Show that

$A=\left[\begin{array}{ll}6 & 5\end{array}\right.$

$\left.\begin{array}{ll}7 & 6\end{array}\right]$

$\therefore A^{2}=\left[\begin{array}{ll}71 & 60\end{array}\right.$

$\left.\begin{array}{ll}84 & 71\end{array}\right]$

If $I_{2}$ is the identity matrix of order 2, then

$A^{2}-12 A+I_{2}=\left[\begin{array}{ll}71 & 60\end{array}\right.$

$84 \quad 71]-12\left[\begin{array}{ll}6 & 5\end{array}\right.$

$\left.\begin{array}{ll}7 & 6\end{array}\right]+\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow A^{2}-12 A+I_{2}=[71-72+1 \quad 60-60+0$

$84-84+0 \quad 71-72+1]$

$\Rightarrow A^{2}-12 A+I_{2}=0$

Thus, $A$ satisfies $x^{2}-12 x+1=0$.

Now,

$A^{2}-12 A+I_{2}=0$

$\Rightarrow I_{2}=12 A-A^{2}$

$\Rightarrow A^{-1} I_{2}=A^{-1}\left(12 A-A^{2}\right)$       $\left[\right.$ Pre $-$ multiplying both sides by $\left.A^{-1}\right]$

$\Rightarrow A^{-1}=12 I_{2}-A$

$\Rightarrow A^{-1}=12\left[\begin{array}{ll}1 & 0\end{array}\right.$

$0 \quad 1]-\left[\begin{array}{ll}6 & 5\end{array}\right.$

$\left.\begin{array}{ll}7 & 6\end{array}\right]$

$\Rightarrow A^{-1}=\left[\begin{array}{ll}12-6 & 0-5\end{array}\right.$

$0-7 \quad 12-6]$

$\Rightarrow A^{-1}=\left[\begin{array}{ll}6 & -5\end{array}\right.$

$\left.\begin{array}{ll}-7 & 6\end{array}\right]$