Question

Let \[ M = \begin{bmatrix} 5 & -6 \\ 3 & -4 \end{bmatrix} \] be a \(2 \times 2\) matrix. If \(\alpha = \det(M^4 - 6I_2)\), then the value of \(\alpha^2\) is ............

Hint:

For any diagonalizable matrix \(M\), \(\det(f(M)) = \prod f(\lambda_i)\), where \(\lambda_i\) are the eigenvalues of \(M\).

Answer 1

Correct Answer: 2500

Step 1: Find eigenvalues of \(M\). 
Characteristic equation: \[ |M - \lambda I| = \begin{vmatrix} 5 - \lambda & -6 \\ 3 & -4 - \lambda \end{vmatrix} = (5 - \lambda)(-4 - \lambda) + 18 = \lambda^2 - \lambda - 2 = 0. \] So the eigenvalues are \(\lambda_1 = 2\), \(\lambda_2 = -1\). 
Step 2: Express determinant in terms of eigenvalues. 
\[ \det(M^4 - 6I) = (\lambda_1^4 - 6)(\lambda_2^4 - 6). \] Compute: \[ \lambda_1^4 = 2^4 = 16, \quad \lambda_2^4 = (-1)^4 = 1. \] \[ \alpha = (16 - 6)(1 - 6) = (10)(-5) = -50. \] 
Step 3: Compute \(\alpha^2\). 
\[ \alpha^2 = (-50)^2 = 2500. \]