Answer
See below
Work Step by Step
Given: $A=\begin{bmatrix}
2 & -1 & 1\\0 & 5 & -1\\1 & 1 & 3
\end{bmatrix}$
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 $\rightarrow \det(A)=2\begin{vmatrix}
5&-1 \\1&3
\end{vmatrix}+1.\begin{vmatrix}
-1 &1 \\5&-1
\end{vmatrix}\\
=2.[5.3-(-1).1]+1.[-1.(-1)-1.5]\\
=28$
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:
$adj(A)=\begin{bmatrix}
\begin{vmatrix}
5&-1\\1&3
\end{vmatrix} & -\begin{vmatrix}
0&-1\\1&3
\end{vmatrix} & \begin{vmatrix}
0&5\\1&1
\end{vmatrix}\\-\begin{vmatrix}
-1&1\\1&3
\end{vmatrix} & \begin{vmatrix}
2&1\\1&3
\end{vmatrix}
& -\begin{vmatrix}
2&-1\\1&1
\end{vmatrix}\\\begin{vmatrix}
-1&1\\5&-1
\end{vmatrix}&\begin{vmatrix}
2&1\\0&-1
\end{vmatrix}&\begin{vmatrix}
2&-1\\0&5
\end{vmatrix}\end{bmatrix}^T\\
=\begin{bmatrix}
16& 4&-4\\-1& 5&2 \\-5&-2&10
\end{bmatrix}$
Consequently, $A^{-1}=\frac{1}{28}\begin{bmatrix}
16& 4&-4\\-1& 5&2 \\-5&-2&10
\end{bmatrix}=\begin{bmatrix}
\frac{4}{7}& \frac{1}{7}&\frac{4}{7}\\\frac{-1}{28}& \frac{5}{28}&\frac{1}{14} \\-\frac{5}{28}&-\frac{3}{28}& \frac{1}{14}
\end{bmatrix}$
