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

Answer: \(A^{-1}=\frac{1}{10}\begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix}\)

Let \(A=\begin{bmatrix}1 & -1 & 1\\ 2 & 1 & -3\\ 1 & 1 & 1\end{bmatrix}\).

\(|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10\ne 0\), so \(A^{-1}\) exists.

Cofactors:

\(A_{11}=(1+3)=4,\ A_{12}=-(2+3)=-5,\ A_{13}=(2-1)=1\)

\(A_{21}=-(-1-1)=2,\ A_{22}=(1-1)=0,\ A_{23}=-(1+1)=-2\)

\(A_{31}=(3-1)=2,\ A_{32}=-(-3-2)=5,\ A_{33}=(1+2)=3\)

\(\text{adj}A=\begin{bmatrix}4 & -5 & 1\\ 2 & 0 & -2\\ 2 & 5 & 3\end{bmatrix}^T=\begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix}\)

\(A^{-1}=\frac{1}{|A|}\text{adj}A=\frac{1}{10}\begin{bmatrix}4 & 2 & 2\\ -5 & 0 & 5\\ 1 & -2 & 3\end{bmatrix}\)

Check: \(A\cdot\text{adj}A=10I\) (e.g. first row \(\times\) first column: \(4+5+1=10\)).