Question

If \[ \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \quad \text{is the eigenvector of the matrix} \quad A = \begin{bmatrix} 8 & 11 & 3 \\ 4 & -1 & 3 \\ -4 & 10 & 6 \end{bmatrix}, \quad \text{then the corresponding eigenvalue is:} \]

• A. \(-12\)
• B. \(0\)
• C. \(1\)
• D. \(9\)

Hint:

To verify an eigenvector and eigenvalue pair, compute \( A\vec{v} \) and check whether the result is a scalar multiple of \( \vec{v} \). If so, that scalar is the eigenvalue.

Answer 1

The Correct Option is D

Step 1: Let the eigenvector be \[ \vec{v} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}. \] Step 2: Multiply the matrix \( A \) with vector \( \vec{v} \): \[ A \vec{v} = \begin{bmatrix} 8 & 11 & 3 \\ 4 & -1 & 3 \\ -4 & 10 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} (8)(2) + (11)(-1) + (3)(1) \\ (4)(2) + (-1)(-1) + (3)(1) \\ (-4)(2) + (10)(-1) + (6)(1) \end{bmatrix} = \begin{bmatrix} 16 - 11 + 3 \\ 8 + 1 + 3 \\ -8 - 10 + 6 \end{bmatrix} = \begin{bmatrix} 8 \\ 12 \\ -12 \end{bmatrix}. \] Step 3: Check if \( A\vec{v} = \lambda \vec{v} \): \[ \lambda \vec{v} = \lambda \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2\lambda \\ -\lambda \\ \lambda \end{bmatrix}. \] Matching with: \[ \begin{bmatrix} 8 \\ 12 \\ -12 \end{bmatrix} \Rightarrow 2\lambda = 8 \Rightarrow \lambda = 4. \] Final Answer: \[ \boxed{4}. \]