<s>[INST] Generator of a solution space In $\mathbb{F}_3$, consider:
$a+b+2c+d=0$
$2a+c+d=0$
$2a+b+c=0$
I got $a=0$, $c=-d$ and $b=d$, so we have $\mathbb{L}=\left \{ t \begin{pmatrix} 0\\1\\-1 \\ 1 \end{pmatrix} : t \in \mathbb{F}_3 
  \right \}$. Does this mean that $(0,1,-1,1)^T$ is the generator of the solution space?
 [/INST] And easy way: form and then bring to echelon form the coefficients' matrix:
$$\begin{pmatrix}1&1&1&1\\
2&0&1&1\\
1&1&1&0\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&1&1\\
0&2&0&0\\
0&0&0&2\end{pmatrix}\implies\begin{cases}d=0=b\\{}\\a=-c=2c\end{cases}$$
so the general solution is
$$\left\{\;\begin{pmatrix}2c\\0\\c\\0\end{pmatrix}\;\right\}=t\begin{pmatrix}2\\0\\1\\0\end{pmatrix}\,,\,\,t\in\Bbb F_3=\text{Span}_{\Bbb F_3}\left\{\;\begin{pmatrix}2\\0\\1\\0\end{pmatrix}\;\right\}$$
</s>