find a basis using gaussian elimination

Find a basis of the following subspace U of $\Re^4$

U:= span $\{ \begin{pmatrix} 1 \\ -1 \\ 3 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 1 \\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\ 5 \\ -7 \\ 0 \end{pmatrix}, \begin{pmatrix} 4 \\ -1 \\ 7 \\ 7 \end{pmatrix} \}$

I've tried doing this using Gaussian elimination using the following row operations:

$\begin{pmatrix} 1 & -1 & 3 & 2 \\2 & 1 & 1 & 3 \\1 & 5 & -7 & 0 \\4 & -1 & 7 & 7\end{pmatrix}$

$\begin{pmatrix} 1 & -1 & 3 & 2 \\0 & 3 & -5 & -1 \\0 & 6 & -10 & -2 \\0 & -3 & 5 & 1\end{pmatrix}$

r2 - 2r1
r3 - r1
r4 - 2r2

$\begin{pmatrix} 1 & -1 & 3 & 2 \\0 & 3 & -5 & -1 \\0 & 0 & 0 & 0 \\0 & -3 & 5 & 1\end{pmatrix}$

r3 - 2r2

So am I right in thinking I can stop now since there is a row of all 0's, and therefore a basis is $\begin{pmatrix} 1 \\ -1 \\ 3 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 3 \\ -5 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ -3 \\ 5 \\ 1 \end{pmatrix}$?

thanks in advance!

Thanks very much James, I did this and got to $\begin{pmatrix} 1 & 0 & 4/3 & 5/3 \\0 & 1 & -5/3 & -1/3 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{pmatrix}$

So is the basis just the first 2 rows, or should the third row be included as well?

I would think it should just be the first 2 rows...

sorry to draw up an old thread.. but my question is of a similar sort

if a set of vectors forms a basis of R4 say, does this mean that the rre form of their matrix would be the 4 x 4 identity matrix?

thanks