Answer
The solution is $(0,0,0,0)$
Work Step by Step
$Ax=0$
$\begin{bmatrix}
2&1 \\
3 &4
\end{bmatrix}.\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
2&1 |0\\
3 &4 |0
\end{bmatrix} \approx^1
\begin{bmatrix}
1&-\frac{1}{2} |0\\
3 &4 |0
\end{bmatrix} \approx^2 \begin{bmatrix}
1&-\frac{1}{2} |0\\
0 &\frac{11}{2} |0
\end{bmatrix} \approx^3 \begin{bmatrix}
1&-\frac{1}{2} |0\\
0 &1 |0
\end{bmatrix} \approx^4 \begin{bmatrix}
1&0|0\\
0 &1 |0
\end{bmatrix}$
This matrix is now in row-echelon form.
1. $M_1(\frac{1}{2})$
2. $A_{12}(-3)$
3. $M_2(\frac{2}{11})$
4. $A_{21}(\frac{1}{2})$
The solution set is:
$x_1=0$
$x_2=0$
Hence, the solution is $(0,0,0,0)$
