Answer
Solution set: $\{(-3,0,1)\}$
Work Step by Step
Begin with the system's augmented matrix.
$\left[\begin{array}{lllll}
0 & 3 & -1 & | & -1\\
1 & 5 & -1 & | & -4\\
-3 & 6 & 2 & | & 11
\end{array}\right]$
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).
$\left[\begin{array}{lllll}
0 & 3 & -1 & | & -1\\
1 & 5 & -1 & | & -4\\
-3 & 6 & 2 & | & 11
\end{array}\right]\left\{\begin{array}{l}
R1\leftrightarrow R2.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 5 & -1 & | & -4\\
0 & 3 & -1 & | & -1\\
-3 & 6 & 2 & | & 11
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
R3+3R1\rightarrow R3
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 5 & -1 & | & -4\\
0 & 3 & -1 & | & -1\\
0 & 21 & -1 & | & -1
\end{array}\right]\left\{\begin{array}{l}
.\\
\times\frac{1}{3}.\\
R3-7R2\rightarrow R3
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 5 & -1 & | & -4\\
0 & 1 & -1/3 & | & -1/3\\
0 & 0 & 6 & | & 6
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow 6z=6
\end{array}\right.$
$z=1$
Back substitute $z=1$ into row/equation 2,
$y-\displaystyle \frac{1}{3}(1)=-\frac{1}{3}$
$y=-\displaystyle \frac{1}{3}+\frac{1}{3}=0$
Back substitute into row/equation 1:
$x+5(0)-1(1)=-4$
$x=-4+1=-3$
Solution set: $\{(-3,0,1)\}$
