Answer
The solution set is $(-5t,-1-2t,t)$
Work Step by Step
$Ax=b$
$\begin{bmatrix}
1 & 0 & 5\\
3 & -2 & 11\\
2 & -2 &6
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
0\\
2\\
2
\end{bmatrix}$
Converting the given system of equations to an augmented matrix and then using Gauss-Jordan elimination to determine the solution set to the given system.
$\begin{bmatrix}
1 & 0 & 5|0\\
3 & -2 & 11|2\\
2 & -2 &6|2
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & 0 & 5|0\\
0 & -2 & -4|2\\
0 & -2 &-4|2
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & 0 & 5|0\\
0 &1 & 2|-1\\
0 & -2 &-4|2
\end{bmatrix} \approx^3\begin{bmatrix}
1 & 0 & 5|0\\
0 &1 & 2|-1\\
0 & 0 &0|0
\end{bmatrix} $
1. $A_{12}(-3),A_{13}(-2)$
2. $M_2(\frac{-1}{2})$
5. $A_{23}(2)$
The equivalent system is
$x_1+5x_3=0$
$x_2+2x_3=-1$
$x_3 = t$
There is one free variable, which we take to be $x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_3 = t$
$x_2+2t=-1 \rightarrow x_2=-1-2t$
$x_1+5t=0 \rightarrow x_1=-5t$
The solution set is $(-5t,-1-2t,t)$
