Question
If \[ A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] find \( \lambda, \mu \) such that \[ A^2 = \lambda A + \mu I. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2\cdot2 + 3\cdot1 & 2\cdot3 + 3\cdot2 \\ 1\cdot2 + 2\cdot1 & 1\cdot3 + 2\cdot2 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \]Step 2: Form RHS
\[ \lambda A + \mu I = \begin{bmatrix} 2\lambda + \mu & 3\lambda \\ \lambda & 2\lambda + \mu \end{bmatrix} \]Step 3: Compare
\[ \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 2\lambda + \mu & 3\lambda \\ \lambda & 2\lambda + \mu \end{bmatrix} \]Step 4: Solve
\[ 3\lambda = 12 \Rightarrow \lambda = 4 \] \[ 2\lambda + \mu = 7 \Rightarrow 8 + \mu = 7 \Rightarrow \mu = -1 \]Final Answer
\[
\lambda = 4, \quad \mu = -1
\]