Answer
Solution set: $\{ ( 2$ , $0$ , $-1 ) \}$
Work Step by Step
$\left[\begin{array}{lllll}
3 & 2 & 3 & | & 3\\
4 & -5 & 7 & | & 1\\
2 & 3 & -2 & | & 6
\end{array}\right]\left\{\begin{array}{l}
\leftarrow R_{1}-R_{3}.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -1 & 5 & | & -3\\
4 & -5 & 7 & | & 1\\
2 & 3 & -2 & | & 6
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow(-4R_{1}+R_{2}).\\
\leftarrow(-2R_{1}+R_{3}).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -1 & 5 & | & -3\\
0 & -1 & -13 & | & 13\\
0 & 5 & -12 & | & 12
\end{array}\right]\left\{\begin{array}{l}
.\\
\times(-1).\\
\leftarrow(5R_{2}+R_{3}).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & -1 & 5 & | & -3\\
0 & 1 & 13 & | & -13\\
0 & 0 & -77 & | & 77
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=-1.
\end{array}\right.$
Back substitute into row/equation 2,
$y+13(-1)=-13$
$y=-13+13=0$
Back substitute into row/equation 1:
$x-(0)+5(-1)=-3$
$x=-3+5=2$
Solution set: $\{ ( 2$ , $0$ , $-1 ) \}$
