Answer
$A^{-1}=\begin{bmatrix}
\frac{-2}{3} &\frac{17}{5} & \frac{-19}{15} \\
\frac{1}{6} & \frac{-4}{5} & \frac{11}{30} \\
\frac{1}{3} & -1& \frac{1}{3}
\end{bmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
3 & 4 & 7 \\
2 & 6 & 1 \\
3 &14 &-1
\end{bmatrix} $
$\det(A)=3\begin{vmatrix}
6 & 1 \\
14 & -1
\end{vmatrix}-4\begin{vmatrix}
2 & 1 \\
3 & -1
\end{vmatrix}+7\begin{vmatrix}
2 & 6 \\
3 & 14
\end{vmatrix}=3.(-20)-4(-5)+7.10=30$
Since $D=30 \ne0$, the matrix is invertible.
$A_{11}=\begin{vmatrix}
6 & 1\\
14 & -1
\end{vmatrix}=-20$
Just by doing that, we will gradually find:
$A_{12}=5$
$A_{13}=10$
$A_{21}=102$
$A_{22}=24$
$A_{23}=-30$
$A_{31}=-38$
$A_{32}=11$
$A_{33}=10$
$\rightarrow adj(A)=\begin{bmatrix}
-20 & 102 & -38 \\
5 & -24 & 11 \\
10& -30 & 10
\end{bmatrix}$
Use the adjoint method to find $A^{-1}:$
$A^{-1}=\frac{1}{\det(A)}adj(A)$
$A^{-1}=\frac{1}{30}\begin{bmatrix}
-20 & 102 & -38 \\
5 & -24 & 11 \\
10& -30 & 10
\end{bmatrix}$
$A^{-1}=\begin{bmatrix}
\frac{-2}{3} &\frac{17}{5} & \frac{-19}{15} \\
\frac{1}{6} & \frac{-4}{5} & \frac{11}{30} \\
\frac{1}{3} & -1& \frac{1}{3}
\end{bmatrix}$
