Solve the following equations by the method of inversion :\(2x-y+z=1\)\(x+2y+3z=8\)\(3x+y-4z=1\)

Answer: \(x=1,\ y=2,\ z=1\)

Matrix form \(AX=B\), where \(A=\begin{bmatrix}2 & -1 & 1\\ 1 & 2 & 3\\ 3 & 1 & -4\end{bmatrix},\ X=\begin{bmatrix}x\\ y\\ z\end{bmatrix},\ B=\begin{bmatrix}1\\ 8\\ 1\end{bmatrix}\).

\(|A|=2(-8-3)+1(-4-9)+1(1-6)=-22-13-5=-40\ne0\), so \(A^{-1}\) exists.

Cofactors:

\(A_{11}=-11,\ A_{12}=-(-4-9)=13,\ A_{13}=1-6=-5\)

\(A_{21}=-(4-1)=-3,\ A_{22}=-8-3=-11,\ A_{23}=-(2+3)=-5\)

\(A_{31}=-3-2=-5,\ A_{32}=-(6-1)=-5,\ A_{33}=4+1=5\)

\(\text{adj}A=\begin{bmatrix}-11 & -3 & -5\\ 13 & -11 & -5\\ -5 & -5 & 5\end{bmatrix}\), so \(A^{-1}=-\frac{1}{40}\begin{bmatrix}-11 & -3 & -5\\ 13 & -11 & -5\\ -5 & -5 & 5\end{bmatrix}\).

\(X=A^{-1}B=-\frac{1}{40}\begin{bmatrix}-11-24-5\\ 13-88-5\\ -5-40+5\end{bmatrix}=-\frac{1}{40}\begin{bmatrix}-40\\ -80\\ -40\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 1\end{bmatrix}\)

Hence \(x=1,\ y=2,\ z=1\). (Check: \(2-2+1=1\), \(1+4+3=8\), \(3+2-4=1\).)