Answer
$\{(3,-2,1)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& +3y &-z&=&-1\\
x& +2y & +3z&=&2\\
3x& +5y &-2z &=&-3
\end{matrix}\right.$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
2 & 3 & -1& -1\\
1 & 2 & 3& 2 \\
3&5&-2&-3
\end{array}\right]$
Perform $R_1\rightarrow \frac{R_1}{2}$.
$\Rightarrow \left[\begin{array}{ccc|c}
2/2 & 3/2 & -1/2& -1/2\\
1 & 2 & 3& 2 \\
3&5&-2&-3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3/2 & -1/2& -1/2\\
1 & 2 & 3& 2 \\
3&5&-2&-3
\end{array}\right]$
Perform $R_2\rightarrow R_2- R_1$ and $R_3\rightarrow R_3-3 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3/2 & -1/2& -1/2\\
1-1 & 2-3/2 & 3-(-1/2)& 2-(-1/2) \\
3-3(1)&5-3(3/2)&-2-3(-1/2)&-3-3(-1/2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3/2 & -1/2& -1/2\\
0 & 1/2 & 7/2& 5/2 \\
0&1/2&-1/2&-3/2
\end{array}\right]$
Perform $R_2\rightarrow R_2(2)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3/2 & -1/2& -1/2\\
0(2) & 1/2(2) & 7/2(2)& 5/2(2) \\
0&1/2&-1/2&-3/2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 3/2 & -1/2& -1/2\\
0 & 1 & 7& 5 \\
0&1/2&-1/2&-3/2
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2(3/2)$ and $R_3\rightarrow R_3- R_2(1/2)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(0)(3/2) & 3/2-(1)(3/2) & -1/2-(7)(3/2) & -1/2-(5)(3/2) \\
0 & 1 & 7& 5 \\
0-(0)(1/2)&1/2-(1)(1/2)&-1/2-(7)(1/2)&-3/2-(5)(1/2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0& -11 & -8 \\
0 & 1 & 7& 5 \\
0&0&-4&-4
\end{array}\right]$
Perform $R_3\rightarrow R_3/(-4)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0& -11 & -8 \\
0 & 1 & 7& 5 \\
0/(-4)&0/(-4)&-4/(-4)&-4 /(-4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0& -11 & -8 \\
0 & 1 & 7& 5 \\
0&0&1&1
\end{array}\right]$
Perform $R_1\rightarrow R_1+11 R_3$ and $R_2\rightarrow R_2-7 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+11(0) & 0+11(0)& -11+11(1) & -8+11(1) \\
0-7(0) & 1-7(0) & 7-7(1)& 5-7(1) \\
0&0&1&1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0& 0 & 3 \\
0 & 1 & 0& -2 \\
0&0&1&1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=3$
and
$\Rightarrow y=-2$.
and
$\Rightarrow z=1$.
The solution set is $\{(x,y,z)\}=\{(3,-2,1)\}$.
