Answer
Solution set: $\{ ( -1$ , $2$ , $-2 ) \}$
Work Step by Step
Augmented matrix:
$\left[\begin{array}{lllll}
3 & -1 & -4 & | & 3\\
2 & -1 & 2 & | & -8\\
1 & 2 & -3 & | & 9
\end{array}\right]\left\{\begin{array}{l}
\leftrightarrow R_{3}.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -3 & | & 9 \\
2 & -1 & 2 & | & -8\\
3 & -1 & -4 & | & 3
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow R_{2}-2R_{1}\\
\leftarrow R_{3}-3R_{1}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -3 & | & 9 \\
0 & -5 & 8 & | & -26\\
0 & -7 & 5 & | & -24
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow(-3R_{2}+2R_{3})\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -3 & | & 9 \\
0 & 1 & -24 & | & 30\\
0 & -7 & 5 & | & -24
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow(7R_{2}+R_{3}).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -3 & | & 9 \\
0 & 1 & -24 & | & 30\\
0 & 0 & -93 & | & 186
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=-2.
\end{array}\right.$
Back substitute into row/equation 2,
$y-14(-2)=-52$
$y=30-28=2$
Back substitute into row/equation 1:
$x+2(2)-3(-2)=9$
$x=9-10=-1$
Solution set: $\{ ( -1$ , $2$ , $-2 ) \}$
