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

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 \( A \), the characteristic polynomial is found by calculating the determinant of \( A - \lambda I \):

\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 1 \\ 0 & k - \lambda & 0 \\ 3 & 0 & -1 - \lambda \end{pmatrix}. \]
The determinant of this matrix is:

\[ \det(A - \lambda I) = (1 - \lambda) \left( (k - \lambda)(-1 - \lambda) \right) - 3 \cdot 0 + 1 \cdot 0. \]
This simplifies to:

\[ \det(A - \lambda I) = (1 - \lambda)(k - \lambda)(-1 - \lambda). \]

Step 2: Use the given eigenvalues.
We know that the eigenvalues of the matrix are \( -2 \), \( 1 \), and \( 2 \), so we can factor the characteristic polynomial as:

\[ (\lambda + 2)(\lambda - 1)(\lambda - 2) = 0. \]
Expanding the factors:

\[ (\lambda + 2)(\lambda^2 - 3\lambda + 2) = \lambda^3 - \lambda^2 - 4\lambda + 4. \]

This is the characteristic polynomial for the matrix \( A \). Now, we compare this expanded polynomial with the determinant expression from earlier.

Step 3: Compare the coefficients.
The coefficient of \( \lambda^2 \) in the expanded characteristic polynomial is \( -1 \).
The coefficient of \( \lambda^2 \) from expanding \( (1 - \lambda)(k - \lambda)(-1 - \lambda) \) is \( k - 1 \).

So we equate:

\[ k - 1 = -1 \Rightarrow k = 1. \]

Thus, the value of \( k \) is \( \boxed{1} \).