If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then $A^2-5 A+7 I=$

0

$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \Rightarrow A^2=A . A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \Rightarrow A^2=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

So  $A^2-5A+7I$

$=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]+\left[\begin{array}{cc}-15 & -5 \\ 5 & -10\end{array}\right]+\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right]$

$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$= 0$