Answer
$A^{-1}=\begin{bmatrix}
\frac{-5}{18} &\frac{1}{18} & \frac{7}{18} \\
\frac{1}{18} & \frac{7}{18} & \frac{-5}{18} \\
\frac{7}{18} & \frac{-5}{18} & \frac{1}{18}
\end{bmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
3 &1 &2
\end{bmatrix} $
$\det(A)=1\begin{vmatrix}
3 & 1 \\
1 & 2
\end{vmatrix}-2\begin{vmatrix}
2 & 1 \\
3 & 2
\end{vmatrix}+\begin{vmatrix}
2 & 3 \\
3 & 1
\end{vmatrix}=1(5-2)+3(2-9)=-18$
Since $D=-18 \ne0$, the matrix is invertible.
$A_{11}=\begin{vmatrix}
3 & 1\\
1 & 2
\end{vmatrix}=5$
Just by doing that, we will gradually find:
$A_{12}=3$
$A_{13}=-7$
$A_{21}=-1$
$A_{22}=-7$
$A_{23}=5$
$A_{31}=-7$
$A_{32}=5$
$A_{33}=-1$
$\rightarrow adj(A)=\begin{bmatrix}
5 & -1 & -7 \\
-1 & -7 & 5 \\
-7 & 5 & -1
\end{bmatrix}$
Use the adjoint method to find $A^{-1}:$
$A^{-1}=\frac{1}{\det(A)}adj(A)$
$A^{-1}=\frac{1}{-18}\begin{bmatrix}
5 & -1 & -7 \\
-1 & -7 & 5 \\
-7 & 5 & -1
\end{bmatrix}$
$A^{-1}=\begin{bmatrix}
\frac{-5}{18} &\frac{1}{18} & \frac{7}{18} \\
\frac{1}{18} & \frac{7}{18} & \frac{-5}{18} \\
\frac{7}{18} & \frac{-5}{18} & \frac{1}{18}
\end{bmatrix}$
