Answer
The solution is $(2s-3t,s,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&-4 &6 |0\\
3&-6 &9 |0\\
1&-2 &3 | 0 \\
5 &-10 &15|0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&-4 &6 |0\\
3&-6 &9 |0\\
1&-2 &3 | 0 \\
5 &-10 &15|0
\end{bmatrix}\approx^1
\begin{bmatrix}
1&-2 &3 |0\\
3&-6 &9 |0\\
2&-4 &6 | 0 \\
5 &-10 &15|0
\end{bmatrix} \approx^2
\begin{bmatrix}
1&-2 &3 |0\\
0&0 &0|0\\
0&0 &0 | 0 \\
0&0 &0|0
\end{bmatrix} $
$1.M_1(\frac{1}{2})$
$2.A_{12}(-3),A_{23}(-2),A_{24}(-5)$
This matrix is now in row-echelon form.
The solution set is:
$x_1-2x_2+3x_3=0$
$x_2=s$
$x_3=t$
There is one free variable, which we take to be $x_2=s,x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_1-2s+3t=0 \rightarrow x_1=2s-3t$
$x_2=s$
$x_3=t$
Hence, the solution is $(2s-3t,s,t)$
