Question
If \[ A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] find \( \lambda \) such that \[ A^2 = 5A + \lambda I. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3\cdot3 + 1(-1) & 3\cdot1 + 1\cdot2 \\ (-1)\cdot3 + 2(-1) & (-1)\cdot1 + 2\cdot2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \]Step 2: Form \(5A + \lambda I\)
\[ 5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} ,\quad \lambda I = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \] \[ 5A + \lambda I = \begin{bmatrix} 15 + \lambda & 5 \\ -5 & 10 + \lambda \end{bmatrix} \]Step 3: Compare with \(A^2\)
\[ \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 15 + \lambda & 5 \\ -5 & 10 + \lambda \end{bmatrix} \]Step 4: Solve
\[ 15 + \lambda = 8 \Rightarrow \lambda = -7 \] (Check: \(10 + \lambda = 3 \Rightarrow \lambda = -7\), consistent)Final Answer
\[
\lambda = -7
\]