= Solution
The <reduced row echelon form> yields
$$
x_1-2x_2-x_4+3x_5=0,\qquad x_3+2x_4-2x_5=0.
$$
Take the <free variables in a linear system> to be $s=x_2$, $t=x_4$, $u=x_5$. Then
$$
x=s\begin{pmatrix}2\\1\\0\\0\\0\end{pmatrix}+t\begin{pmatrix}1\\0\\-2\\1\\0\end{pmatrix}+u\begin{pmatrix}-3\\0\\2\\0\\1\end{pmatrix}.
$$
Thus
$$
\boxed{\text{null-space basis }\{(2,1,0,0,0)^{\mathsf T},(1,0,-2,1,0)^{\mathsf T},(-3,0,2,0,1)^{\mathsf T}\},\qquad\dim\operatorname{Null}(A)=3.}
$$
Each listed <vector> lies in the <null space>, and every solution has the displayed form. Their coordinates in positions 2, 4 and 5 prove <linear independence>. The dimension also follows from the <rank-nullity theorem>: $5-2=3$.
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