If \(\mathrm{AX}=\mathrm{B}\), where \(\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right], \mathrm{B}=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]\), then \(2 x+y-z=0\)

\(\begin{aligned} & {\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 2 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]} \\ & \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \text { and } \mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1 \\ & {\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}4 \\ -8 \\ -2\end{array}\right]}\end{aligned}\)
\(\therefore x-y+z=4\quad\ldots(1)\)
\(3 y-5 z=-8\quad\ldots(2)\)
\(2 y=-2\quad\ldots(3)\)
\(\therefore y=-1\quad \ldots[\operatorname{from}(3)] \)
\( \therefore -3-5 z=-8 \Rightarrow z=1\quad \ldots[\operatorname{from}(2)] \)
\( \therefore x-(-1)+1=4 \Rightarrow x=2\quad \ldots[\operatorname{from}(1)] \)
\( \therefore 2 x+y-z=2(2)-1-1=2\)