Quiz Chapter 10: Eigenvalues and Eigenvectors – Introduction to Matrix Algebra

1. The eigenvalues of $\begin{bmatrix} 5&6&17 \\ 0&-19&23 \\ 0&0&37 \\ \end{bmatrix}$ are

-19, \, 5, \, 37

19, \, -5, \, -37

2, \, -3, \, 7

3, \, -5, \, 37

2. If $\begin{bmatrix} -4.5\\-4\\1\\ \end{bmatrix}$ is an eigenvector of $\begin{bmatrix} 8&-4&2 \\ 4&0&2 \\ 0&-2&-4 \\ \end{bmatrix}$ , the eigenvalue corresponding to the eigenvector is

1

4

-4.5

6

3. The eigenvalues of the following matrix $\begin{bmatrix} 3&2&9 \\ 7&5&13 \\ 6&17&19 \\ \end{bmatrix}$ are given by solving the cubic equation

\lambda^{3} - 27\lambda^{2} + 167\lambda - 285

\lambda^{3} - 27\lambda^{2} - 122\lambda - 313

\lambda^{3} + 27\lambda^{2} + 167\lambda + 285

\lambda^{3} + 23.23\lambda^{2} - 158.3\lambda + 313

4. The eigenvalues of a $4 \times 4$ matrix $\left[ A \right]$ are given as $2, \, -3, \, 13,$ and $7$ . The |det $(A)$ | then is

546

19

25

cannot be determined

5. If one of the eigenvalues of $\left[ A \right]_{n \times n}$ is zero, it implies

The solution to

\left[ A \right] \left[ X \right] = \left[ C \right]

system of equations is unique

The determinant of

\left[ A \right]

is zero

The solution to

\left[ A \right] \left[ X \right] = \left[ 0 \right]

system of equations is trivial

The determinant of

\left[ A \right]

is non-zero

6. Given that matrix $\left[ A \right] = \begin{bmatrix} 8&-4&2 \\ 4&0&2 \\ 0&-2&-3 \\ \end{bmatrix}$ has an eigenvalue value of $4$ with the corresponding eigenvectors of $\left[ x \right] = \begin{bmatrix} -4.5 \\ -4 \\ 1 \\ \end{bmatrix}$ , then $\left[ A \right]^{5} \left[ X \right]$ is

\begin{bmatrix} -18 \\ -16 \\ 4 \\ \end{bmatrix}

\begin{bmatrix} -4.5 \\ -4 \\ 1 \\ \end{bmatrix}

\begin{bmatrix} -4608 \\ -4096 \\ 1024 \\ \end{bmatrix}

\begin{bmatrix} -0.004395 \\ -0.003906 \\ 0.0009766 \\ \end{bmatrix}