Answer
The solution is $(0,0,0)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1& 2 &3\\
2 & -1 & 0\\
1 &1 &1
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1& 2 &3 | 0\\
2 & -1 & 0 |0\\
1 &1 &1 |0
\end{bmatrix} \approx^1 \begin{bmatrix}
1& 2 &3 | 0\\
0& -5 & -6 |0\\
0 &-1 &-2 |0
\end{bmatrix} \approx^2\begin{bmatrix}
1& 2 &3 | 0\\
0 &-1 &-2 |0\\
0& -5 & -6 |0
\end{bmatrix} \approx^3 \begin{bmatrix}
1& 2 &3 | 0\\
0 &1 &2 |0\\
0& -5 & -6 |0
\end{bmatrix} \approx^4 \begin{bmatrix}
1& 0 &-1 | 0\\
0 &1 &2 |0\\
0& 0 &4 |0
\end{bmatrix} \approx^5 \begin{bmatrix}
1& 0 &-1 | 0\\
0 &1 &2 |0\\
0& 0 &1 |0
\end{bmatrix} \approx^6 \begin{bmatrix}
1& 0 &0 | 0\\
0 &1 &0 |0\\
0& 0 &1 |0
\end{bmatrix}$
This matrix is now in row-echelon form.
$1. A_{12}(-2),A_{13}(-1)$
$2.P_{23}$
$3.M_2(-1)$
$4. A_{21}(-2),A_{23}(5)$
$5.M_{3}(\frac{1}{4})$
$6. A_{31}(1),A_{32}(-2)$
The solution set is:
$x_1=0$
$x_2=0$
$x_3=0$
Hence, the solution is $(0,0,0)$
