Question
If \[ A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \] show that \[ A^2 – 5A + 7I = O \] and hence find \(A^4\).
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \]Step 2: Verify Identity
\[ A^2 – 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} – \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]Step 3: Use Identity
\[ A^2 = 5A – 7I \]Step 4: Find \(A^4\)
\[ A^4 = (A^2)^2 = (5A – 7I)^2 \] \[ = 25A^2 – 70A + 49I \]Step 5: Substitute \(A^2\)
\[ = 25(5A – 7I) – 70A + 49I \] \[ = 125A – 175I – 70A + 49I \] \[ = 55A – 126I \]Final Answer
\[
A^4 = 55A – 126I
\]