Answer
The solution is $(1-2r+3s-t,r,2-4s+3t,s,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
1 &2& 1 &1 & -2 |3\\
0&0 1 &&4 &-3| 2\\
2& 4 & -1 &-10 & 5|0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
1 &2& 1 &1 & -2 |3\\
0&0 1 &&4 &-3| 2\\
2& 4 & -1 &-10 & 5|0
\end{bmatrix} \approx^1\begin{bmatrix}
1 &2& 1 &1 & -2 |3\\
0&0 &1 &4 &-3| 2\\
0& 0 & -3 &-12 & 9|-6
\end{bmatrix} \approx^2 \begin{bmatrix}
1 &2& 1 &1 & -2 |3\\
0&0 &1 &4 &-3| 2\\
0& 0 & 0 &0 & 0|0
\end{bmatrix}$
$x_1+2x_2+x_3+x_4-2x_5=3$
$x_3+4x_4-3x_5=2$
$x_4=s$
$ x_5 = t$
There are three free variables, which we take to be $x_2=r,x_4=s, x_5 = t$, where s and t can assume any complex values. Applying back substitution yields:
$x_2=r$
$x_4=s$
$ x_5 = t$
$x_3+4s-3t=2 \rightarrow x_3=2-4s+3t$
$x_1+2r+2-4s+3t+s-2t=3 \rightarrow x_1=1-2r+3s-t$
The solution is $(1-2r+3s-t,r,2-4s+3t,s,t)$
