For the system \(x-y+z=4,2 x+y-3 z=0\), \(x+y+z=2\), the values of \(x, y, z\) respectively are given by

Given equation in matrix form
\(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
It is of the form \(\mathrm{AX}=\mathrm{B}\)
Now, \(|A|=\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right|=10 \neq 0\)
\(\begin{aligned} \therefore \quad & A^{-1} \text { exist } \\ & \operatorname{adj} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right] \\ A^{-1} & =\frac{1}{|A|} \text { adj } A \\ & =\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\end{aligned}\)
\(\begin{array}{ll} & \text {Now, } A X=B \\ \therefore \quad & A^{-1}(A X)=A^{-1} B \\ \therefore \quad & X=A^{-1} B\end{array}\)
\(\begin{aligned} \therefore \quad X & =\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right] \\ & =\frac{1}{10}\left[\begin{array}{c}16+0+4 \\ -20+0+10 \\ 4+0+6\end{array}\right] \\ & \therefore=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right] \\ \therefore \quad & x=2, y=-1, z=1\end{aligned}\)