Question
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ find $A^2 - 5A + 7 I.$

Answer

$A^2=A \cdot A  $
$ =\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]  $
$ =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]  $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]  $
$ \Rightarrow A^2-5 A+71  $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]  $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] $
$ =\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]  $
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ =0 $

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