Answer
The solution is $(-\frac{1}{2}s+\frac{1}{2}t,\frac{1}{2}s-\frac{3}{2}t,s,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
4&2 &-1 & -1 |0\\
3&1 &-2 &3 |0\\
5 &-1 &-2 & 1|0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
4&2 &-1 & -1 |0\\
3&1 &-2 &3 |0\\
5 &-1 &-2 & 1|0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&-3 &1 & -4 |0\\
3&1 &-2 &3 |0\\
5 &-1 &-2 & 1|0
\end{bmatrix} \approx^2
\approx^3 \begin{bmatrix}
1&-3 &1 & -4 |0\\
0&10&-5 &15 |0\\
0 &14&-7 & 21|0
\end{bmatrix} \approx^4 \begin{bmatrix}
1&-3 &1 & -4 |0\\
0&2&-1 &3 |0\\
0&2&-1 &3|0
\end{bmatrix}\approx^5\begin{bmatrix}
1&-3 &1 & -4 |0\\
0&1&-\frac{1}{2} &\frac{3}{2} |0\\
0&0&0 &0|0
\end{bmatrix} $
This matrix is now in row-echelon form.
The solution set is:
$x_1-3x_2+x_3-4x_4=0$
$x_2-\frac{1}{2}x_3+\frac{3}{2}x_4=0$
$x_3=s$
$x_4=t$
There is one free variable, which we take to be $x_3=s,x_4 = t$, where s and t can assume any complex value. Applying back substitution yields:
$x_1-3(\frac{1}{2}s-\frac{3}{2}t)+s-4t=0 \rightarrow x_1=-\frac{1}{2}s+\frac{1}{2}t$
$x_2-\frac{1}{2}s+\frac{3}{2}t=0 \rightarrow x_2=\frac{1}{2}s-\frac{3}{2}t$
$x_3=s$
$x_4=t$
Hence, the solution is $(-\frac{1}{2}s+\frac{1}{2}t,\frac{1}{2}s-\frac{3}{2}t,s,t)$
