Answer
Solution set:$\quad\{(-3,4,-2)\}$
Work Step by Step
Start with the augmented matrix.
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
4 & -2 & -8 & | & -4\\
3 & 4 & 2 & | & 3
\end{array}\right]$
Swapping R1 and R3 brings a $-1$ into the first diagonal position,
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
4 & -2 & -8 & | & -4\\
3 & 4 & 2 & | & 3
\end{array}\right]\left\{\begin{array}{l}
.\\
-4R1+R2\rightarrow R2.\\
-3R1+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
0 & -6 & -4 & | & -16\\
0 & 1 & 5 & | & -6
\end{array}\right]\left\{\begin{array}{l}
.\\
R2\leftrightarrow R3.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
0 & 1 & 5 & | & -6\\
0 & -6 & -4 & | & -16
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
6R2+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
0 & 1 & 5 & | & -6\\
0 & 0 & 26 & | & -52
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\div 26
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & 3\\
0 & 1 & 5 & | & -6\\
0 & 0 & 1 & | & -2
\end{array}\right]$
$z=-2,$
Back-substitute into row 2:
$y+5(-2)=-6$
$y=-6+10=4$
Back-substitute into row $1$:
$x+4+2=3$
$x=3-6$
$x=-3$
Solution set:$\quad\{(-3,4,-2)\}$
