Answer
Solution set: $\{ ( 1$ , $1$ , $2 ) \}$
Work Step by Step
Rewrite the system in standard form.
$\left\{\begin{array}{l}
2x+y-z=1\\
2x-3y+z=1\\
x+y+z=4
\end{array}\right.$
Augmented matrix:
$\left[\begin{array}{lllll}
2 & 1 & -1 & | & 1\\
2 & -3 & 1 & | & 1\\
1 & 1 & 1 & | & 4
\end{array}\right]\left\{\begin{array}{l}
\leftrightarrow R_{3}.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & 1 & | & 4\\
2 & -3 & 1 & | & 1\\
2 & 1 & -1 & | & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow R_{2}-2R_{1}\\
\leftarrow R_{3}-2R_{1}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & 1 & | & 4\\
0 & -5 & -1 & | & -7 \\
0 & -1 & -3 & | & -7
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftrightarrow(-R_{3}).\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & 1 & | & 4\\
0 & 1 & 3 & | & 7\\
0 & -5 & -1 & | & -7
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow R_{3}+5R_{2}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & 1 & | & 4\\
0 & 1 & 3 & | & 7\\
0 & 0 & 14 & | & 28
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\div 14.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & 1 & | & 4\\
0 & 1 & 3 & | & 7\\
0 & 0 & 1 & | & 2
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=2.
\end{array}\right.$
Back substitute into row/equation 2,
$y+3(2)=7$
$y=7-6=1$
Back substitute into row/equation 1:
$x+(1)+(2)=4$
$x=4-3=1$
Solution set: $\{ ( 1$ , $1$ , $2 ) \}$
