Answer
The solution is $(3t,2t,t)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
2&1 &-8 |0\\
3& -2 & -5 |0\\
5&-6 &-3 | 0 \\
3 &-5 &1
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
2&1 &-8 |0\\
3& -2 & -5 |0\\
5&-6 &-3 | 0 \\
3 &-5 &1
\end{bmatrix} \approx^1
\begin{bmatrix}
3& -2 & -5 |0\\
2&1 &-8 |0\\
5&-6 &-3 | 0 \\
3 &-5 &1
\end{bmatrix} \approx^2
\begin{bmatrix}
1& -3 & 3 |0\\
2&1 &-8 |0\\
5&-6 &-3 | 0 \\
3 &-5 &1
\end{bmatrix}
\approx^3
\begin{bmatrix}
1& -3 & 3 |0\\
0&7&-14 |0\\
0&9 &-18 | 0 \\
0 &4 &-8
\end{bmatrix}
\approx^4 \begin{bmatrix}
1& -3 & 3 |0\\
0&1&-2 |0\\
0&9 &-18 | 0 \\
0 &4 &-8
\end{bmatrix} \approx^5 \begin{bmatrix}
1& 0 & -3 |0\\
0&1&-2 |0\\
0&0 &0 | 0 \\
0 &0 &0|0
\end{bmatrix}$
$1.P_{12}$
$2.A_{21}(-1)$
$3.A_{12}(-2),A_{13}(-5),A_{14}(-3)$
$4.M_2(\frac{1}{7})$
$5.A_{21}(3),A_{23}(-9),A_{24}(-4)$
This matrix is now in row-echelon form.
The solution set is:
$x_1-3x_3=0$
$x_2-2x_3=0$
$x_3=t$
There is one free variable, which we take to be $x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_1-3t=0 \rightarrow x_1=t$
$x_2-2t=0 \rightarrow x_2=2t$
$x_3=t$
Hence, the solution is $(3t,2t,t)$
