Determining the dimension and a basis for a vector space I have the following problem:

Let $W$ be a vector space of all solutions to these homogenous equations:
$$\begin{matrix} x &+& 2y &+& 2z &-& s &+& 3t &=& 0 \\
    x &+& 2y &+& 3z &+& s &+& t &=& 0 \\
    3x &+& 6y &+& 8z &+& s &+& 5t &=& 0\end{matrix}$$
Find the dimension of the space $W$ and determine a basis for $W$.

I tried solving the above kernel to get the solutions.
The matrix:
$$\left(\begin{matrix} 1 & 2 & 2 & -1 & 3 \\ 1 & 2 & 3 & 1 & 1 \\ 3 & 6 & 8 & 1 & 5\end{matrix}\right)$$
When performing Gauss-Jordan on it, I get the matrix rank to be $3$:
$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 4 & 10 & 0 & 0 & 0\end{matrix}\right)$$
So I get lost at this point. I don't know how to get the dimension nor how to determine a basis for it.
Can anyone point out the next thing I should do and whether I started off good?
 A: Subtracting the first row $4$ times from the third, we get the matrix
$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 0 & 10 & 4 & 0 & 0\end{matrix}\right)$$
Subtracting the second row $2$ times from the third, we get the matrix
$$\left(\begin{matrix} 1 & 0 & -1 & 0 & 0 \\ 0 & 5 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{matrix}\right).$$
This means the original set of equations is equivalent to
$$\begin{align} x - z &= 0 \\ 5y + 2z &= 0.\end{align}$$
So for any $\lambda, \mu, \rho$ we get $z = \lambda$, $s = \mu$, $r = \rho$, $x = \lambda$ and $y = \frac{-2}{5} \lambda$ as a solution. In particular:


*

*$(\lambda, \mu, \rho) = (1,0,0)$ gives the solution $(x,y,z,s,t) = (1, \frac{-2}{5}, 1, 0, 0)$.

*$(\lambda, \mu, \rho) = (0,1,0)$ gives the solution $(x,y,z,s,t) = (0,0,0,1,0)$.

*$(\lambda, \mu, \rho) = (0,0,1)$ gives the solution $(x,y,z,s,t) = (0,0,0,0,1)$.


This means that $W$ is spanned by the three vectors:
$$W = \left\langle \left(\begin{matrix} 1 \\ -2/5 \\ 1 \\ 0 \\ 0\end{matrix}\right), \left(\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \\ 0\end{matrix}\right), \left(\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 1\end{matrix}\right)\right\rangle.$$
So there are three basis vectors. Can you then determine the dimension of $W$?
A: After clearing the bottom row, beneath the pivots $1$ and $5$, you'll have all zeros there. Thus you can take any $z,s,t$ and then $x$ and $y$ will be determined by $z$. Thus your basis is $\{e_1-\tfrac25e_2+e_3,e_4,e_5\}$.
