= Solution
Denote the five rows by $r_1,\ldots,r_5$. They satisfy
$$
r_1=r_2+r_4,\qquad r_5=r_2+r_4-r_3.
$$
Thus $r_2,r_3,r_4$ span the <row space>. They are <linearly independent>: in a <linear combination> $\alpha r_2+\beta r_3+\gamma r_4=0$, coordinate 1 gives $\alpha=0$, coordinate 2 then gives $\beta=0$, and coordinate 4 gives $\gamma=0$. Therefore
$$
\boxed{\mathcal B_{\rm row}=\{(1,1,0,0),(0,1,1,0),(0,0,1,1)\},\qquad\operatorname{rank}A=3.}
$$
As an alternative <basis>, the nonzero rows of the <reduced row echelon form> are
$$
\operatorname{rref}(A)=\begin{pmatrix}
1&0&0&1\\0&1&0&-1\\0&0&1&1\\0&0&0&0\\0&0&0&0
\end{pmatrix}.
$$
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