Question

If $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$, then show that $A^2 - 4A + 7I = 0$.

Hint:

When working with matrix equations, perform matrix multiplications carefully and ensure that each term is calculated and simplified correctly.

Answer 1

We are given the matrix $A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}$, and we need to show that: \[ A^2 - 4A + 7I = 0 \] First, calculate $A^2$: \[ A^2 = A \times A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \] Performing matrix multiplication: \[ A^2 = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 2 \\ -1 \times 2 + 2 \times (-1) & -1 \times 3 + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \] Now calculate $4A$: \[ 4A = 4 \times \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} \] Next, calculate $7I$ where $I$ is the identity matrix: \[ 7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \] Now, compute $A^2 - 4A + 7I$: \[ A^2 - 4A + 7I = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \] Simplifying: \[ A^2 - 4A + 7I = \begin{bmatrix} 1 - 8 + 7 & 12 - 12 + 0 \\ -4 + 4 + 0 & 1 - 8 + 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] Thus, $A^2 - 4A + 7I = 0$ as required.