Question 1

Consider the difference equation $\mathbf { x } _ { \mathbf { k } + 1 } = \mathrm { Ax } _ { \mathbf { k } }$, where $\mathrm { A }$ has eigenvalues and corresponding eigenvectors $\mathbf { v } _ { 1 }$, $\mathbf { v } _ { 2 }$, and $\mathbf { v } _ { 3 }$ given below. Find the general solution of this difference equation if $x _ { 0 }$ is given as below.

-$\lambda _ { 1 } = 1.3 , \lambda _ { 2 } = 0.8 , \lambda _ { 3 } = 0.6 , \mathbf { v } _ { 1 } = \left[ \begin{array} { r } - 6 \ 6 \ 1 \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { l } 2 \ 1 \ 2 \end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } 1 \ 2 \ - 2 \end{array} \right]$, and $\mathbf { x } _ { 0 } = \left[ \begin{array} { r } - 33 \ 33 \ - 7 \end{array} \right]$
A)
$\mathrm{x}_{\mathrm{k}}=(1.3)^{\mathrm{k}_{\mathbf{v1}}}-3(0.8)^{\mathrm{k}_{\mathbf{}}} \mathbf{v}_{2}+3(0.6)^{\mathrm{k}_{\mathbf{v3}}}$
B)
$\mathrm{x}_{\mathrm{k}}=(1.3)^{\mathrm{k}_{\mathbf{v1}}}+(0.8)^{\mathrm{k}_{\mathbf{v2}}}+(0.6)^{\mathrm{k}_{\mathbf{v3}}}$

C) $\mathbf { x } _ { \mathrm { k } } = 5 ( 1.3 ) ^ { \mathrm { k } _ { \mathbf { v1 } } } - 3 ( 0.8 ) ^ { \mathrm { k } _ { \mathbf {v2} } } + ( 0.6 ) ^ { \mathrm { k } _ { \mathbf { v3 } } }$

D) $\mathbf { x } _ { \mathrm { k } } = 5 ( 1.3 ) ^ { \mathrm { k } _ { \mathbf { v1 } } } - 3 ( 0.8 ) ^ { \mathrm { k } _ { \mathbf { v2 } } } + 3 ( 0.6 ) ^ { \mathrm { k } _ { \mathbf { v3 } } }$

Accepted Answer

C

Question 2

Solve the initial value problem.
-$$A = \left[ \begin{array} { r r } 
16 & - 6 \
63 & - 23
\end{array} \right]$$
A)
$- 2 , \left[ \begin{array} { l } 1 \ 3 \end{array} \right]$

B)
$- 5 , \left[ \begin{array} { l } - 2 \ - 7 \end{array} \right]$

C)
$- 2 , \left[ \begin{array} { l } - 2 \ - 7 \end{array} \right]$

D)

$- 5 , \left[ \begin{array} { l } 1 \ 3 \end{array} \right]$

Accepted Answer

A

Question 3

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue.
-$A = \left[ \begin{array} { r r r } - 4 & 0 & 0 \ - 10 & 6 & 0 \ - 30 & 16 & - 2 \end{array} \right] , \lambda = - 4$
A)
$\left\{ \left[ \begin{array} { r } 1 \ - 1 \ - 7 \end{array} \right] \right\}$

B)
$\left\{ \left[ \begin{array} { l } 1 \ 1 \ 0 \end{array} \right] , \left[ \begin{array} { r } 1 \ 0 \ - 7 \end{array} \right] \right\}$

C)
$\left\{ \left[ \begin{array} { l } 1 \ 1 \ 7 \end{array} \right] \right\}$

D)

$\left\{ \left[ \begin{array} { l } 1 \ 1 \ 0 \end{array} \right] , \left[ \begin{array} { l } 1 \ 0 \ 7 \end{array} \right] \right\}$

Accepted Answer

C

Question 4

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities.
-$\lambda ^ { 5 } - 24 \lambda ^ { 4 } + 189 \lambda ^ { 3 } - 486 \lambda ^ { 2 }$
A) 0 (multiplicity 2), 9 (multiplicity 2), 6 (multiplicity 1 )
B) 0 (multiplicity 2), $- 9$ (multiplicity 2), -6 (multiplicity 1)
C) 0 (multiplicity 2), $- 9$ (multiplicity 2), 6 (multiplicity 1 )
D) 0 (multiplicity 1), 9 (multiplicity $3 ), 6$ (multiplicity 1 )

Accepted Answer

The characteristic polynomial is missing its constant term, which implies it is divisible by $\lambda^2$, indicating $\lambda = 0$ has multiplicity 2. Factoring the polynomial as $\lambda^2(\lambda^3 - 24\lambda^2 + 189\lambda - 486)$ and solving for the roots of the cubic part will give the other eigenvalues. The correct factorization of the cubic part gives $\lambda = 9$ (multiplicity 2) and $\lambda = 6$ (multiplicity 1), making option A correct.

Question 5

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that 
 $$A = P D P^{ - 1}$$
-$$A = \left[ \begin{array} { l l l } 
2 & 0 & 0 \
1 & 2 & 0 \
0 & 0 & 2
\end{array} \right]$$
A) $P = \left[ \begin{array} { l l l } 1 & 0 & 0 \ 2 & 2 & 0 \ 0 & 1 & 1 \end{array} \right] , D = \left[ \begin{array} { l l l } 2 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{array} \right]$
B)
$$P = \left[ \begin{array} { r r r } 
1 & 0 & - 1 \
2 & 2 & 0 \
1 & 1 & 1
\end{array} \right] , D = \left[ \begin{array} { l l l } 
2 & 0 & 1 \
1 & 2 & 1 \
0 & 0 & 2
\end{array} \right]$$
C) Not diagonalizable
D)
$$\mathrm { P } = \left[ \begin{array} { r r r } 
1 & 2 & 1 \
0 & 2 & 1 \
- 1 & 0 & 1
\end{array} \right] , \mathrm { D } = \left[ \begin{array} { l l l } 
2 & 0 & 0 \
0 & 2 & 0 \
0 & 0 & 2
\end{array} \right]$$

Accepted Answer

The answer of Diagonalize the matrix A, if possible. That...

Question 6

Find the eigenvalues of the given matrix.
-$$\left[ \begin{array} { r r } 
56 & 10 \
- 275 & - 49
\end{array} \right]$$
A) 1,6
B) $- 1 , - 6$
C) -1
D) 6 )6

Accepted Answer

The characteristic equation of the matrix is:$$\det(A - \lambda I) = 0$$$$(56 - \lambda)(-49 - \lambda) - (10)(-275) = 0$$$$\lambda^2 - 5\lambda - 1260 = 0$$$$(\lambda - 6)(\lambda + 210) = 0$$Therefore, the eigenvalues are $\lambda = 6$ and $\lambda = -210$.

Question 7

Find the characteristic equation of the given matrix.
-$$A = \left[ \begin{array} { r r r r } 
9 & 6 & 4 & 3 \
0 & 7 & - 6 & 8 \
0 & 0 & 5 & 9 \
0 & 0 & 0 & 9
\end{array} \right]$$
A) $( 9 - \lambda ) ^ { 2 } ( 7 - \lambda ) ( 5 - \lambda ) = 0$
B) $( 9 - \lambda ) ( 6 - \lambda ) ( 4 - \lambda ) ( 3 - \lambda ) = 0$
C) $( 9 - \lambda ) ( 7 - \lambda ) ( 5 - \lambda ) = 0$
D) $( 9 - \lambda ) ( 9 - \lambda ) ( 8 - \lambda ) ( 3 - \lambda ) = 0$

Accepted Answer

The characteristic equation of a matrix is found by taking the determinant of $A - \lambda I$, where $I$ is the identity matrix and $\lambda$ is a scalar. For a triangular matrix (including diagonal matrices), the determinant is the product of the diagonal elements minus $\lambda$. Thus, the characteristic equation of the given matrix is $(9-\lambda)^2(7-\lambda)(5-\lambda)=0$, as the diagonal elements are 9, 7, 5, and 9.

Question 8

$$\text { Find the matrix of the linear transformation } \mathrm { T } : \mathrm { V } \rightarrow \mathrm { W } \text { relative to } \mathrm { B } \text { and } \mathrm { C } \text {. }$$
-Suppose $B = \left\{ b _ { 1 } , b _ { 2 } \right\}$ is a basis for $V$ and $C = \left\{ c _ { 1 } , c _ { 2 } , c _ { 3 } \right\}$ is a basis for $W$. Let $T$ be defined by
$$\begin{array} { l } 
\mathrm { T } \left( \mathbf { b } _ { 1 } \right) = 5 c _ { 1 } + 6 c _ { 2 } - 5 c _ { 3 } \
T \left( b _ { 2 } \right) = 5 c _ { 1 } + 12 c _ { 2 } + 7 c _ { 3 }
\end{array}$$
A)
$\left[\begin{array}{rr}
5 & 5 \
6 & 12 \
-5 & 7
\end{array}\right]$

B)
$$\left[ \begin{array} { r r } 
5 & 5 \
6 & 12 \
- 5 & 7
\end{array} \right]$$

C)
$\left[\begin{array}{rr}
5 & 0 \
6 & 6 \
-5 & 12
\end{array}\right]$

D)
$\left[\begin{array}{rrr}
5 & 6 & -5 \
5 & 12 & 7
\end{array}\right]$

Accepted Answer

The matrix of the linear transformation $T$ relative to $B$ and $C$ is$$\left[\begin{array}{rr}5 & 5 \6 & 12 \-5 & 7\end{array}\right]$$because$$\begin{aligned}\mathrm { T } \left( \mathbf { b } _ { 1 } \right) &=5 c _ { 1 } + 6 c _ { 2 } - 5 c _ { 3 } \&=5\left[ \begin{array} { l } 1 \0 \0\end{array} \right] + 6\left[ \begin{array} { l } 0 \1 \0\end{array} \right] - 5\left[ \begin{array} { l } 0 \0 \1\end{array} \right] \&=\left[ \begin{array} { r } 5 \6 \-5\end{array} \right] \\mathrm { T } \left( \mathbf { b } _ { 2 } \right) &=5 c _ { 1 } + 12 c _ { 2 } + 7 c _ { 3 } \&=5\left[ \begin{array} { l } 1 \0 \0\end{array} \right] + 12\left[ \begin{array} { l } 0 \1 \0\end{array} \right] + 7\left[ \begin{array} { l } 0 \0 \1\end{array} \right] \&=\left[ \begin{array} { r } 5 \12 \7\end{array} \right]\end{aligned}$$

Question 9

Find the characteristic equation of the given matrix.
-$$A = \left[ \begin{array} { r r r r } 
8 & - 1 & 5 & 2 \
0 & 3 & 1 & - 1 \
0 & 0 & - 6 & - 6 \
0 & 0 & 0 & - 5
\end{array} \right]$$
A) $( 8 - \lambda ) ( 3 - \lambda ) ( - 6 - \lambda ) ( - 5 - \lambda ) = 0$
B) $( 8 - \lambda ) ( - 1 - \lambda ) ( 5 - \lambda ) ( 2 - \lambda ) = 0$
C) $( - 5 - \lambda ) ( - 6 - \lambda ) ( - 1 - \lambda ) ( 2 - \lambda ) = 0$
D) $( 2 - \lambda ) ( - 1 - \lambda ) ( - 6 - \lambda ) ( - 5 - \lambda ) = 0$

Accepted Answer

The characteristic equation of a matrix $A$ is found by calculating the determinant of $A - \lambda I$, where $I$ is the identity matrix and $\lambda$ are the eigenvalues. For a triangular matrix (including diagonal matrices), the eigenvalues are the entries on the main diagonal. Thus, the characteristic equation is $(8 - \lambda)(3 - \lambda)(-6 - \lambda)(-5 - \lambda) = 0$.

Question 10

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities.
-$$\text { Find a formula for } A ^ { k } \text {, given that } A = P D P - 1 \text {, where } P \text { and } D \text { are given below. }$$

Accepted Answer

The answer of The characteristic polynomial of a 5 ×...

Question 11

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue.
-$$A = \left[ \begin{array} { r r r } 
1 & 6 & 6 \
6 & 1 & - 6 \
- 6 & 6 & 13
\end{array} \right] , \lambda = 7$$

A)$ \left\{\left[\begin{array}{l}1 \ 0 \ 1\end{array}\right],\left[\begin{array}{r}0 \ 1 \ -1\end{array}\right]\right\} $

B)$ \left\{\left[\begin{array}{r}1 \ 0 \ -1\end{array}\right],\left[\begin{array}{l}0 \ 1 \ 1\end{array}\right]\right\} $

C)$ \left\{\left[\begin{array}{r}0 \ 1 \ -1\end{array}\right]\right\} $
 
D)$ \left\{\left[\begin{array}{r}1 \ 0 \ -1\end{array}\right]\right\} $

Accepted Answer

To find a basis for the eigenspace corresponding to $\lambda = 7$, we solve $(A - \lambda I)x = 0$. Subtracting $7$ from the diagonal entries of $A$, we get:$$\left[ \begin{array}{ccc}-6 & 6 & 6 \6 & -6 & -6 \-6 & 6 & 6\end{array} \right]$$Solving this system leads to the eigenvectors $\left[\begin{array}{r}1 \ 0 \ 1\end{array}\right]$ and $\left[\begin{array}{r}0 \ 1 \ -1\end{array}\right]$, which form a basis for the eigenspace corresponding to $\lambda = 7$.

Question 12

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities.
-$$\lambda ^ { 5 } + 17 \lambda ^ { 4 } + 72 \lambda ^ { 3 }$$
A) 0 (multiplicity 1), 8 (multiplicity 1), 9 (multiplicity 1)
B) 0 (multiplicity $1 ), - 9$ (multiplicity $1 ), - 8$ (multiplicity 1 )
C) 0 (multiplicity 3), 8 (multiplicity 1), 9 (multiplicity 1)
D) 0 (multiplicity 3), $- 9$ (multiplicity $1 ), - 8$ (multiplicity 1 )

Accepted Answer

The characteristic polynomial can be factored as $\lambda^3(\lambda + 9)(\lambda + 8)$, indicating eigenvalues 0 (multiplicity 3), -9 (multiplicity 1), and -8 (multiplicity 1).

Question 13

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that 
 $$A = P D P^{ - 1}$$
-$A=\left[\begin{array}{rrrr}
9 & 0 & 0 & 0 \
0 & 9 & 0 & 0 \
-16 & 4 & 1 & 16 \
0 & 0 & 0 & 9
\end{array}\right]$

A)$P=\left[\begin{array}{rrrr}
2 & 0 & -2 & 1 \
0 & 2 & 1 & 0 \
1 & 0 & 0 & 1 \
0 & 0 & 1 & 0
\end{array}\right], D=\left[\begin{array}{llll}
9 & 0 & 0 & 0 \
0 & 9 & 0 & 0 \
0 & 0 & 9 & 0 \
0 & 0 & 0 & 1
\end{array}\right]$

B) Not diagonalizable

C)$P=\left[\begin{array}{rrrr}
4 & -2 & 1 & 0 \
8 & -2 & 0 & 0 \
1 & 0 & 1 & 1 \
0 & 1 & 1 & 0
\end{array}\right], D=\left[\begin{array}{llll}
9 & 0 & 0 & 0 \
0 & 9 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
\end{array}\right]$

D)$P=\left[\begin{array}{rrrr}
2 & 0 & 1 & 0 \
0 & 2 & 0 & 0 \
-2 & 1 & 0 & 1 \
1 & 0 & 1 & 0
\end{array}\right], D=\left[\begin{array}{llll}
9 & 0 & 0 & 0 \
0 & 9 & 0 & 0 \
0 & 0 & 9 & 0 \
0 & 0 & 0 & 1
\end{array}\right]$

Accepted Answer

The answer of Diagonalize the matrix A, if possible. That...

Question 14

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that 
 $$A = P D P^{ - 1}$$
-$$A = \left[ \begin{array} { r r r } 
- 11 & 3 & - 9 \
0 & - 5 & 0 \
6 & - 3 & 4
\end{array} \right]$$

A) $P = \left[ \begin{array} { r r r } 1 & 5 & - 1 \ 5 & 3 & 0 \ 1 & 3 & 1 \end{array} \right] , D = \left[ \begin{array} { r r r } - 5 & 1 & 0 \ 0 & - 5 & 0 \ 0 & 0 & - 2 \end{array} \right]$
B) 
$$P = \left[ \begin{array} { r r r } 
1 & 0 & - 1 \
5 & 3 & 0 \
1 & 1 & 1
\end{array} \right] , D = \left[ \begin{array} { r r r } 
- 5 & 0 & 0 \
0 & - 5 & 0 \
0 & 0 & - 2
\end{array} \right]$$
C)
$$\mathrm { P } = \left[ \begin{array} { r r r } 
1 & 0 & - 1 \
0 & 3 & 0 \
1 & 1 & 1
\end{array} \right] , \mathrm { D } = \left[ \begin{array} { r r r } 
- 5 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & - 2
\end{array} \right]$$
D)
$$\mathrm { P } = \left[ \begin{array} { r r r } 
1 & 0 & - 1 \
5 & 3 & 0 \
1 & 1 & 1
\end{array} \right] , \mathrm { D } = \left[ \begin{array} { r r r } 
- 5 & 0 & - 2 \
0 & - 5 & 0 \
0 & - 5 & - 2
\end{array} \right]$$

Accepted Answer

To diagonalize a matrix, we find its eigenvalues and corresponding eigenvectors. The given matrix A has eigenvalues -5, -5, and -2. The corresponding eigenvectors form the columns of P, and the eigenvalues are placed on the diagonal of D. Choice B correctly represents the diagonalization of A with the correct P and D matrices, where P contains the eigenvectors and D contains the eigenvalues on its diagonal.

Question 15

Use the inverse power method to determine the smallest eigenvalue of the matrix A.
-Assume that the eigenvalues are roughly 1.3, 2.1, and 15. $$A = \left[ \begin{array} { r r r } 
1 & 0.5 & 0 \
0 & 1.5 & 0 \
- 1 & 1 & 2
\end{array} \right]$$
A) $1.5$
B) 2
C) $1.3$
D) 5

Accepted Answer

The answer of Use the inverse power method to determine...

Question 16

Find the eigenvalues of the given matrix.
-$$\left[ \begin{array} { r r } 
- 14 & - 6 \
36 & 16
\end{array} \right]$$
A) $- 2,4$
B) $- 2 , - 4$
C) $- 4$
D) $- 2$

Accepted Answer

To find the eigenvalues of the matrix, we solve the characteristic equation $\det(A - \lambda I) = 0$, where $A$ is the given matrix and $I$ is the identity matrix. For the given matrix $\left[ \begin{array} { r r } - 14 & - 6 \ 36 & 16 \end{array} \right]$, the characteristic equation becomes $\det\left( \begin{array} { c c } -14 - \lambda & -6 \ 36 & 16 - \lambda \end{array} \right) = 0$. This simplifies to $(-\lambda - 14)(16 - \lambda) - (-6)(36) = \lambda^2 - 2\lambda - 40 = 0$. Solving this quadratic equation gives $\lambda = -2, 4$.

Question 17

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue.
-$$A = \left[ \begin{array} { l l } 
- 13 & 2 \
- 40 & 5
\end{array} \right] , \lambda = - 5$$
A)
$\left[ \begin{array} { l } - 4 \ 1 \end{array} \right]$

B)

$\left[ \begin{array} { l } 1 \ 4 \end{array} \right]$

C)
$\left[ \begin{array} { r } 1 \ - 4 \end{array} \right]$

D)

$\left[ \begin{array} { l } 4 \ 1 \end{array} \right]$

Accepted Answer

The answer of For the given matrix and eigenvalue, find...

Question 18

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue.
-$A = \left[ \begin{array} { r r } - 18 & - 5 \ 60 & 17 \end{array} \right] , \lambda = 2$
A)
$\left[ \begin{array} { r } 1 \ 17 \end{array} \right]$

B)
$\left[ \begin{array} { r } - 4 \ 1 \end{array} \right]$

C)
$\left[ \begin{array} { r } 1 \ - 4 \end{array} \right]$

D)
$\left[ \begin{array} { l } 1 \ 0 \end{array} \right]$

Accepted Answer

To find an eigenvector for $\lambda = 2$, solve $(A - 2I)\mathbf{v} = 0$. The matrix becomes $\left[ \begin{array} { r r } -20 & -5 \ 60 & 15 \end{array} \right]$. Solving gives the eigenvector $\left[ \begin{array} { r } 1 \ -4 \end{array} \right]$.

Question 19

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that 
 $$A = P D P^{ - 1}$$
-$$A = \left[ \begin{array} { r r r r } 
6 & 0 & 0 & 0 \
0 & 6 & 0 & 0 \
1 & - 4 & - 6 & 0 \
- 1 & 2 & 0 & - 6
\end{array} \right]$$

A) $P = \left[ \begin{array} { r r r r } - 12 & - 24 & 0 & 0 \ 6 & 6 & 0 & 0 \ 1 & 0 & 1 & 0 \ 0 & 1 & 0 & 1 \end{array} \right] , D = \left[ \begin{array} { r r r r } - 6 & 0 & 0 & 0 \ 0 & - 6 & 0 & 0 \ 0 & 0 & 6 & 0 \ 0 & 0 & 0 & 6 \end{array} \right]$

B) $P = \left[ \begin{array} { r r r r } - 12 & 6 & 1 & 0 \ - 24 & 6 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{array} \right] , D = \left[ \begin{array} { r r r r } - 6 & 0 & 0 & 0 \ 0 & - 6 & 0 & 0 \ 0 & 0 & 6 & 0 \ 0 & 0 & 0 & 6 \end{array} \right]$

C) Not diagonalizable

D)
$$P = \left[ \begin{array} { r r r r } 
- 12 & - 24 & 0 & 0 \
- 6 & - 6 & 0 & 0 \
1 & 0 & 1 & 0 \
0 & 1 & 0 & 1
\end{array} \right] , D = \left[ \begin{array} { r r r r } 
6 & 0 & 0 & 0 \
0 & 6 & 0 & 0 \
0 & 0 & - 6 & 0 \
0 & 0 & 0 & - 6
\end{array} \right]$$

Accepted Answer

The answer of Diagonalize the matrix A, if possible. That...

Question 20

$$\text { Find the matrix of the linear transformation } \mathrm { T } : \mathrm { V } \rightarrow \mathrm { W } \text { relative to } \mathrm { B } \text { and } \mathrm { C } \text {. }$$
-Suppose $B = \left\{ b _ { 1 } , b _ { 2 } , b _ { 3 } \right\}$ is a basis for $V$ and $C = \left\{ c _ { 1 } , c _ { 2 } \right\}$ is a basis for $W$. Let $T$ be defined by
$$\begin{array} { l } 
\mathrm { T } \left( \mathbf { b } _ { 1 } \right) = 3 \mathbf { c } _ { 1 } + \mathrm { c } _ { 2 } \
\mathrm {~T} \left( \mathbf { b } _ { 2 } \right) = 8 \mathbf { c } _ { 1 } - 8 \mathbf { c } _ { 2 } \
\mathrm {~T} \left( \mathbf { b } _ { 3 } \right) = 3 \mathbf { c } _ { 1 } + 2 \mathbf { c } _ { 2 }
\end{array}$$
A)
$\left[ \begin{array} { r r r } 4 & 0 & 5 \ 1 & - 8 & 2 \end{array} \right]$

B)
$\left[ \begin{array} { r r r } 3 & 8 & 3 \ 1 & - 8 & 2 \end{array} \right]$

C)
$\left[ \begin{array} { r r } 3 & 1 \ 8 & - 8 \ 3 & 2 \end{array} \right]$

D)

$\left[ \begin{array} { l l } 3 & 4 \ 8 & 0 \ 3 & 5 \end{array} \right]$

Accepted Answer

The matrix of a linear transformation relative to bases $B$ and $C$ is constructed by expressing the images of the basis vectors of $V$ (in this case, $b_1, b_2, b_3$) as linear combinations of the basis vectors of $W$ (in this case, $c_1, c_2$). The coefficients of these linear combinations form the columns of the matrix. Given $T(b_1) = 3c_1 + c_2$, $T(b_2) = 8c_1 - 8c_2$, and $T(b_3) = 3c_1 + 2c_2$, the matrix is $\left[ \begin{array} { r r r } 3 & 8 & 3 \ 1 & - 8 & 2 \end{array} \right]$.

Quiz 5: Eigenvalues and Eigenvectors

Solve the initial value problem. - $A = \left[ \begin{array} { r r } 16 & - 6 \\63 & - 23\end{array} \right]$

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue. - $A = \left[ \begin{array} { r r r } - 4 & 0 & 0 \\ - 10 & 6 & 0 \\ - 30 & 16 & - 2 \end{array} \right] , \lambda = - 4$

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - $\lambda ^ { 5 } - 24 \lambda ^ { 4 } + 189 \lambda ^ { 3 } - 486 \lambda ^ { 2 }$

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that $A = P D P^{ - 1}$ - $A = \left[ \begin{array} { l l l } 2 & 0 & 0 \\1 & 2 & 0 \\0 & 0 & 2\end{array} \right]$

Find the eigenvalues of the given matrix. - $\left[ \begin{array} { r r } 56 & 10 \\- 275 & - 49\end{array} \right]$

Find the characteristic equation of the given matrix. - $A = \left[ \begin{array} { r r r r } 9 & 6 & 4 & 3 \\0 & 7 & - 6 & 8 \\0 & 0 & 5 & 9 \\0 & 0 & 0 & 9\end{array} \right]$

Find the characteristic equation of the given matrix. - $A = \left[ \begin{array} { r r r r } 8 & - 1 & 5 & 2 \\0 & 3 & 1 & - 1 \\0 & 0 & - 6 & - 6 \\0 & 0 & 0 & - 5\end{array} \right]$

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - $\text { Find a formula for } A ^ { k } \text {, given that } A = P D P - 1 \text {, where } P \text { and } D \text { are given below. }$

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue. - $A = \left[ \begin{array} { r r r } 1 & 6 & 6 \\6 & 1 & - 6 \\- 6 & 6 & 13\end{array} \right] , \lambda = 7$

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - $\lambda ^ { 5 } + 17 \lambda ^ { 4 } + 72 \lambda ^ { 3 }$

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that $A = P D P^{ - 1}$ - $A=\left[\begin{array}{rrrr}9 & 0 & 0 & 0 \\0 & 9 & 0 & 0 \\-16 & 4 & 1 & 16 \\0 & 0 & 0 & 9\end{array}\right]$

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that $A = P D P^{ - 1}$ - $A = \left[ \begin{array} { r r r } - 11 & 3 & - 9 \\0 & - 5 & 0 \\6 & - 3 & 4\end{array} \right]$

Use the inverse power method to determine the smallest eigenvalue of the matrix A. -Assume that the eigenvalues are roughly 1.3, 2.1, and 15. $A = \left[ \begin{array} { r r r } 1 & 0.5 & 0 \\0 & 1.5 & 0 \\- 1 & 1 & 2\end{array} \right]$

Find the eigenvalues of the given matrix. - $\left[ \begin{array} { r r } - 14 & - 6 \\36 & 16\end{array} \right]$

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - $A = \left[ \begin{array} { l l } - 13 & 2 \\- 40 & 5\end{array} \right] , \lambda = - 5$

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - $A = \left[ \begin{array} { r r } - 18 & - 5 \\ 60 & 17 \end{array} \right] , \lambda = 2$

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that $A = P D P^{ - 1}$ - $A = \left[ \begin{array} { r r r r } 6 & 0 & 0 & 0 \\0 & 6 & 0 & 0 \\1 & - 4 & - 6 & 0 \\- 1 & 2 & 0 & - 6\end{array} \right]$

Quiz 5: Eigenvalues and Eigenvectors

Solve the initial value problem. - A=[16−663−23]A = \left[ \begin{array} { r r } 16 & - 6 \\63 & - 23\end{array} \right]A=[1663​−6−23​]

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - λ5−24λ4+189λ3−486λ2\lambda ^ { 5 } - 24 \lambda ^ { 4 } + 189 \lambda ^ { 3 } - 486 \lambda ^ { 2 }λ5−24λ4+189λ3−486λ2

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that A=PDP−1A = P D P^{ - 1}A=PDP−1 - A=[200120002]A = \left[ \begin{array} { l l l } 2 & 0 & 0 \\1 & 2 & 0 \\0 & 0 & 2\end{array} \right]A=​210​020​002​​

Find the eigenvalues of the given matrix. - [5610−275−49]\left[ \begin{array} { r r } 56 & 10 \\- 275 & - 49\end{array} \right][56−275​10−49​]

Find the characteristic equation of the given matrix. - A=[964307−6800590009]A = \left[ \begin{array} { r r r r } 9 & 6 & 4 & 3 \\0 & 7 & - 6 & 8 \\0 & 0 & 5 & 9 \\0 & 0 & 0 & 9\end{array} \right]A=​9000​6700​4−650​3899​​

Find the characteristic equation of the given matrix. - A=[8−152031−100−6−6000−5]A = \left[ \begin{array} { r r r r } 8 & - 1 & 5 & 2 \\0 & 3 & 1 & - 1 \\0 & 0 & - 6 & - 6 \\0 & 0 & 0 & - 5\end{array} \right]A=​8000​−1300​51−60​2−1−6−5​​

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue. - A=[16661−6−6613],λ=7A = \left[ \begin{array} { r r r } 1 & 6 & 6 \\6 & 1 & - 6 \\- 6 & 6 & 13\end{array} \right] , \lambda = 7A=​16−6​616​6−613​​,λ=7

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - λ5+17λ4+72λ3\lambda ^ { 5 } + 17 \lambda ^ { 4 } + 72 \lambda ^ { 3 }λ5+17λ4+72λ3

Use the inverse power method to determine the smallest eigenvalue of the matrix A. -Assume that the eigenvalues are roughly 1.3, 2.1, and 15. A=[10.5001.50−112]A = \left[ \begin{array} { r r r } 1 & 0.5 & 0 \\0 & 1.5 & 0 \\- 1 & 1 & 2\end{array} \right]A=​10−1​0.51.51​002​​

Find the eigenvalues of the given matrix. - [−14−63616]\left[ \begin{array} { r r } - 14 & - 6 \\36 & 16\end{array} \right][−1436​−616​]

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - A=[−132−405],λ=−5A = \left[ \begin{array} { l l } - 13 & 2 \\- 40 & 5\end{array} \right] , \lambda = - 5A=[−13−40​25​],λ=−5

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - A=[−18−56017],λ=2A = \left[ \begin{array} { r r } - 18 & - 5 \\ 60 & 17 \end{array} \right] , \lambda = 2A=[−1860​−517​],λ=2

Solve the initial value problem. - $A = \left[ \begin{array} { r r } 16 & - 6 \\63 & - 23\end{array} \right]$

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - $\lambda ^ { 5 } - 24 \lambda ^ { 4 } + 189 \lambda ^ { 3 } - 486 \lambda ^ { 2 }$

Diagonalize the matrix A, if possible. That is, find an invertible matrix P and a diagonal matrix D such that $A = P D P^{ - 1}$ - $A = \left[ \begin{array} { l l l } 2 & 0 & 0 \\1 & 2 & 0 \\0 & 0 & 2\end{array} \right]$

Find the eigenvalues of the given matrix. - $\left[ \begin{array} { r r } 56 & 10 \\- 275 & - 49\end{array} \right]$

Find the characteristic equation of the given matrix. - $A = \left[ \begin{array} { r r r r } 9 & 6 & 4 & 3 \\0 & 7 & - 6 & 8 \\0 & 0 & 5 & 9 \\0 & 0 & 0 & 9\end{array} \right]$

Find the characteristic equation of the given matrix. - $A = \left[ \begin{array} { r r r r } 8 & - 1 & 5 & 2 \\0 & 3 & 1 & - 1 \\0 & 0 & - 6 & - 6 \\0 & 0 & 0 & - 5\end{array} \right]$

For the given matrix A, find a basis for the corresponding eigenspace for the given eigenvalue. - $A = \left[ \begin{array} { r r r } 1 & 6 & 6 \\6 & 1 & - 6 \\- 6 & 6 & 13\end{array} \right] , \lambda = 7$

The characteristic polynomial of a 5 × 5 matrix is given below. Find the eigenvalues and their multiplicities. - $\lambda ^ { 5 } + 17 \lambda ^ { 4 } + 72 \lambda ^ { 3 }$

Use the inverse power method to determine the smallest eigenvalue of the matrix A. -Assume that the eigenvalues are roughly 1.3, 2.1, and 15. $A = \left[ \begin{array} { r r r } 1 & 0.5 & 0 \\0 & 1.5 & 0 \\- 1 & 1 & 2\end{array} \right]$

Find the eigenvalues of the given matrix. - $\left[ \begin{array} { r r } - 14 & - 6 \\36 & 16\end{array} \right]$

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - $A = \left[ \begin{array} { l l } - 13 & 2 \\- 40 & 5\end{array} \right] , \lambda = - 5$

For the given matrix and eigenvalue, find an eigenvector corresponding to the eigenvalue. - $A = \left[ \begin{array} { r r } - 18 & - 5 \\ 60 & 17 \end{array} \right] , \lambda = 2$