Find p such that A² = pA

Given:

\[ A = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \]

Step 1: Compute A²

\[ A^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \]

\[ A^2 = \begin{bmatrix} 2(2)+(-2)(-2) & 2(-2)+(-2)(2) \\ (-2)(2)+2(-2) & (-2)(-2)+2(2) \end{bmatrix} = \begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix} \]

Step 2: Compare with pA

\[ pA = p \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 2p & -2p \\ -2p & 2p \end{bmatrix} \]

Step 3: Equate Elements

\[ 2p = 8 \Rightarrow p = 4 \]

Final Answer:

\[ \boxed{p = 4} \]