Answer
$\{(-4,2,-1)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
3x& -y &+z&=&-15\\
x& +2y & -z&=&1\\
2x& +3y &-2z &=&0
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
3 & -1 & 1& -15\\
1 & 2 & -1& 1 \\
2&3&-2&0
\end{array}\right]$
Perform $R_1\leftrightarrow R_2$
Swap row one and row two.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2 & -1& 1 \\
3 & -1 & 1& -15\\
2&3&-2&0
\end{array}\right]$
Perform $R_2\rightarrow R_2-3\times R_1$ and $R_3\rightarrow R_3-2 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2 & -1& 1 \\
3-3(1) & -1-3(2) & 1-3(-1)& -15-3(1)\\
2-2(1)&3-2(2)&-2-2(-1)&0-2(1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2 & -1& 1 \\
0 & -7 & 4& -18\\
0&-1&0&-2
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{-7}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2 & -1& 1 \\
0/(-7) & -7 /(-7)& 4/(-7)& -18/(-7)\\
0&-1&0&-2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2 & -1& 1 \\
0 & 1& -4/7& 18/7\\
0&-1&0&-2
\end{array}\right]$
Perform $R_1\rightarrow R_1-2R_2$ and $R_3\rightarrow R_3+ R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-2(0) & 2-2(1) & -1-2(-4/7)& 1-2(18/7) \\
0 & 1& -4/7& 18/7\\
0+0&-1+1&0-4/7&-2+18/7
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 1/7& -29/7 \\
0 & 1& -4/7& 18/7\\
0&0&-4/7&4/7
\end{array}\right]$
Perform $R_3\rightarrow R_3(-7/4)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 1/7& -29/7 \\
0 & 1& -4/7& 18/7\\
0(-7/4)&0(-7/4)&-4/7(-7/4)&4/7(-7/4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 1/7& -29/7 \\
0 & 1& -4/7& 18/7\\
0&0&1&-1
\end{array}\right]$
Perform $R_1\rightarrow R_1-(1/7) R_3$ and $R_2\rightarrow R_2+(4/7) R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(1/7)(0) & 0-(1/7)(0) & 1/7-(1/7)(1)& -29/7-(1/7)(-1) \\
0+(4/7)(0) & 1+(4/7)(0)& -4/7+(4/7)(1)& 18/7+(4/7)(-1)\\
0&0&1&-1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& -4 \\
0 & 1& 0& 2\\
0&0&1&-1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-4$
and
$\Rightarrow y=2$.
and
$\Rightarrow z=-1$.
The solution set is $\{(x,y,z)\}=\{(-4,2,-1)\}$.
