Question
If \[ A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \] find \[ A^2 – 5A + 14I. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 3\cdot3 + (-5)(-4) & 3(-5) + (-5)(2) \\ (-4)\cdot3 + 2(-4) & (-4)(-5) + 2\cdot2 \end{bmatrix} = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} \]Step 2: Form Expression
\[ A^2 – 5A + 14I = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} – \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} + \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} \]Step 3: Simplify
\[ = \begin{bmatrix} 29 – 15 + 14 & -25 + 25 + 0 \\ -20 + 20 + 0 & 24 – 10 + 14 \end{bmatrix} = \begin{bmatrix} 28 & 0 \\ 0 & 28 \end{bmatrix} \]Final Answer
\[
A^2 – 5A + 14I =
\begin{bmatrix}
28 & 0 \\
0 & 28
\end{bmatrix}
= 28I
\]