Find a basis for $\displaystyle \mathbb{R}^3$ which contains a basis of $\displaystyle im(A)$, where

$\displaystyle A=\begin{pmatrix}1&2&3&4\\ 2&-4&6&-2\\-1 & 2& -3 & 1 \end{pmatrix}$

After row reduction

$\displaystyle U\sim\begin{pmatrix}1&0&3&1.5\\ 0&1&0&1.25\\0 & 0& 0 & 0 \end{pmatrix}$

So I've found the basis for the image of A

$\displaystyle \textrm{basis(im(A))}=\left\{ \begin{pmatrix}1\\2\\-1\end{pmatrix}, \begin{pmatrix}2\\-4\\2\end{pmatrix}\right\}$

How would I find the basis for $\displaystyle \mathbb{R}^3$ which contains the image of A

EDIT: Ok so I've just seen the way my textbook does it, and they row reduce the matrix A alongside the identity matrix and then choose the columns from A and the identity matrix (which are the pivot columns in the reduced form).

Why does this method work?

$\displaystyle (A|I)=\begin{pmatrix}1&2&3&4&1&0&0\\ 2&-4&6&-2&0&1&0\\-1 & 2& -3 & 1&0&0&1 \end{pmatrix}$