Find the inverse of matrix \(\begin{bmatrix}1 & 2 & 3\\1 & 1 & 5\\2 & 4 & 7\end{bmatrix}\) by adjoint method.

Answer: \(A^{-1}=\begin{bmatrix}13 & 2 & -7\\-3 & -1 & 2\\-2 & 0 & 1\end{bmatrix}\)

\(|A|=1(7-20)-2(7-10)+3(4-2)=-13+6+6=-1\ne 0\), so \(A^{-1}\) exists.

Cofactors:
\(A_{11}=7-20=-13,\ A_{12}=-(7-10)=3,\ A_{13}=4-2=2\)
\(A_{21}=-(14-12)=-2,\ A_{22}=7-6=1,\ A_{23}=-(4-4)=0\)
\(A_{31}=10-3=7,\ A_{32}=-(5-3)=-2,\ A_{33}=1-2=-1\)

\(\operatorname{adj}A=\begin{bmatrix}-13 & -2 & 7\\3 & 1 & -2\\2 & 0 & -1\end{bmatrix}\) (transpose of the cofactor matrix).

\(A^{-1}=\frac{1}{|A|}\operatorname{adj}A=-1\cdot\operatorname{adj}A=\begin{bmatrix}13 & 2 & -7\\-3 & -1 & 2\\-2 & 0 & 1\end{bmatrix}\).