if

Solution:

Given : $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$

Now,

$A^{2}=A A$

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

$\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]$

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

$A^{2}-5 A+7 I$

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

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

$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{cc}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$

Hence proved.

Given : $A^{2}-5 A+7 I=0$

$\Rightarrow A^{2}=5 A-7 I$

$\Rightarrow A^{3}=A(5 A-7 I) \quad$ (Multilpying by $A$ on both sides)

$\Rightarrow A^{3}=5 A^{2}-7 A I$

$\Rightarrow A^{3}=5(5 A-7 I)-7 A \quad$ [From eq. (1)]

$\Rightarrow A^{3}=25 A-35 I-7 A$

$\Rightarrow A^{3}=18 A-35 I$

$\Rightarrow A^{4}=(18 A-35 I) A$                   (Multilpying by $A$ on both sides)

$\Rightarrow A^{4}=18 A^{2}-35 A$

$\Rightarrow A^{4}=18(5 A-7 I)-35 A$            [From eq. (1)]

$\Rightarrow A^{4}=90 A-126 I-35 A$

$\Rightarrow A^{4}=55 A-126 I$

$\Rightarrow A^{4}=55\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-126\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{4}=\left[\begin{array}{cc}165 & 55 \\ -55 & 110\end{array}\right]-\left[\begin{array}{cc}126 & 0 \\ 0 & 126\end{array}\right]$

$\Rightarrow A^{4}=\left[\begin{array}{cc}165-126 & 55-0 \\ -55-0 & 110-126\end{array}\right]$

$\Rightarrow A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$