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