If, show that

It is given that $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$

$\therefore A^{2}=A \cdot A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$

$=\left[\begin{array}{lr}3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2)\end{array}\right]$

$=\left[\begin{array}{lr}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right]$

$\therefore$ L.H.S. $=A^{2}-5 A+7 I$

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

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

$=\left[\begin{array}{rr}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{rr}7 & 0 \\ 0 & 7\end{array}\right]$

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

$=O=$ R.H.S.

$\therefore A^{2}-5 A+7 I=O$