Answer
Solution set = $\{( 8 ,\ 2 ,\ 0 )\}$
Work Step by Step
Reduce the augmented matrix $[A|B]$ to reduced row echelon form and interpret the result
$\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
2 & 0 & -3 & | & 16\\
0 & 2 & 1 & | & 4
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}=r_{2}-2r_{1}\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
0 & 2 & -3 & | & 4\\
0 & 2 & 1 & | & 4
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}=r_{3}-r_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
0 & 2 & -3 & | & 4\\
0 & 0 & 4 & | & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}=\frac{1}{4}r_{3}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
0 & 2 & -3 & | & 4\\
0 & 0 & 1 & | & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}=r_{2}+3r_{1}\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
0 & 2 & 0 & | & 4\\
0 & 0 & 1 & | & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}=\frac{1}{2}r_{2}\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & 0 & | & 6\\
0 & 1 & 0 & | & 2\\
0 & 0 & 1 & | & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}=r_{1}+r_{2}.\\
.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & 0 & | & 8\\
0 & 1 & 0 & | & 2\\
0 & 0 & 1 & | & 0
\end{array}\right]$
The system is consistent and has a single solution.
$x=8,$
$y=2$
$z=0$
Solution set = $\{( 8 ,\ 2 ,\ 0 )\}$
