Answer
The solution is $(t-s,-3+s+t,s,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&-1 &3 &-1 |3\\
3 &2 &1 &-5| -6\\
1& -2 & 3 & 1 |6
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&-1 &3 &-1 |3\\
3 &2 &1 &-5| -6\\
1& -2 & 3 & 1 |6
\end{bmatrix} \approx^1\begin{bmatrix}
0&3 &-3 &-3 |-9\\
0 &8 &-8 &-8| -24\\
1& -2 & 3 & 1 |6
\end{bmatrix} \approx^2\begin{bmatrix}
1& -2 & 3 & 1 |6\\
0 &8 &-8 &-8| -24\\
0&3 &-3 &-3 |-9
\end{bmatrix}\approx^3 \begin{bmatrix}
1& -2 & 3 & 1 |6\\
0 &1 &-1 &-1| -3\\
0 &1 &-1 &-1| -3
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 0 & 1 & -1 |0\\
0 &1 &-1 &-1| -3\\
0 &0 &0 &0| 0
\end{bmatrix}$
$1.A_{31}(-2), A_{32}(-3)$
$2.P_{13}$
$3.M_2(\frac{1}{8}),M_3(\frac{1}{3})$
$4.A_{23}(1),A_{23}(2)$
This matrix is now in row-echelon form. The equivalent system is:
$x_1+x_3-x_4=0$
$x_2-x_3-x_4=-3$
$x_3=s$
$x_4=t$
There are two free variables, which we take to be $x_3 = s, x_4=t$, where s and t can assume any complex values. Applying back substitution yields:
$x_3=s$
$x_4=t$
$x_1+s-t=0 \rightarrow x_1=t-s$
$x_2-s-t=-3 \rightarrow x_2=-3+s+t$
The solution is $(t-s,-3+s+t,s,t)$
