Answer
$A^{-1}=\begin{bmatrix}
\frac{51}{116} &\frac{2}{29} & \frac{31}{116} \\
\frac{-8}{29} & \frac{-5}{29} & \frac{6}{29} \\
\frac{27}{58} &\frac{3}{29}& \frac{-13}{58}
\end{bmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
2 & 5 & 7 \\
4 & -3 & 2 \\
6 &9 &11
\end{bmatrix} $
$\det(A)=-2\begin{vmatrix}
-3 & 2 \\
9 &11
\end{vmatrix}-5\begin{vmatrix}
4 & 2 \\
6 & 11
\end{vmatrix}+7\begin{vmatrix}
4 & -3 \\
6 & 9
\end{vmatrix}=2.51-5.32+7.54=-116$
Since $D=-116 \ne0$, the matrix is invertible.
$A_{11}=\begin{vmatrix}
-3 & 2\\
9 & 11
\end{vmatrix}=-51$
Just by doing that, we will gradually find:
$A_{12}=-32$
$A_{13}=54$
$A_{21}=8$
$A_{22}=-20$
$A_{23}=12$
$A_{31}=31$
$A_{32}=24$
$A_{33}=-26$
$\rightarrow adj(A)=\begin{bmatrix}
-51 &8& 31\\
-32 & -20 & 24 \\
54& 12& -26
\end{bmatrix}$
Use the adjoint method to find $A^{-1}:$
$A^{-1}=\frac{1}{\det(A)}adj(A)$
$A^{-1}=\frac{1}{-116}\begin{bmatrix}
-51 &8& 31\\
-32 & -20 & 24 \\
54& 12& -26
\end{bmatrix}$
$A^{-1}=\begin{bmatrix}
\frac{51}{116} &\frac{2}{29} & \frac{31}{116} \\
\frac{-8}{29} & \frac{-5}{29} & \frac{6}{29} \\
\frac{27}{58} &\frac{3}{29}& \frac{-13}{58}
\end{bmatrix}$
