Solve this

$A=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$

Now,

$C_{11}=\left|\begin{array}{ll}0 & 1 \\ 4 & 3\end{array}\right|=-4, C_{12}=-\left|\begin{array}{ll}1 & 1 \\ 4 & 3\end{array}\right|=1$ and $C_{13}=\left|\begin{array}{ll}1 & 0 \\ 4 & 4\end{array}\right|=4$

$C_{21}=-\left|\begin{array}{cc}-3 & -3 \\ 4 & 3\end{array}\right|=-3, C_{22}=\left|\begin{array}{cc}-4 & -3 \\ 4 & 3\end{array}\right|=0$ and $C_{23}=-\left|\begin{array}{cc}-4 & -3 \\ 4 & 4\end{array}\right|=4$

$C_{31}=\left|\begin{array}{cc}-3 & -3 \\ 0 & 1\end{array}\right|=-3, C_{32}=-\left|\begin{array}{cc}-4 & -3 \\ 1 & 1\end{array}\right|=1$ and $C_{33}=\left|\begin{array}{cc}-4 & -3 \\ 1 & 0\end{array}\right|=3$

$\therefore \operatorname{adj} A=\left[\begin{array}{ccc}-4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3\end{array}\right]^{T}=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]=A$

Hence proved.