Answer
The solution is $(3t,-t,-t,t)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1& 1&1&-1\\
-1 &0 & -1 & 2\\
1 &3 &2 &2
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1& 1&1&-1|0\\
-1 &0 & -1 & 2|0\\
1 &3 &2 &2 |0
\end{bmatrix} \approx^1 \begin{bmatrix}
1& 1&1&-1|0\\
0 &1 & 0 & 1|0\\
0 &2 &3 &1 |0
\end{bmatrix} \approx^2\begin{bmatrix}
1& 0&1&-2|0\\
0 &1 & 0 & 1|0\\
0 &0&1 &1 |0
\end{bmatrix} \approx^3\begin{bmatrix}
1& 0&0&-3|0\\
0 &1 & 0 & 1|0\\
0 &0&1 &1 |0
\end{bmatrix} $
This matrix is now in row-echelon form.
$1. A_{12}(1),A_{13}(-1)$
$2. A_{21}(-1),A_{23}(-2)$
$3. A_{31}(-1)$
The solution set is:
$x_1-3x_4=0$
$x_2+x_4=0$
$x_3+x_4=0$
$x_4=t$
There is one free variable, which we take to be $x_4 = t$, where t can assume any complex value. Applying back substitution yields:
$x_1-3t=0 rightarrow x_3=3t$
$x_2+t=0 rightarrow x_3=-t$
$x_3+t=0 \rightarrow x_3=-t$
$x_4=t$
Hence, the solution is $(3t,-t,-t,t)$
