Answer
The solution is $(-t,-3t,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&-1 &-1 |0\\
5& -1 & 2 |0\\
1 &1 &4 | 0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
1&1 &4 |0\\
5 &-1&2 | 0\\
2& -1 & -1 |0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&1 &4 |0\\
0& -6 & -18 |0\\
0&-3&-9
\end{bmatrix} \approx^2
\begin{bmatrix}
1&1 &4 |0\\
0& -6& -18 |0\\
0&-3&-9
\end{bmatrix}
\approx^3
\begin{bmatrix}
1&1 &4 |0\\
0&1 & 3 |0\\
0&-3&-9|0
\end{bmatrix}
\approx^4 \begin{bmatrix}
1&0 &1 |0\\
0& 1 & 3 |0\\
0&0&0
\end{bmatrix}$
$1.P_{13}(-1)$
$2.A_{12}(-5),A_{13}(-2)$
$3.M_2(-\frac{1}{6})$
$4.A_{21}(-1),A_{23}(3)$
This matrix is now in row-echelon form.
The solution set is:
$x_1+x_3=0$
$x_2+3x_3=0$
$x_3=1$
There is one free variable, which we take to be $x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_1+t=0 \rightarrow x_1=-t$
$x_2+3t=0 \rightarrow x_2=-3t$
$x_3=1$
Hence, the solution is $(-t,-3t,t)$
