Answer
$\{(1,2,3)\}$.
Work Step by Step
The given system of equations is
$x+y+z=6$
$x-z=-2$
$y+3z=11$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 6\\
1 &0 & -1& -2 \\
0&1&3&11
\end{array}\right]$
Perform $R_2\rightarrow R_2- R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 6\\
1-1 &0-1 & -1-1& -2-6 \\
0&1&3&11
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 6\\
0 &-1 & -2& -8 \\
0&1&3&11
\end{array}\right]$
Perform $R_2\Rightarrow R_2/(-1)$
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 6\\
0/(-1) &-1/(-1) & -2/(-1)& -8/(-1) \\
0&1&3&11
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 1& 6\\
0 &1 & 2& 8 \\
0&1&3&11
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3- R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & 1-2& 6-8\\
0 &1 & 2& 8 \\
0-0&1-1&3-2&11-8
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -1& -2\\
0 &1 & 2& 8 \\
0&0&1&3
\end{array}\right]$
Perform $R_1\rightarrow R_1+R_3$ and $R_2\rightarrow R_2-2 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+0 & 0+0 & -1+1& -2+3\\
0-2(0) &1-2(0) & 2-2(1)& 8-2(3) \\
0&0&1&3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0 &1 & 0& 2 \\
0&0&1&3
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=1$.
and
$\Rightarrow y=2$.
and
$\Rightarrow z=3$.
The solution set is $\{(x,y,z)\}=\{(1,2,3)\}$.
