Question

An eigenvector of the matrix

\[ A= \begin{bmatrix} 3 & 2 \\ 1 & 2 \end{bmatrix} \]

is

• A. \( \begin{bmatrix} 7\\ 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} -2\\ 1 \end{bmatrix} \)
• C. \( \begin{bmatrix} -1\\ 2 \end{bmatrix} \)
• D. \( \begin{bmatrix} 5\\ -5 \end{bmatrix} \)

Hint:

To verify an eigenvector, simply check \[ \boxed{ AX=\lambda X. } \] If the resulting vector is a scalar multiple of the original vector, then it is an eigenvector.

Answer 1

The Correct Option is D

Step 1: Use the eigenvector definition. A vector \(X\) is an eigenvector if \[ AX=\lambda X \] for some scalar \(\lambda\).

Step 2: Check option (D). Let \[ X= \begin{bmatrix} 5 \\ -5 \end{bmatrix}. \] Then, \[ AX= \begin{bmatrix} 3 & 2 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 5 \\ -5 \end{bmatrix} = \begin{bmatrix} 15-10 \\ 5-10 \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \end{bmatrix}. \] Thus, \[ AX=1\cdot X. \] Hence, \[ \boxed{ \begin{bmatrix} 5 \\ -5 \end{bmatrix} } \] is an eigenvector corresponding to the eigenvalue \[ \lambda=1. \] Therefore, \[ \boxed{(D)} \] is the correct answer.