Answer
$A^{-1}=\begin{bmatrix}
\frac{1}{4} &0 & \frac{-1}{4} & 0\\
\frac{-4}{19} & \frac{-7}{38} & \frac{31}{38} & \frac{3}{19} \\
\frac{-17}{76} &\frac{-11}{38}& \frac{111}{76} & \frac{2}{19} \\
\frac{-1}{76} & \frac{15}{38} & \frac{-65}{76} & \frac{-1}{19}
\end{bmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
5 & -1 & 2 & 1 \\
3 & -1 & 4 & 5 \\
1 & -1 & 2 &1 \\
5 & 9 & -3 & 2
\end{bmatrix} $
We will gradually find:
$C_{11}=-38$
$C_{12}=32$
$C_{13}=34$
$C_{14}=2$
$C_{21}=0$
$C_{22}=28$
$C_{23}=44$
$C_{24}=-60$
$C_{31}=38$
$C_{32}=-124$
$C_{33}=-222$
$C_{34}=130$
$C_{41}=0$
$C_{42}=-24$
$C_{43}=-16$
$C_{44}=8$
$\det(A)=5.(-38)-1.32+2(34)+1.2=$
Since $D=-152\ne0$, the matrix is invertible.
$ adj(A)=\begin{bmatrix}
-38 &0& 38 & 0\\
32 & 28 & -124 & -24\\
34& 44& -222 & -16\\
2 & -60 & 130 & 8
\end{bmatrix}$
Use the adjoint method to find $A^{-1}:$
$A^{-1}=\frac{1}{\det(A)}adj(A)$
$A^{-1}=\frac{1}{-152}\begin{bmatrix}
-38 &0& 38 & 0\\
32 & 28 & -124 & -24\\
34& 44& -222 & -16\\
2 & -60 & 130 & 8
\end{bmatrix}$
$A^{-1}=\begin{bmatrix}
\frac{1}{4} &0 & \frac{-1}{4} & 0\\
\frac{-4}{19} & \frac{-7}{38} & \frac{31}{38} & \frac{3}{19} \\
\frac{-17}{76} &\frac{-11}{38}& \frac{111}{76} & \frac{2}{19} \\
\frac{-1}{76} & \frac{15}{38} & \frac{-65}{76} & \frac{-1}{19}
\end{bmatrix}$