Question

Given matrix \( A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \), the eigenvalue corresponding to the eigenvector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) is \(\underline{\hspace{2cm}}\).

Hint:

When calculating eigenvalues, always use the equation \( A \mathbf{v} = \lambda \mathbf{v} \), and solve for \( \lambda \).

Answer 1

Correct Answer: 3

To find the eigenvalue corresponding to the eigenvector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \), we use the equation: \\ \[ A \mathbf{v} = \lambda \mathbf{v} \] Substituting the matrix \( A \) and the eigenvector \( \mathbf{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \): \\ \[ \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] Performing the matrix multiplication: \\ \[ \begin{bmatrix} 2 \times 1 + (-1) \times (-1) \\ -1 \times 1 + 2 \times (-1) \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] \[ \begin{bmatrix} 2 + 1 \\ -1 - 2 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] \[ \begin{bmatrix} 3 \\ -3 \end{bmatrix} = \lambda \begin{bmatrix} 1 \\ -1 \end{bmatrix} \] Thus, \( \lambda = 3 \). Therefore, the eigenvalue is \( \boxed{3} \).