Find k such that A2 = kA
Given:
\[ A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \]
Step 1: Compute A2
\[ A^2 = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \]
\[ A^2 = \begin{bmatrix} 1(1) + (-1)(-1) & 1(-1) + (-1)(1) \\ (-1)(1) + 1(-1) & (-1)(-1) + 1(1) \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \]
Step 2: Compare with kA
\[ kA = k \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \]
Step 3: Equate Elements
\[ \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \]
\[ k = 2 \]
Final Answer:
\[ k = 2 \]