= Solution
The <reduced row echelon form> gives
$$
x_1=-2x_2-\frac{17}{5}x_4,\qquad x_3=\frac35x_4.
$$
Taking the <free variables> as $x_2=s$ and $x_4=5t$, the <null space> is
$$
\ker A=\left\{s\begin{pmatrix}-2\\1\\0\\0\end{pmatrix}+t\begin{pmatrix}-17\\0\\3\\5\end{pmatrix}:s,t\in\mathbb R\right\}.
$$
The two displayed <vectors> are <linearly independent>, as their second and fourth coordinates show, and each satisfies $Ax=0$. Thus
$$
\boxed{\text{a null-space basis is }\{(-2,1,0,0)^{\mathsf T},(-17,0,3,5)^{\mathsf T}\},\quad\dim\ker A=2.}
$$
This also agrees with the <rank-nullity theorem>: $4-2=2$.
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