Question 53

Let $M = \begin{pmatrix} -9 & -6 & -2 & -4 \\ -8 & -6 & -3 & -1 \\ 20 & 15 & 8 & 5 \\ 32 & 21 & 7 & 12 \end{pmatrix}$. We test each candidate vector:

1. For $v_A = (-1, 1, 0, 1)^T$:

   $$M v_A = \begin{pmatrix} 9 - 6 - 0 - 4 \\ 8 - 6 - 0 - 1 \\ -20 + 15 + 0 + 5 \\ -32 + 21 + 0 + 12 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = 1 \cdot v_A$$

   Since $M v_A = 1 \cdot v_A$, $v_A$ is an eigenvector with eigenvalue $\lambda = 1$.

2. For $v_B = (1, 0, -1, 0)^T$:

   $$M v_B = \begin{pmatrix} -9 + 0 + 2 - 0 \\ -8 + 0 + 3 - 0 \\ 20 + 0 - 8 + 0 \\ 32 + 0 - 7 + 0 \end{pmatrix} = \begin{pmatrix} -7 \\ -5 \\ 12 \\ 25 \end{pmatrix}$$

   This is not a scalar multiple of $v_B$, so $v_B$ is not an eigenvector.

3. For $v_C = (-1, 0, 2, 2)^T$:

   $$M v_C = \begin{pmatrix} 9 - 0 - 4 - 8 \\ 8 - 0 - 6 - 2 \\ -20 + 0 + 16 + 10 \\ -32 + 0 + 14 + 24 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ 6 \\ 6 \end{pmatrix} = 3 \begin{pmatrix} -1 \\ 0 \\ 2 \\ 2 \end{pmatrix} = 3 \cdot v_C$$

   Since $M v_C = 3 \cdot v_C$, $v_C$ is an eigenvector with eigenvalue $\lambda = 3$.

4. For $v_D = (0, 1, -3, 0)^T$:

   $$M v_D = \begin{pmatrix} 0 - 6 + 6 - 0 \\ 0 - 6 + 9 - 0 \\ 0 + 15 - 24 + 0 \\ 0 + 21 - 21 + 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -9 \\ 0 \end{pmatrix} = 3 \begin{pmatrix} 0 \\ 1 \\ -3 \\ 0 \end{pmatrix} = 3 \cdot v_D$$

   Since $M v_D = 3 \cdot v_D$, $v_D$ is an eigenvector with eigenvalue $\lambda = 3$.

Therefore, (A), (C), and (D) are eigenvectors.