If A = $\begin{bmatrix$}[r] 2 & -2 \ -2 & 2 \end{bmatrix}, then A² = pA, then the value of p is

1. 2

2. 4

3. -2

4. -4

Given,

$\text{A}^2 = pA \\[0.5em] \Rightarrow \begin{bmatrix$}[r] 2 & -2 \ -2 & 2 \end{bmatrix} \begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} = p\begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2 \times 2 + (-2) \times (-2) & 2 \times (-2) + (-2) \times 2 \ -2 \times 2 + 2 \times (-2) & (-2) \times (-2) + 2 \times 2 \end{bmatrix} = p\begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 4 + 4 & -4 - 4 \ -4 - 4 & 4 + 4 \end{bmatrix} = p\begin{bmatrix}[r] 2 & -2 \ -2 & 2 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 8 & -8 \ -8 & 8 \end{bmatrix} = \begin{bmatrix}[r] 2p & -2p \ -2p & 2p \end{bmatrix} \\[0.5em]

By definition of equality of matrices we get,

⇒ 2p = 8

∴ p = 4.

∴ Option 2 is the correct option.