Answer
$\{(-3,0,1)\}$.
Work Step by Step
The given system of equations is
$3y-z=-1$
$x+5y-z=-4$
$-3x+6y+2z=11$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
0 & 3 & -1& -1\\
1 &5 & -1& -4 \\
-3&6&2&11
\end{array}\right]$
Perform $R_1\leftrightarrow R_2$.
Swap the 2nd and the 1st rows.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &5 & -1& -4 \\
0 & 3 & -1& -1\\
-3&6&2&11
\end{array}\right]$
Perform $R_3\rightarrow R_3+3( R_1)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &5 & -1& -4 \\
0 & 3 & -1& -1\\
-3+3(1)&6+3(5)&2+3(-1)&11+3(-4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &5 & -1& -4 \\
0 & 3 & -1& -1\\
0&21&-1&-1
\end{array}\right]$
Perform $R_2\Rightarrow R_2/(3)$
Change the sign of the second row and swap the 3rd and 2nd row.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &5 & -1& -4 \\
0/(3) & 3/(3) & -1/(3)& -1/(3)\\
0&21&-1&-1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &5 & -1& -4 \\
0 & 1 & -1/3& -1/3\\
0&21&-1&-1
\end{array}\right]$
Perform $R_1\rightarrow R_1-5 R_2$ and $R_3\rightarrow R_3-21 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-5(0) &5-5(1) & -1-5(-1/3)& -4-5(-1/3) \\
0 & 1 & -1/3& -1/3\\
0-21(0)&21-21(1)&-1-21(-1/3)&-1-21(-1/3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 2/3& -7/3 \\
0 & 1 & -1/3& -1/3\\
0&0&6&6
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{6}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 2/3& -7/3 \\
0 & 1 & -1/3& -1/3\\
0/(6)&0/(6)&6/(6)&6/(6)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 2/3& -7/3 \\
0 & 1 & -1/3& -1/3\\
0&0&1&1
\end{array}\right]$
Perform $R_1\rightarrow R_1-(2/3)R_3$ and $R_2\rightarrow R_2+(1/3) R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(2/3)(0) &0-(2/3)(0) & 2/3-(2/3)(1)& -7/3-(2/3)(1) \\
0+(1/3) (0) & 1+(1/3) (0) & -1/3+(1/3) (1)& -1/3+(1/3) (1)\\
0&0&1&1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 0& -3 \\
0 & 1 & & 0\\
0&0&1&1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-3$.
and
$\Rightarrow y=0$.
and
$\Rightarrow z=1$.
The solution set is $\{(x,y,z)\}=\{(-3,0,1)\}$.
