Best way to compute the rank of $A$ Let 
$$A=\begin{bmatrix}
 1&2&3&4&5\\6&7&8&9&10\\11&12&13&14&15\\16&17&18&19&20\\21&22&23&24&25\end{bmatrix}.$$
Which would be the best way to compute its rank?
I first thought about computing the determinant but then it seemed better to find its echelon form which would give me the rank. Is there a more efficient way?
I found out that the last three rows were spanned by the first two (So the rank is 2) but it was kind of a lucky thing so I can't expect to get it every time like that. 

You can notice that if $a=[1\ 2\ 3\ 4 \ 5]$ and $u=[1\ 1\ 1\ 1\ 1]$, then the matrix is
$$
\begin{bmatrix}
a \\
a + 5u \\
a + 10u \\
a + 15u \\
a + 20u
\end{bmatrix}
$$
and it is clear that the row space is generated by $a$ and $u$.
