= Solution
Swap the first two rows, then perform $R_2\gets R_2+R_1$ and $R_3\gets R_3-2R_1$. These <elementary row operations> give
$$
A\sim\begin{pmatrix}1&-2&2&3&-1\\0&0&5&10&-10\\0&0&1&2&-2\end{pmatrix}.
$$
Scale row 2 by $1/5$, subtract it from row 3, and finally subtract twice row 2 from row 1. The <reduced row echelon form> is
$$
R=\begin{pmatrix}1&-2&0&-1&3\\0&0&1&2&-2\\0&0&0&0&0\end{pmatrix}.
$$
<Elementary row operations> preserve the <row space>. The two nonzero rows of $R$ are <linearly independent>, since their leading entries lie in different columns. Hence
$$
\boxed{\text{row-space basis }\{(1,-2,0,-1,3),(0,0,1,2,-2)\},\qquad\operatorname{rank}A=2.}
$$
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