Answer
The solution is $(2+\frac{1}{2}s-\frac{1}{2}t,s,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
6& -3 &3 |12\\
2& -1 &1| 4\\
-4 & 2 & -2|8
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
6& -3 &3 |12\\
2& -1 &1| 4\\
-4 & 2 & -2|8
\end{bmatrix} \approx^1\begin{bmatrix}
1& -\frac{1}{2} & -\frac{1}{2} |2\\
2& -1 &1| 4\\
-4 & 2 & -2|-8
\end{bmatrix} \approx^2\begin{bmatrix}
1& -\frac{1}{2} & -\frac{1}{2} |2\\
0& 0 &0| 0\\
0 & 0 & 0|0
\end{bmatrix}$
The equivalent system is:
$x_1-\frac{1}{2} x_2+\frac{1}{2} x_3=2$
$x_2=s$
$x_3 = t$
There are two free variables, which we take to be $x_2=s,x_3 = t$, where t and s can assume any complex value. Applying back substitution yields:
$x_1-\frac{1}{2}s+\frac{1}{2}t=2 \rightarrow x_1=2+\frac{1}{2}s-\frac{1}{2}t$
$x_2=s$
$x_3 = t$
The solution is $(2+\frac{1}{2}s-\frac{1}{2}t,s,t)$
