Answer
$\{(3,0,-4)\}$.
Work Step by Step
The given system of equations is
$x+3y=3$
$y+2z=-8$
$x-z=7$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3 & 0& 3\\
0 & 1 & 2& -8 \\
1&0&-1&7
\end{array}\right]$
Perform $R_3\rightarrow R_3- R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3 & 0& 3\\
0 & 1 & 2& -8 \\
1-1&0-3&-1-0&7-3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3 & 0& 3\\
0 & 1 & 2& -8 \\
0&-3&-1&4
\end{array}\right]$
Perform $R_1\rightarrow R_1- 3R_2$ and $R_3\rightarrow R_3+3 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 -3(0)& 3 -3(1) & 0 -3(2)& 3 -3(-8)\\
0 & 1 & 2& -8 \\
0 +3(0)&-3+3(1)&-1+3(2)&4+3(-8)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -6& 27\\
0 & 1 & 2& -8 \\
0 &0&5&-20
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{5}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -6& 27\\
0 & 1 & 2& -8 \\
0/(5) &0/(5) &5/(5) &-20/(5)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -6& 27\\
0 & 1 & 2& -8 \\
0 &0 &1 &-4
\end{array}\right]$
Perform $R_1\rightarrow R_1+6 R_3$ and $R_2\rightarrow R_2-2 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+6(0) & 0+6(0) & -6+6(1) & 27+6(-4) \\
0-2(0) & 1-2(0) & 2-2(1)& -8-2(-4) \\
0 &0 &1 &-4
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 &3 \\
0 & 1 & 0& 0 \\
0 &0 &1 &-4
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=3$
and
$\Rightarrow y=0$.
and
$\Rightarrow z=-4$.
The solution set is $\{(x,y,z)\}=\{(3,0,-4)\}$.
