Answer
The solution is $(0,0,0)$
Work Step by Step
The augmented matrix of the system is:
$\begin{bmatrix}
3&3 &-1 |0\\
2 &1 &1 | 0\\
5& -4 & 1 |0
\end{bmatrix}$
with reduced row-echelon form:
$\begin{bmatrix}
3&3 &-1 |0\\
2 &1 &1 | 0\\
5& -4 & 1 |0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&1 &-2 |0\\
0 &-1 &5 | 0\\
0& -9 & 11 |0
\end{bmatrix} \approx^2
\begin{bmatrix}
1&1 &-2 |0\\
0 &1 &-5 | 0\\
0& -9 & 11 |0
\end{bmatrix}
\approx^3
\begin{bmatrix}
1&0 &3 |0\\
0 &1 &-5 | 0\\
0& 0 & -34 |0
\end{bmatrix}
\approx^4 \begin{bmatrix}
1&0 &3 |0\\
0 &1 &-5 | 0\\
0& 0 & 1|0
\end{bmatrix} \approx^5 \begin{bmatrix}
1&0 &0 |0\\
0 &1 &0 | 0\\
0& 0 & 1|0
\end{bmatrix}$
$1.A_{21}(-1), A_{12}(-2),A_{13}(-5)$
$2.M_2(-1)$
$3.A_{21}(-1),A_{23}(9)$
$4.M_4(-\frac{1}{34})$
$5.A_{31}(-3),A_{32}(5)$
This matrix is now in row-echelon form.
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
The solution is $(0,0,0)$
