Find the inverse of the matrix

We have,

$A=\left[\begin{array}{cc}a & b \\ c & \frac{1+b c}{a}\end{array}\right]$

So, $\operatorname{adj}(A)=\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]$

and $|A|=1$

$\therefore A^{-1}=\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]$

Now, $a A^{-1}=\left(a^{2}+b c+1\right) I-a A$

LHS $=a A^{-1}=a\left[\begin{array}{cc}\frac{1+b c}{a} & -b \\ -c & a\end{array}\right]=\left[\begin{array}{cc}1+b c & -b a \\ -c a & a^{2}\end{array}\right]$

$\mathrm{RHS}=\left(a^{2}+b c+1\right) I-a A$

$=a^{2}+b c+1\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}a^{2} & b a \\ c a & 1+b c\end{array}\right]$

$=\left[\begin{array}{cc}a^{2}+b c+1 & 0 \\ 0 & a^{2}+b c+1\end{array}\right]-\left[\begin{array}{cc}a^{2} & b a \\ c a & 1+b c\end{array}\right]$

$=\left[\begin{array}{cc}1+b c & -b a \\ -c a & a^{2}\end{array}\right]=$ LHS

Hence proved.