Question
Given $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$, evaluate $A^2-4 A$
✓
Answer
$\begin{aligned} & A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ & A^2-4 A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]-4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}1+8 & 1+3 \\ 8+24 & 8+9\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}9-4 & 4-4 \\ 32-32 & 17-12\end{array}\right] \\ & =\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right] .\end{aligned}$
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