Question
If \[ A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \] find \(k\) such that \[ A^2 = kA – 2I_2. \]
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 RHS
\[ kA – 2I = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} – \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k – 2 & -2k \\ 4k & -2k – 2 \end{bmatrix} \]Step 3: Compare
\[ \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k – 2 & -2k \\ 4k & -2k – 2 \end{bmatrix} \]Step 4: Solve
\[ 3k – 2 = 1 \Rightarrow k = 1 \] (Check: \(-2k = -2\), \(4k = 4\), consistent)Final Answer
\[
k = 1
\]