Question
If \[ A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \quad I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] show that \[ A^2 – 5A + 7I_2 = O. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3\cdot3 + 1(-1) & 3\cdot1 + 1\cdot2 \\ (-1)\cdot3 + 2(-1) & (-1)\cdot1 + 2\cdot2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \]Step 2: Form Expression
\[ A^2 – 5A + 7I_2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \]Step 3: Simplify
\[ = \begin{bmatrix} 8 – 15 + 7 & 5 – 5 + 0 \\ -5 + 5 + 0 & 3 – 10 + 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]Final Result
\[
A^2 – 5A + 7I_2 = O
\]
Hence proved.