Is my method for finding the Basis for the Row(A) correct? 
$$A=\begin{pmatrix}-1&0&2\\3&2&0\\0&1&3 \end{pmatrix}$$
Find a basis for $\text{Row} (A)$

$$Row(A)=\text{span} \left[\begin{pmatrix}-1\\0\\2\end{pmatrix},\begin{pmatrix}3\\2\\0\end{pmatrix},\begin{pmatrix}0\\1\\3\end{pmatrix}\right]$$
$$\begin{pmatrix}-1&3&0\\0&2&1\\2&0&3 \end{pmatrix}$$
Upon performing Reduced Row Echleon Form (RREF)
I get:
$$\begin{pmatrix}1&-3&0\\0&2&1\\0&0&0 \end{pmatrix}$$
The first and third column has a pivot so the basis must be the corresponding vectors so therefore:
$$\text{basis:  } Row(A)=\text{span} \left[\begin{pmatrix}-1\\0\\2\end{pmatrix},\begin{pmatrix}0\\1\\3\end{pmatrix}\right]$$
Is this flawless, and is there a faster method? Anyway I can improve my answer?
 A: Elementary row operations preserve the row space of $A$ (but change the column space). If you want to find a basis for $\mathrm{row}(A)$, perform elementary row operations on $A$ until you reach the reduced row echelon form. The non-zero rows of the reduced row echelon form will form a basis for $\mathrm{row}(A)$. In fact, you don't even have to perform row operations until you reach the reduced row echelon form - you can stop whenever it is clear what is the basis for $\mathrm{row}(A)$. 
In your example,
$$ \left( \begin{matrix} -1 & 0 & 2 \\ 3 & 2 & 0 \\ 0 & 1 & 3 \end{matrix} \right) \xrightarrow{R_1 = (-1) \cdot R_1} 
  \left( \begin{matrix} 1 & 0 & -2 \\ 3 & 2 & 0 \\ 0 & 1 & 3 \end{matrix} \right) \xrightarrow{R_2 = R_2 - 3R_1}
  \left( \begin{matrix} 1 & 0 & -2 \\ 0 & 2 & 6 \\ 0 & 1 & 3 \end{matrix} \right) $$
and from the last matrix it is already clear that
$$ \mathrm{row}(A) = \mathrm{span} \{ (1, 0, -2), (0, 1, 3) \}. $$
If you want to find the column space of $A$, you can do the same with column operations (which preserve the column space but change the row space).
