Question

If \[ A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \] find the value of \( k \) such that \[ A^2 - 8A + kI = 0. \]

Hint:

To solve \( A^2 - 8A + kI = 0 \), compute \( A^2 \) and simplify each element step by step. Use substitution and matrix algebra rules.

Answer 1

Step 1: Compute \( A^2 \): \[ A^2 = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -7 & 49 \end{bmatrix}. \] Step 2: Substitute \( A^2 \) and \( A \) into \( A^2 - 8A + kI = 0 \): \[ \begin{bmatrix} 1 & 0 \\ -7 & 49 \end{bmatrix} - 8 \cdot \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix} + k \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0. \] Simplify: \[ \begin{bmatrix} 1 - 8 + k & 0 \\ -7 + 8 & 49 - 56 + k \end{bmatrix} = 0 \quad \Rightarrow \quad \begin{bmatrix} k - 7 & 0 \\ 1 & k - 7 \end{bmatrix} = 0. \] Equate \( k - 7 = 0 \), so \( k = 7 \).