Answer
Solution set = $\{( -3 ,\ 2 ,\ 1 )\}$
Work Step by Step
Reduce the augmented matrix $[A|B]$ to reduced row echelon form and interpret the result
$\left[\begin{array}{lllll}
2 & 1 & 0 & | & -4\\
0 & -2 & 4 & | & 0\\
3 & 0 & -2 & | & -11
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}=r_{3}-r_{1}.\\
.\\
R_{3}=2r_{3}-3r_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & -1 & -2 & | & -7\\
0 & -2 & 4 & | & 0\\
0 & -3 & -4 & | & -10
\end{array}\right]\rightarrow\left(\begin{array}{l}
R_{1}=-0.5r_{2}+r_{1}.\\
R_{2}=-0.5r_{2}.\\
R_{3}=-1.5r_{2}+r_{3}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & -4 & | & -7\\
0 & 1 & -2 & | & 0\\
0 & 0 & -10 & | & -10
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}=-\frac{1}{10}r_{3}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & -4 & | & -7\\
0 & 1 & -2 & | & 0\\
0 & 0 & 1 & | & 1
\end{array}\right]\rightarrow\left(\begin{array}{l}
r_{1}=r_{1}+4r_{3}.\\
R_{2}=r_{2}+2r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & 0 & | & -3\\
0 & 1 & 0 & | & 2\\
0 & 0 & 1 & | & 1
\end{array}\right]$
The system is consistent and has a single solution.
$x=-3,$
$y=2$
$z=1$
Solution set = $\{( -3 ,\ 2 ,\ 1 )\}$