Question

Let \( A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & k & 0 \\ 3 & 0 & -1 \end{pmatrix} \). If the eigenvalues of \( A \) are -2, 1, and 2, then the value of \( k \) is __________. (Answer in integer)

Hint:

When solving for eigenvalues, always expand the characteristic polynomial and compare it with the determinant expression. The coefficients will help you solve for unknown parameters like \( k \).

Answer 1

Correct Answer: 1

Step 1: Write the characteristic equation.
The characteristic equation for a matrix \( A \) is given by: \[ \det(A - \lambda I) = 0, \] where \( \lambda \) represents the eigenvalues of the matrix \( A \), and \( I \) is the identity matrix. For the matrix \( A \), we have: \[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 1 \\ 0 & k - \lambda & 0 \\ 3 & 0 & -1 - \lambda \end{pmatrix}. \] The determinant of this matrix is: \[ \det(A - \lambda I) = (k - \lambda) \begin{vmatrix} 1 - \lambda & 1 \\ 3 & -1 - \lambda \end{vmatrix}. \] \[ = (k - \lambda)\left[(1 - \lambda)(-1 - \lambda) - 3\right]. \] \[ = (k - \lambda)\left(\lambda^2 - 1 - 3\right) = (k - \lambda)(\lambda^2 - 4). \] Thus, the characteristic polynomial is: \[ (k - \lambda)(\lambda - 2)(\lambda + 2). \] Step 2: Use the given eigenvalues.
The eigenvalues are given as \( -2, 1, 2 \). Hence, the characteristic polynomial must be: \[ (\lambda + 2)(\lambda - 1)(\lambda - 2). \] Comparing with: \[ (k - \lambda)(\lambda - 2)(\lambda + 2), \] we see that: \[ k - \lambda = \lambda - 1. \] Step 3: Solve for \( k \).
Comparing coefficients: \[ k = 1. \] Final Answer:
\[ \boxed{1} \]