Question

The sum of the eigenvalues of the matrix \( A = \begin{bmatrix} 1 & 2
3 & 4 \end{bmatrix}^2 \) is \_\_\_\_\_ (rounded off to the nearest integer).

Hint:

The trace of a matrix equals the sum of its eigenvalues. For powers of matrices, calculate carefully using the characteristic equation.

Answer 1

Given: \[ A = \begin{bmatrix} 1 & 2
3 & 4 \end{bmatrix}^2 \] Step 1: Find \( A^2 \): \[ A = \begin{bmatrix} 1 & 2
3 & 4 \end{bmatrix}, \quad A^2 = \begin{bmatrix} 7 & 10
15 & 22 \end{bmatrix} \] Step 2: Eigenvalues of \( A^2 \): The characteristic equation of \( A^2 \) is: \[ |A - \lambda I| = 0 \] \[ \lambda^2 - 29\lambda + 154 = 0 \] Solving this gives: \[ \lambda_1 = 28.8615, \quad \lambda_2 = 0.1385 \] Step 3: Sum of eigenvalues: \[ \lambda_1 + \lambda_2 = 28.8615 + 0.1385 = 29 \] Final Answer: The sum of the eigenvalues is \( 29 \). % Quick tip