Find a reduced echelon basis from a reduced echelon matrix. The reduced row matrix was this ---> $\begin{pmatrix}1&2&0&1&0\\0&0&1&3&0\\0&0&0&0&1\\0&0&0&0&0&\end{pmatrix} = 0$
So i computed the basis to be such that 
$\begin{pmatrix}x\\y\\z\\t\\u\end{pmatrix} = y\begin{pmatrix}-2\\1\\0\\0\\0\end{pmatrix} + t\begin{pmatrix}-1\\0\\-3\\1\\0\end{pmatrix}$
So there is a basis $(f_1, f_2)$ where  $f_1$ = $(-2, 1, 0, 0, 0)$ and $f_2$ = $(-1, 0, -3, 1, 0)$. I did it up to there but in addition there is a note saying that a reduced echelon basis is therefore $(−f_2, 2f_2 + f_1)$. Where does this come from? How did they come about this reduced echelon basis?

I suppose they simply made a rref from this:
$$
\begin{pmatrix}
 -1 & 0 &-3 & 1 & 0 \\
 -2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 0 & 3 &-1 & 0 \\
 -2 & 1 & 0 & 0 & 0 \\
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 0 & 3 &-1 & 0 \\
  0 & 1 & 6 &-2 & 0 \\
\end{pmatrix}
$$
i.e.
$$
\begin{pmatrix}
 f_2 \\
 f_1 \\
\end{pmatrix}\sim
\begin{pmatrix}
-f_2 \\
 f_1 \\
\end{pmatrix}\sim
\begin{pmatrix}
-f_2 \\
 f_1-2f_2 \\
\end{pmatrix}
$$
