Question
If \[ A = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}, \quad I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] show that \[ A^2 – 2A + 3I_2 = O. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 2\cdot2 + 3(-1) & 2\cdot3 + 3\cdot0 \\ (-1)\cdot2 + 0(-1) & (-1)\cdot3 + 0\cdot0 \end{bmatrix} = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} \]Step 2: Form Expression
\[ A^2 – 2A + 3I_2 = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} – \begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \]Step 3: Simplify
\[ = \begin{bmatrix} 1 – 4 + 3 & 6 – 6 + 0 \\ -2 + 2 + 0 & -3 + 0 + 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]Final Result
\[
A^2 – 2A + 3I_2 = O
\]
Hence proved.