Question

If the matrix \( A = \begin{pmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{pmatrix} \) has three distinct eigenvalues and one of its eigenvectors is \( \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \), then which of the following can be another eigenvector of \( A \)?

• A. \( \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \)
• B. \( \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \)
• C. \( \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \)
• D. \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \)

Hint:

To check if a vector is an eigenvector of a matrix, multiply the matrix by the vector. If the result is a scalar multiple of the original vector, it is an eigenvector.

Answer 1

The Correct Option is D

Given the matrix \( A \) and one of its eigenvectors \( \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \), we need to check which of the given vectors are eigenvectors of \( A \).

For an eigenvector \( v \), we have the relation: \[ A v = \lambda v \] where \( \lambda \) is the corresponding eigenvalue.

The matrix \( A \) is: \[ A = \begin{pmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{pmatrix} \] Let’s check option (4) \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \) by multiplying it with \( A \): \[ A \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3(1) + (-1)(-2) + 1(1) \\ -1(1) + 5(-2) + (-1)(1) \\ 1(1) + (-1)(-2) + 3(1) \end{pmatrix} = \begin{pmatrix} 3 + 2 + 1 \\ -1 - 10 - 1 \\ 1 + 2 + 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -12 \\ 6 \end{pmatrix} \] This simplifies to: \[ A \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 6 \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \] This shows that \( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \) is indeed an eigenvector of \( A \), corresponding to the eigenvalue \( \lambda = 6 \).

Therefore, the correct answer is option (4).