(a\) \(b\) \(c\) \(d\) ], \twomat[\(w\) \(x\) \(y\) \(z\) ]\in M_{{}_{2\times2}}$. Then

$f\left(\twomat[\(a\) \(b\) \(c\) \(d\) ] + \twomat[\(w\) \(x\) \(y\) \(z\) ]\right)$ $=$ $f\left(\twomat[\(a+w\) \(b+x\) \(c+y\) \(d+z\) ]\right)$ $=$ $\cvf[\(a+w\) \(b+x\) \(c+y\) \(d+z\) ]$ $=$ $\cvf[\(a\) \(b\) \(c\) \(d\) ] + \cvf[\(w\) \(x\) \(y\) \(z\) ]$ $=$ $f\left(\twomat[\(a\) \(b\) \(c\) \(d\) ]\right) + f\left(\twomat[\(w\) \(x\) \(y\) \(z\) ]\right).$

[2] For each of the following, determine whether or not the given set of vectors is a basis for $\Rss 3 .$

[2]()

$\left\{\cvt[1 0 1 ],\cvt[1 -2 1 ], \cvt[2 1 0 ]\right\}$

Yes.

$\threematb[1 1 2 0 -2 1 1 1 0 ]$ $\rref$ $\idmatthree$

The columns are linearly independent.

Three linearly independent vectors form a basis for $\Rss 3 .$

$\left\{\cvt[1 2 3 ],\cvt[-1 0 2 ], \cvt[3 2 -1 ]\right\}$

No.

Note that $\cvt[1 2 3 ] - 2 \cvt[-1 0 2 ]$ $=$ $\cvt[3 2 -1 ].$

Since the vectors are linearly dependent, they do not form a basis.

[2] Give the dimension of each of the following vector spaces.

[3]()

$P_{{}_{3}}$

$4$

$M_{{}_{3\times3}}$

$9$

$C([a,b])$

$\infty$

[2] Find an orthogonal set of vectors with the same span as the set given below.

$\left\{\vv_{{}_{1}} = \cvt[1 0 2 ],\vv_{{}_{2}} = \cvt[1 1 1 ]\right\}$

Set $\vw_{{}_{1}}$ $=$ $\vv_{{}_{1}}$ and $\vw_{{}_{2}}$ $=$ $\vv_{{}_{2}} - \fric<\vv_{{}_{2}}\cdot\vw_{{}_{1}},\Vert\vw_{{}_{1}}\Vert^{2}>\vw_{{}_{1}}$ $=$ $\cvt[1 1 1 ] - \fric<3,5>\cvt[1 0 2 ]$ $=$ $\cvt[\raisebox{2pt}{\tiny\(\fric<2,5>\)} 1 \raisebox{2pt}{\tiny\(\tm\fric<1,5>\)} ]$ $=$ $\fric<1,5>\cvt[2 5 -1 ].$

Then $\left\{\vw_{{}_{1}}, \vw_{{}_{2}}\right\}$ is an orthogonal set of vectors with the same span as the given set.

[6] Give the complete solution of each of the following.

[2]()

$x$	$+$	$y$	$+$	$z$	$=$	$2$
$2x$	$-$	$y$	$-$	$z$	$=$	$1$
$\tm x$	$+$	$y$	$+$	$2z$	$=$	$2$

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & 2\\ 2 & -1 & -1 & 1\\ -1 & 1 & 2 & 2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2 \end{tabular}\hspace{-4pt}\right]$

Complete Solution: $\cvt[1 -1 2 ]$

$x$	$+$	$y$	$+$	$z$	$=$	$2$
$2x$	$+$	$2y$	$+$	$z$	$=$	$5$
$4x$	$+$	$4y$	$+$	$3z$	$=$	$9$

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & 2\\ 2 & 2 & 1 & 5\\ 4 & 4 & 3 & 9 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 0 & 3\\ 0 & 0 & 1 & -1\\ 0 & 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

Pivot variables: $x,$ $z$

Free variable: $y$

Particular solution: $\cvt[3 0 -1 ]$

Special solution: $\cvt[-1 1 0 ]$

Complete solution: $\left\{ \cvt[3 0 1 ] + y\cvt[-1 1 0 ]:y\in\R\right\}$

$x$	$+$	$y$	$+$	$z$	$=$	$2$
$2x$	$+$	$2y$	$+$	$z$	$=$	$5$
$4x$	$+$	$4y$	$+$	$3z$	$=$	$8$

$\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 1 & 2\\ 2 & 2 & 1 & 5\\ 4 & 4 & 3 & 8 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr:r} \vspace{-12pt}\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

No solution.

[4] For each of the following, give the cofactor matrix and the adjoint.

[2]()

$\twomat[1 2 3 4 ]$

Cofactor matrix: $\twomat[4 -3 -2 1 ]$

Adjoint: $\twomattran[4 -3 -2 1 ]$

$\threematb[1 0 -2 2 -1 4 0 2 -3 ]$

Cofactor matrix: $\threematb[-5 6 2 -4 -3 -2 -2 6 -1 ]$

Adjoint: $\threematbtran[-5 6 2 -4 -3 -2 -2 6 -1 ]$

[4] Suppose that $A$ is an $m\times n$ matrix and define $T_{{}_{A}}:\Rss n \to \Rss m$ by $T_{{}_{A}}(\vx)$ $=$ $A\vx$.

Prove that $T_{{}_{A}}$ is a linear transformation.

See class notes.

Prove that $\ker T_{{}_{A}}$ $=$ $N(A).$

Note that for any $\vx\in\Rss n ,$ the following are equivalent.

$\vx\in\ker T_{{}_{A}}$

$T_{{}_{A}}(\vx)$ $=$ $\vzero$

$A\vx$ $=$ $\vzero$

$\vx\in\N(A)$

[2] Define $T: \Rss3 \to \Rss 3$ by $T\left(\cvt[\(x\) \(y\) \(z\) ]\right)$ $=$ $\cvt[\(x+y\) \(\tm z\) \(y\) ]$. Find a matrix $A$ such that $T(\vx)$ $=$ $A\vx$ for all $\vx\in\Rss 3 .$

Set $A$ $=$ $\threematb[1 1 0 0 0 -1 0 1 0 ]$. Then for any $\cvt[\(x\) \(y\) \(z\) ]\in\Rss 3 ,$ $\threematb[1 1 0 0 0 -1 0 1 0 ]\cvt[\(x\) \(y\) \(z\) ]$ $=$ $\cvt[\(x+y\) \(\tm z\) \(y\) ]$ $=$ $T\left(\cvt[\(x\) \(y\) \(z\) ]\right).$

Total Points:

(Exam 1)

Exam 1 Math 3720 Summer 2023 Name:

[70] Set $\vv$ $=$ $\rvt[1 \tm2 1 ],$ $\vw$ $=$ $\rvt[3 0 \tm1 ],$ and let $\theta$ be the angle between $\vv$ and $\vw$.

Compute each of the following.

[4]()

$\vv\cdot\vw$ $=$ $2$

$\Vert\vv\Vert$ $=$ $\sqrt{6}$

$\Vert\vw\Vert$ $=$ $\sqrt{10}$

[3]()

$\cos\theta$ $=$ $\fric<2,\sqrt{60}>$ $=$ $\fric<1,\sqrt{15}>$

$\comp(\vv,\vw)$ $=$ $\fric<2,\sqrt{6}>$

If possible, find $\alpha$ and $\beta$ such that $\alpha\vv + \beta\vw$ $=$ $\rvt[0 \tm4 3 ]$. If this is not possible, explain why it is not.

This is not possible. Suppose that $\alpha,\beta\in\R$ such that $\alpha\vv + \beta\vw$ $=$ $\rvt[0 \tm4 3 ]$. From the second components, we see that $\alpha$ $=$ $2$. Then from the third components, we see that $\beta$ $=$ $\tm 1$. However, $2\vv - \vw$ $=$ $\rvt[1 \tm4 3 ].$

[10] Suppose that $\vu,\vv,\vw,\vx\in\Rss n$. Prove that $(\vu + \vv)\cdot(\vw + \vx)$ $=$ $\vu\cdot\vw + \vu\cdot\vx + \vv\cdot\vw + \vv\cdot\vx.$

See class notes.

[10] Find a vector in $\Rss3$ that is orthogonal to $\rvt[2 1 -1 ].$

There are infinitely many solutions. One solution is $\rvt[-1 2 0 ].$

[10] Which of the following is always true for any $\vv,\vw\in\Rss n ?$

[3]()

$\Vert\vv + \vw\Vert$ $=$ $\Vert\vv\Vert + \Vert\vw\Vert$

$\Vert\vv + \vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\vw\Vert$

$\Vert\vv + \vw\Vert$ $=$ $\Vert\vv - \vw\Vert$

[2]()

$\Vert\vv - \vw\Vert$ $=$ $\Vert\vv\Vert - \Vert\vw\Vert$

$\Vert\vv - \vw\Vert$ $\leq$ $\Vert\vv\Vert - \Vert\vw\Vert$

By the Triangle Inequality, $\Vert\vv + \vw\Vert$ $\leq$ $\Vert\vv\Vert + \Vert\vw\Vert.$

[30] Set $A$ $=$ $\twothreemat[3 -1 4 0 1 8 ],$ $B$ $=$ $\twothreemat[1 2 -1 -4 1 -6 ],$ and $C$ $=$ $\threetwomat[1 0 0 1 -2 3 ]$. For each of the following, either perform the calculation or explain why it is not possible.

$2A + B$ $=$ $\twothreemat[6 -2 8 0 2 16 ] + \twothreemat[1 2 -1 -4 1 -6 ]$ $=$ $\twothreemat[7 0 7 -4 3 10 ]$

$AC$ $=$ $\twothreemat[3 -1 4 0 1 8 ] \threetwomat[1 0 0 1 -2 3 ]$ $=$ $\twomat[-5 11 -16 25 ]$

$CA$ $=$ $\threetwomat[1 0 0 1 -2 3 ] \twothreemat[3 -1 4 0 1 8 ]$ $=$ $\threematb[3 -1 4 0 1 8 -6 5 16]$

[20] Set $A$ $=$ $\threematb[0 3 1 2 1 6 2 -1 -1]$ and $B$ $=$ $\threematb[2 3 0 -2 4 3 0 0 1]$. Compute each of the following.

$\vr_{{}_{3}}(AB)$ $=$ $\vr_{{}_{3}}(A)B$ $=$ $\rvt[2 -1 -1 ]\threematb[2 3 0 -2 4 3 0 0 1]$ $=$ $\rvt[6 2 -4 ]$

$\vc_{{}_{2}}(AB)$ $=$ $A\vc_{{}_{2}}(B)$ $=$ $\threematb[0 3 1 2 1 6 2 -1 -1]\cvt[3 4 0 ]$ $=$ $\cvt[12 10 2 ]$

[10] Set $A$ $=$ $\threematb[0 1 0 1 0 0 0 0 1]$. Compute $A^{{}^{20}}.$

$A^{{}^{2}}$ $=$ $\threematb[0 1 0 1 0 0 0 0 1] \threematb[0 1 0 1 0 0 0 0 1]$ $=$ $\threematb[1 0 0 0 1 0 0 0 1]$ $=$ $I_{{}_{3}}$

$A^{{}^{20}}$ $=$ $\left(A^{{}^{2}}\right)^{{}^{10}}$ $=$ $\left(I_{{}_{3}}\right)^{{}^{10}}$ $=$ $I_{{}_{3}}$

[30] For each of the following, find the inverse or show that it does not exist.

[3]()

$\twomat[1 2 -1 -1 ]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 2 & 1 & 0\\ -1 & -1 & 0 & 1 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 2 & 1 & 0 \\ ... ...{-8pt} \begin{tabular}{l} \\ \tiny\(R_{{}_{1}} + R_{{}_{2}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rr:rr} \vspace{-12pt}\\ 1 & 0 & -1 & -2 \\... ...-8pt} \begin{tabular}{l} \tiny\(R_{{}_{1}} - 2R_{{}_{2}}\)\\ \\ \end{tabular}$

$\twomat[-1 -2 1 1 ]$

$\twomat[1 2 -1 -2 ]$

Does not exist.

$R_{{}_{2}}$ $=$ $\tm R_{{}_{1}}$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ \raisebox{2pt}{{\Tin... ... 1 & \raisebox{2pt}{{\Tiny\(\fric<1,2>\)}}\\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ \raisebox{2pt}{{... ...ox{2pt}{{\Tiny\(\fric<1,2>\)}} & 0 & 0 & 1\\ \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 2 & 3 & 2 & 0... ...\)\\ \tiny\(2R_{{}_{2}}\)\\ \tiny\(R_{{}_{3}} - R_{{}_{1}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 5 & 2 & -... ...iny\(R_{{}_{1}} - 2R_{{}_{2}}\)\\ \\ \tiny\(\tm R_{{}_{3}}\)\\ \end{tabular}$

$\left[\hspace{-4pt}\begin{tabular}{rrr:rrr} \vspace{-12pt}\\ 1 & 0 & 0 & -3 & ... ...}_{1}} - 5R_{{}_{3}}\)\\ \tiny\(R_{{}_{2}} + R_{{}_{3}}\)\\ \\ \end{tabular}$

$\threematb[-3 -4 5 1 2 -1 1 0 -1]$

[10] Prove that the product of an invertible matrix and a singular matrix is singular.

See class notes.

[10] Suppose that $A\vx$ $=$ $\vb$ where $\inv A;$ $=$ $\threematb[1 1 -1 0 1 1 1 1 1 ]$ and $\vb$ $=$ $\cvt[-2 1 3 ]$. Find $\vx.$

$A\vx$ $=$ $\vb$

$\inv A;A\vx$ $=$ $\inv A;\vb$

$\vx$ $=$ $\inv A;\vb$ $=$ $\threematb[1 1 -1 0 1 1 1 1 1 ]\cvt[-2 1 3 ]$ $=$ $\cvt[-4 4 2 ]$

[10] Give the transpose of the following matrix.

[2]()

$\twofourmatb[1 2 3 4 5 6 7 8]$

$\fourtwomatb[1 5 2 6 3 7 4 8]$

[10] Suppose that $A$ and $B$ are symmetric matrices and $AB$ $=$ $BA$. Prove that $AB$ is symmetric.

See class notes.

[20] Calculate the following determinants.

[2]()

$\twomat\vert 1 -1 3 2 \vert$ $=$ $5$

$\threematd\vert 1 0 -1 2 1 -8 0 1 0 \vert$ $=$ $\tm1\cdot\twomat\vert 1 -1 2 -8 \vert$ $=$ $6$

[10] Suppose that $A$ is an invertible matrix. Prove that $\vert A\vert\cdot\vert\inv A;\vert$ $=$ $1.$

Note that $\vert A\vert\cdot\vert\inv A;\vert$ $=$ $\vert A\inv A;\vert$ $=$ $\vert I\vert$ $=$ $1.$

(Exam 2)

Exam 2 Math 3720 Summer 2023 Name:

[10] Let $A$ be an $m\times n$ matrix. Prove that $N(A)\leq\Rss n .$

See class notes.

[10] Do the following vectors form a basis for $\Rss 2 ?$

$\cv[2 1 ],$ $\cv[1 1 ]$

Yes. Two linearly independent vectors span $\Rss 2 .$

[10] Do the following vectors form a basis for $\Rss3 ?$

$\cvt[1 1 0 ],$ $\cvt[-2 0 1 ],$ $\cvt[0 2 1 ]$

No. Note that $2\cvt[1 1 0 ] + \cvt[-2 0 1 ]$ $=$ $\cvt[0 2 1 ]$. Therefore the vectors are linarly dependent and do not form a basis.

[10] Note that $\left\{\vv_{{}_{1}} = \cvt[1 1 1 ], \vv_{{}_{2}} = \cvt[1 0 -1 ], \vv_{{}_{3}} = \cvt[1 -2 1 ]\right\}$ is an orthogonal set of vectors. Express $\vw$ $=$ $\cvt[2 1 6 ]$ as a linear combination of $\vv_{{}_{1}},$ $\vv_{{}_{2}},$ and $\vv_{{}_{3}}.$

[2]()

$\fric<\vv_{{}_{1}}\cdot\vw,\vv_{{}_{1}}\cdot\vv_{{}_{1}}>$ $=$ $3$

$\fric<\vv_{{}_{2}}\cdot\vw,\vv_{{}_{2}}\cdot\vv_{{}_{2}}>$ $=$ $\tm2$

$\fric<\vv_{{}_{2}}\cdot\vw,\vv_{{}_{2}}\cdot\vv_{{}_{2}}>$ $=$ $1$

$3\cvt[1 1 1 ] - 2 \cvt[1 0 -1 ] + \cvt[1 -2 1 ]$ $=$ $\cvt[2 1 6 ]$

[10] Give an orthogonal set of vectors with the same span as the vectors below. Note that two of the given vectors are orthogonal.

[2]()

$\vv_{{}_{1}}$ $=$ $\cvf[1 1 1 1 ],$ $\vv_{{}_{2}}$ $=$ $\cvf[1 -1 1 -1 ],$ $\vv_{{}_{3}}$ $=$ $\cvf[1 1 1 0 ]$

Set $\vw$ $=$ $\vv_{{}_{3}} - \left[\fricp<\vv_{{}_{1}}\cdot\vv_{{}_{3}},\vv_{{}_{1}}\cdot\vv_... ...vv_{{}_{2}}\cdot\vv_{{}_{3}},\vv_{{}_{2}}\cdot\vv_{{}_{2}}> \vv_{{}_{2}}\right]$

$=$ $\cvf[1 1 1 0 ] - \left(\fric<3,4>\cvf[1 1 1 1 ] - \fric<1,4>\cvf[1 -1 1 -1 ]\right)$

$=$ $\cvf[0 \raisebox{2pt}{\Tiny\(\fric<1,2>\)} 0 \raisebox{2pt}{\Tiny\(\tm\fric<1,2>\)} ]$

[10] Suppose that $A$ is an orthogonal matrix. Prove that $\inv A;$ $=$ $\tran A;$. See class notes.

[80] Give the four fundamental subspaces of each of the following.

[2]()

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrrr} \vspace{-12pt}\\ 1 & -3 & 0 & 0 & 2\\ 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 \end{tabular}\hspace{-4pt}\right]$

$R(A)$ $=$ span$\left\{\cvfive[1 -3 0 0 2 ],\cvfive[0 0 1 0 1 ],\cvfive[0 0 0 1 0 ]\right\}$

$N(A)$ $=$ span$\left\{\cvfive[3 1 0 0 0 ],\cvfive[-2 0 -1 0 1 ]\right\}$

$C(A)$ $=$ $\Rss3$

$N\left(\tran A;\right)$ $=$ $\left[C(A)\right]^{{}^{\perp}}$ $=$ $\left\{\cvt[0 0 0 ] \right\}$

$A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & -1\\ 0 & 1 & 0\\ 2 & 1 & -2 \end{tabular}\hspace{-4pt}\right]$

$\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 1 & -1\\ 0 & 1 & 0\\ 2 & 1 & -2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$R(A)$ $=$ span$\left\{\cvt[1 0 -1 ],\cvt[0 1 0 ]\right\}$

$N(A)$ $=$ span$\left\{\cvt[1 0 1 ]\right\}$

$r(A)$ $=$ $2$

$C(A)$ $=$ span$\left\{\cvt[1 0 2 ],\cvt[1 1 1 ]\right\}$

$\tran A;$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 1 & 1 & 1\\ -1 & 0 & -2 \end{tabular}\hspace{-4pt}\right]$ $\rref$ $\left[\hspace{-4pt}\begin{tabular}{rrr} \vspace{-12pt}\\ 1 & 0 & 2\\ 0 & 1 & -1\\ 0 & 0 & 0 \end{tabular}\hspace{-4pt}\right]$

$N\left(\tran A;\right)$ $=$ span$\left\{\cvt[-2 1 1 ]\right\}$

[20] Set $\vW$ $=$ span$\left(\cvf[1 1 1 1 ], \cvf[1 -1 1 -1 ]\right)$.

Find $\vW^{{}^{\perp}}.$

[2]()

Set $A$ $=$ $\twofourmatb[1 1 1 1 1 -1 1 -1].$

$\twofourmatb[1 1 1 1 1 -1 1 -1]$ $\rref$ $\twofourmatb[1 0 1 0 0 1 0 1]$

$\vW^{{}^{\perp}}$ $=$ $N(A)$ $=$ span$\left\{\cvf[-1 0 1 0 ],\cvf[0 -1 0 1 ]\right\}$

Find vectors $\vw$ and $\vw^{{}^{\perp}}$ such that $\vw\in\vW,$ $\vw^{{}^{\perp}}\in\vW^{{}^{\perp}},$ and $\vw + \vw^{{}^{\perp}}$ $=$ $\cvf[1 2 3 4 ].$

Set $\vw_{{}_{1}}$ $=$ $\cvf[1 1 1 1 ],$ $\vw_{{}_{2}}$ $=$ $\cvf[1 -1 1 -1 ],$ $\vw_{{}_{3}}$ $=$ $\cvf[-1 0 1 0 ],$ $\vw_{{}_{4}}$ $=$ $\cvf[0 -1 0 1 ],$ and $\vv$ $=$ $\cvf[1 2 3 4 ]$. Then

$\vw$ $=$ $\fricp<\vw_{{}_{1}}\cdot\vv,\vw_{{}_{1}}\cdot\vw_{{}_{1}}> \vw_{{}_{1}} + \fricp<\vw_{{}_{2}}\cdot\vv,\vw_{{}_{2}}\cdot\vw_{{}_{2}}> \vw_{{}_{2}}$ $=$ $\cvf[2 3 2 3 ]$ and $\vw^{{}^{\perp}}$ $=$ $\fricp<\vw_{{}_{3}}\cdot\vv,\vw_{{}_{3}}\cdot\vw_{{}_{3}}> \vw_{{}_{3}} + \fricp<\vw_{{}_{4}}\cdot\vv,\vw_{{}_{4}}\cdot\vw_{{}_{4}}> \vw_{{}_{4}}$ $=$ $\cvf[-1 -1 1 1 ].$

[15] Suppose that a $30 \times 20$ matrix has rank 12. What is the nullity?

Since rank$+$   nullity $=$ $20,$ nullity $=$ $8.$

(Final Exam)

Final Exam Math 3720 Summer 2023 Name:

Directions: Show all of your work and justify all of your answers.

Give the complete solution of each of the following.

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[15]

$x$	$+$	$y$	$=$	$\tm3$
$\tm2x$	$-$	$y$	$=$	$\tm 1$

[10]

$2x$	$-$	$4y$	$=$	$4$
$x$	$-$	$2y$	$=$	$2$

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[10]

$x$	$+$	$y$	$+$	$z$	$=$	$\tm 1$
$\tm x$	$+$	$2y$	$-$	$z$	$=$	$\tm5$
$3x$	$+$	$y$	$-$	$2z$	$=$	$1$

[10]

$x$	$+$	$2y$	$+$	$z$	$=$	$3$
$\tm x$	$+$	$4y$	$-$	$z$	$=$	$12$
$2x$	$+$	$10y$	$+$	$2z$	$=$	$21$

Suppose that $A$ $=$ $\left[\hspace{-4pt}\begin{tabular}{rrrr} \vspace{-12pt}\\ 1 & -2 & 0 & 1 \\ ... ... & \(\sqrt{2}\) & 1\\ 0 & 0 & 0 & \(\pi\) \end{tabular}\hspace{-4pt}\right],$ $\vx$ $=$ $\cvf[\(x_{{}_{1}}\) \(x_{{}_{2}}\) \(x_{{}_{3}}\) \(x_{{}_{4}}\) ],$ and $\vb$ $=$ $\cvf[\(\sqrt{3}\) 1 3 0 ].$

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[10] Calculate $\vert A\vert.$

[10] Calculate $\left\vert A_{{}_{c_{{}_{3}}}}(\vb)\right\vert.$

[10] If $A\vx$ $=$ $\vb,$ what is $x_{{}_{3}}?$

[10] Suppose that $\vV$ and $\vW$ are vector spaces and $T:\vV\to\vW$ is a linear transformation. Prove that $\ker T\leq \vV.$

[10] Suppose that Define $T:\Rss n \to \R$ by $T(\vx)$ $=$ $\Vert\vx\Vert$. Is $T$ a linear transformation?

[10] Suppose that $A$ is an invertible matrix with eigenvalue $\lambda$ and corresponding eigenvector $\vx$. Prove that $\fric<1,\lambda>$ is an eigenvalue of $\inv A;$ with corresponding eigenvector $\vx.$

For each of the following, find the eigenvalues, a basis for each eigenspace, the algebraic multiplicity of each eigenvalue, and the geometric multiplicity of each eigenvalue.

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[10] $\twomat[1 8 1 3 ]$

[10] $\idmatfour$