If $A=\left[\begin{array}{r}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3\\ -4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right],\text{ then adj }\left(3A^2+12A\right)$ is equal to 

$\because A^2=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]$

$\therefore 3A^2+12A=3\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]+12\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

                               $=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$

$\Rightarrow \mathrm{adj}\left(3A^2+12A\right)=\left[\begin{array}{l}51\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}63\\ 84\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}72\end{array}\right]$