= Solution
The reduced <row echelon form> gives
$$
x_1+x_3+x_4=0,\qquad x_2-x_3+x_4=0.
$$
Choose the free variables $x_3=s,x_4=t$. Then
$$
x=s\begin{pmatrix}-1\\1\\1\\0\end{pmatrix}+t\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix}.
$$
The coefficient <vectors> are <linearly independent>, because their third and fourth entries give independent coordinates. Consequently
$$
\boxed{\operatorname{Null}(A)\text{ has basis }\left\{\begin{pmatrix}-1\\1\\1\\0\end{pmatrix},\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix}\right\},\qquad\dim\operatorname{Null}(A)=2.}
$$
Substitution in the original <matrix> verifies $Ax=0$ for both <basis vectors>. The <rank-nullity theorem> supplies the independent dimension check $2+2=4$.
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