Answer
The solution is $(0,0,0)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1& 3&0\\
-2 & -3 & 0\\
1 &4 &0
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1& 3&0|0\\
-2 & -3 & 0|0\\
1 &4 &0|0
\end{bmatrix} \approx^1\begin{bmatrix}
1& 3&0|0\\
0 & 3 & 0|0\\
0 &1 &0|0
\end{bmatrix} \approx^2\begin{bmatrix}
1& 3&0|0\\
0 & 1 & 0|0\\
0 &3 &0|0
\end{bmatrix} \approx^3 \begin{bmatrix}
1& 0&0|0\\
0 & 1 & 0|0\\
0 &0 &0|0
\end{bmatrix} $
This matrix is now in row-echelon form.
$1. A_{12}(2),A_{13}(-1)$
$2.P_{23}$
$3.A_{21}(-3),A_{23}(-3)$
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
Hence, the solution is $(0,0,0)$
