Answer
$\{(2,0,-1)\}$.
Work Step by Step
The given system of equations is
$3x+2y+3z=3$
$4x-5y+7z=1$
$2x+3y-2z=6$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
3 & 2 & 3& 3\\
4 &-5 & 7& 1 \\
2&3&-2&6
\end{array}\right]$
Perform $R_1\Rightarrow R_1/(3)$
$\Rightarrow \left[\begin{array}{ccc|c}
3/(3) & 2/(3) & 3/(3)& 3/(3)\\
4 &-5 & 7& 1 \\
2&3&-2&6
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2/3 & 1& 1\\
4 &-5 & 7& 1 \\
2&3&-2&6
\end{array}\right]$
Perform $R_2\rightarrow R_2-4 R_1$ and $R_3\rightarrow R_3-2 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2/3 & 1& 1\\
4-4(1) &-5-4(2/3) & 7-4(1)& 1-4(1) \\
2-2(1)&3-2(2/3)&-2-2(1)&6-2(1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2/3 & 1& 1\\
0 &-23/3 &3& -3 \\
0&5/3&-4&4
\end{array}\right]$
Perform $R_2\Rightarrow R_2(-3/23)$
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2/3 & 1& 1\\
0(-3/23) &-23/3(-3/23) &3(-3/23)& -3(-3/23) \\
0&5/3&-4&4
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 2/3 & 1& 1\\
0 &1 &-9/23& 9/23 \\
0&5/3&-4&4
\end{array}\right]$
Perform $R_1\rightarrow R_1-(2/3) R_2$ and $R_3\rightarrow R_3-(5/3) R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(2/3)(0) & 2/3-(2/3)(1) & 1-(2/3)(-9/23)& 1-(2/3)(9/23)\\
0 &1 &-9/23& 9/23 \\
0-(5/3)(0)&5/3-(5/3)(1)&-4-(5/3)(-9/23)&4-(5/3)(9/23)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 29/23& 17/23\\
0 &1 &-9/23& 9/23 \\
0&0&-77/23&77/23
\end{array}\right]$
Perform $R_3\rightarrow R_3(-23/77)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 29/23& 17/23\\
0 &1 &-9/23& 9/23 \\
0(-23/77)&0(-23/77)&-77/23(-23/77)&77/23(-23/77)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 29/23& 17/23\\
0 &1 &-9/23& 9/23 \\
0&0&1&-1
\end{array}\right]$
Perform $R_1\rightarrow R_1-(29/23)R_3$ and $R_2\rightarrow R_2+(9/23) R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-(29/23)(0) &0-(29/23)(0) & 29/23-(29/23)(1)& 17/23-(29/23)(-1)\\
0+(9/23)(0) &1+(9/23)(0) &-9/23+(9/23)(1)& 9/23+(9/23)(-1) \\
0&0&1&-1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 &0 & 0& 2\\
0 &1 &0& 0 \\
0&0&1&-1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=2$.
and
$\Rightarrow y=0$.
and
$\Rightarrow z=-1$.
The solution set is $\{(x,y,z)\}=\{(2,0,-1)\}$.
