Question
If \[ A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] prove that \[ A^2 – A + 2I = 0. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 3\cdot3 + (-2)\cdot4 & 3(-2) + (-2)(-2) \\ 4\cdot3 + (-2)\cdot4 & 4(-2) + (-2)(-2) \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \]Step 2: Form Expression
\[ A^2 – A + 2I = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} – \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \]Step 3: Simplify
\[ = \begin{bmatrix} 1 – 3 + 2 & -2 + 2 + 0 \\ 4 – 4 + 0 & -4 + 2 + 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]Final Result
\[
A^2 – A + 2I =
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
\]
Hence proved.