[SOLVED] Matrix problem, a bit clueless

Find the reduced row equivalent to $\displaystyle A$, $\displaystyle R$ where $\displaystyle A=\left( \begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 2 \\ 1 & 3 & 2 \end{array} \right)$. I won't show my calculus but I found $\displaystyle R$ to be $\displaystyle \left( \begin{array}{ccc} 1 & 0 & \frac{4}{5} \\ 0 & 1 & \frac{2}{5} \\ 0 & 0 & 0 \end{array} \right)$.

Now the problem states : Find all the $\displaystyle X$ $\displaystyle \in \mathbb{R}(3\times 1)$ such that $\displaystyle RX=0$. So I had the following system $\displaystyle \big[ \begin{array}{c} x_1+\frac{4}{5}x_3=0 \\ x_2+\frac{2}{5}x_3=0 \end{array}$ I think this implies infinity solutions (but I'm not sure) so I thought I made an error. So I restarted the calculus of $\displaystyle R$ without following what I've done for my first try and I found out that $\displaystyle R=\left( \begin{array}{ccc} 1 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & \frac{5}{2} & 1 \end{array} \right)$. Is that possible?

The problem says "the reduced row equivalent to $\displaystyle A$" so I made at least an error... And also if this calculus was right, I would have the same problem solving for $\displaystyle x_1$, $\displaystyle x_2$ and $\displaystyle x_3$. Please help me!