Let \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\) and \(B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]\) such that \(\mathrm{AX}=\mathrm{B}\), then \(\mathrm{X}=\)

\(\begin{aligned}
& \mathrm{AX}=\mathrm{B} \\
& \Rightarrow\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]
\end{aligned}\)
Applying \(\mathrm{R}_2 \rightarrow \mathrm{R}_2-2 \mathrm{R}_1\) and \(\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1\),
\(\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 3 & -5 \\
0 & 2 & 0
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{c}
4 \\
-8 \\
-2
\end{array}\right]\)
\(\begin{array}{ll}
\therefore \quad & x_1-x_2+x_3=4 ...(i)\\
& 3 x_2-5 x_3=-8 ...(ii)\\
\therefore & 2 x_2=-2 \Rightarrow x_2=-1
\end{array}\)
From (ii),
\(3(-1)-5 x_3=-8 \Rightarrow x_3=1\)
From (i),
\(\begin{aligned}
& x_1+1+1=4 \Rightarrow x_1=2 \\
\therefore \quad & {\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right] }
\end{aligned}\)