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