Answer
Solution set: $\{(3,-1,-1)\}$
Work Step by Step
Begin with the system's augmented matrix.
$\left[\begin{array}{lllll}
1 & 3 & 0 & | & 0\\
2 & 1 & 1 & | & 1\\
3 & -1 & -1 & | & 11
\end{array}\right]$
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and Os below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
1 & 3 & 0 & | & 0\\
2 & 1 & 1 & | & 1\\
3 & -1 & -1 & | & 11
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R1+R2\rightarrow R2\\
-3R1+R3\rightarrow R3
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 3 & 0 & | & 0\\
0 & -2 & 1 & | & 1\\
0 & -10 & -1 & | & 11
\end{array}\right]\left\{\begin{array}{l}
.\\
\times(-\frac{1}{2})\\
-5R2+R3\rightarrow R3
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 3 & 0 & | & 0\\
0 & 1 & -1/2 & | & -1/2\\
0 & 0 & -6 & | & 6
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow-6z=6
\end{array}\right.$
$z=-1$
Back substitute $z=-1$ into row 2,
$y-\displaystyle \frac{1}{2}(-1)=-\frac{1}{2}$
$y=-\displaystyle \frac{1}{2}-\frac{1}{2}=-1$
Back substitute into equation 1:
$x+3(-1)=0$
$x=3$
Solution set: $\{(3,-1,-1)\}$
