Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
0 & 0 & 0 & 1\\0 & 1 & 3 & -3\\-2 & -3& -5&2\\4&-4&4&6
\end{bmatrix}$
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 $\rightarrow \det(A)=-1\begin{vmatrix}
0&1&3\\-2& -3& -5\\ 4&-4&4
\end{vmatrix}\\
=-1[2(1.4-3.(-4))+4(1.(-5)-3.(-3))\\
=-48$
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:
$adj(A)=\begin{bmatrix}
88&32&16&-4\\-24& 12&12 &6 \\-40&-20& -4 & -2\\-48 & 0 &0 &0
\end{bmatrix}$
Consequently, $A^{-1}=\frac{-1}{48}\begin{bmatrix}
88&32&16&-4\\-24& 12&12 &6 \\-40&-20& -4 & -2\\-48 & 0 &0 &0
\end{bmatrix}=\begin{bmatrix}
\frac{-11}{6}& \frac{-2}{3}&\frac{-1}{3} &\frac{1}{12}\\\frac{1}{2}& \frac{-1}{4}& -\frac{1}{4} & -\frac{1}{8} \\\frac{5}{6}&\frac{5}{12}& \frac{1}{12} &\frac{1}{24} \\1 &0 & 0 &0
\end{bmatrix}$
