If \(A=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\) then find \(A^{-1}\) by Adjoint method.

Answer: \(A^{-1}=\dfrac14\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\)

\(|A|=2(4-1)+1(-2+1)+1(1-2)=6-1-1=4\ne0\), so \(A^{-1}\) exists.

Cofactors \(A_{ij}=(-1)^{i+j}M_{ij}\):

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

\(A_{21}=-(-2+1)=1\), \(A_{22}=4-1=3\), \(A_{23}=-(-2+1)=1\)

\(A_{31}=1-2=-1\), \(A_{32}=-(-2+1)=1\), \(A_{33}=4-1=3\)

Cofactor matrix \(=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\); its transpose is the same, so \(\operatorname{adj}A=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\).

\(A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A=\dfrac14\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\).

(Check: \(A\cdot\operatorname{adj}A=4I\).)