Solve this

We have, $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$

Now,

$\operatorname{adj}(A)=\left[\begin{array}{cc}1 & -5 \\ -2 & 4\end{array}\right]$

and $|A|=-6$

$\therefore A^{-1}=-\frac{1}{6}\left[\begin{array}{cc}1 & -5 \\ -2 & 4\end{array}\right]$

Now, $A-3 I=I+3 A^{-1}$

$\mathrm{LHS}=A-3 I=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]-3\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 5 \\ 2 & -2\end{array}\right]$

 RHS $=2\left(I+3 A^{-1}\right)=2\left\{\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-3 \times \frac{1}{6}\left[\begin{array}{cc}1 & -5 \\ -2 & 4\end{array}\right]\right\}=2\left[\begin{array}{cc}0.5 & 2.5 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}1 & 5 \\ 2 & -2\end{array}\right]=$ LHS

Hence proved.