Question

Suppose that 2 is an eigenvalue of the matrix 

Then the value of \( \alpha \) is equal to (Answer in integer):

Hint:

To find the eigenvalue of a matrix, subtract the eigenvalue from the diagonal elements of the matrix, then compute the determinant. Set the determinant equal to zero and solve for the unknown.

Answer 1

Given that 2 is an eigenvalue of the matrix \( A \), we know that the determinant of \( A - 2I \) must be zero, where \( I \) is the identity matrix. The matrix \( A - 2I \) is:

\[ A - 2I = \begin{bmatrix} 0 & 3 & -\alpha \\ 0 & 1 & 0 \\ 1 & -1 & 3 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 3 & -\alpha \\ 0 & -1 & 0 \\ 1 & -1 & 1 \end{bmatrix} \]

Now, we compute the determinant of \( A - 2I \):

\[ \text{det}(A - 2I) = \begin{vmatrix} -2 & 3 & -\alpha \\ 0 & -1 & 0 \\ 1 & -1 & 1 \end{vmatrix} \]

Expanding along the first row:

\[ \text{det}(A - 2I) = (-2) \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} - 3 \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} + (-\alpha) \begin{vmatrix} 0 & -1 \\ 1 & -1 \end{vmatrix} \]

Simplifying the minors:

\[ \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} = (-1)(1) - (0)(-1) = -1 \]

\[ \begin{vmatrix} 0 & 0 \\ 1 & 1 \end{vmatrix} = (0)(1) - (0)(1) = 0 \]

\[ \begin{vmatrix} 0 & -1 \\ 1 & -1 \end{vmatrix} = (0)(-1) - (-1)(1) = 1 \]

Substituting back:

\[ \text{det}(A - 2I) = (-2)(-1) - 3(0) + (-\alpha)(1) = 2 - \alpha \]

For 2 to be an eigenvalue, we set the determinant to zero:

\[ 2 - \alpha = 0 \quad \Rightarrow \quad \alpha = 2 \]

Thus, the value of \( \alpha \) is \( 2 \).