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