Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
0 & -3 & 2 & 2\\0 & 1 & 1 & 1\\1 & 2& 3&-4\\1&0&0&5
\end{bmatrix}$
We will use Cofactor Expansion Theorem to find the determinant of this matrix.
$\det(A)=\sum^n_{j=1}a_{kj}C_{kj}\\
\det(A)=\sum^n_i=1 a_{ik}C_{ik}$
Then $\rightarrow \det(A)=1\begin{vmatrix}
-3&2&2 \\1& 1&1 \\0&0&5
\end{vmatrix}-1.\begin{vmatrix}
-3 & 2 & 2\\1&1&1\\2 & 3 & -4
\end{vmatrix}\\
=1[-3(1.(-5)-1.0)-1.(2.5-2.0)+0.(2.1-2.1)]-[-3.(1.(-4)-1.3)-1.(2.(-4)-2.3)+2.(2.1-2.1)]\\
=-60$
By using adjoint method, we have
$A^{-1}=\frac{1}{\det(A)}adj(A)=\frac{1}{\det(A)}C^T$
where $C$ is cofactor matrix.
Obtain:
$adj(A)=\begin{bmatrix}
\begin{vmatrix}
1&1 &1\\2&3&-4\\0&0&5
\end{vmatrix} & -\begin{vmatrix}
0&& 1&1\\1&3 &-4\\1&0&5
\end{vmatrix} & \begin{vmatrix}
0&1&1\\1&2&-4\\1&0&5
\end{vmatrix} &- \begin{vmatrix}
0&1&1\\1&2&3\\1&0&0
\end{vmatrix}\\-\begin{vmatrix}
-3&2&2\\2&3 &-4\\0&0&5
\end{vmatrix} & \begin{vmatrix}
0 &2&2\\1&3 &-4\\1&0&5
\end{vmatrix}
& -\begin{vmatrix}
0-3&2\\1&2&-4\\1&0&5
\end{vmatrix}&\begin{vmatrix}
0&-3&2\\1&2&3\\1&0&0
\end{vmatrix}\\\begin{vmatrix}
-3&2 & 2\\1&1&1\\0&0&5
\end{vmatrix}&-\begin{vmatrix}
0&2&2\\0&1 & 1\\1&0&5
\end{vmatrix}&\begin{vmatrix}
0&-3 & 2\\0&1 & 1\\1&0&5
\end{vmatrix}&\begin{vmatrix}
0&-3&2\\0&1&1\\1&0&5
\end{vmatrix}\\-\begin{vmatrix}
-3&2&2\\1&1&1\\2&3&-4
\end{vmatrix}& \begin{vmatrix}
0&2&2\\0&1&1\\1&3&-4
\end{vmatrix} & -\begin{vmatrix}
0& -3&2\\0 &1& 1\\1&2& -4
\end{vmatrix} & \begin{vmatrix}
0&-3&2\\0&1&1\\1&2&3
\end{vmatrix}\end{bmatrix}^T\\
=\begin{bmatrix}
5&65&-25&-35\\12& -24&0 &0 \\-11&-23& -5 & 5\\-1 & -13 &5 &-5
\end{bmatrix}$
Consequently, $A^{-1}=\frac{-1}{60}\begin{bmatrix}
5&65&-25&-35\\12& -24&0 &0 \\-11&-23& -5 & 5\\-1 & -13 &5 &-5
\end{bmatrix}=\begin{bmatrix}
\frac{-1}{12}& \frac{-13}{12}&\frac{5}{12} &\frac{7}{12}\\\frac{-1}{12}& \frac{2}{5}& 0 & 0 \\\frac{11}{60}&\frac{23}{60}& \frac{1}{12} &-\frac{1}{12} \\\frac{1}{60} &\frac{13}{60} & \frac{-1}{12} &\frac{1}{12}
\end{bmatrix}$
