If
[tex]\mathbf A=\begin{bmatrix}6&-12\\-2&4\end{bmatrix}[/tex]
then notice that the columns satisfy
[tex]\begin{bmatrix}6\\-2\end{bmatrix}=-\dfrac12\begin{bmatrix}-12\\4\end{bmatrix}\implies\begin{bmatrix}6\\-2\end{bmatrix}+\dfrac12\begin{bmatrix}-12\\4\end{bmatrix}=\mathbf 0[/tex]
which means the columns are linearly dependent and thus only span a subspace of [tex]\mathbb R^2[/tex].
Whether you actually meant to write
[tex]\mathbf A=\begin{bmatrix}6&-2\\-12&4\end{bmatrix}[/tex]
would not alter the answer - the columns do not span [tex]\mathbb R^2[/tex] - but this time
[tex]\begin{bmatrix}6\\-12\end{bmatrix}=-\dfrac13\begin{bmatrix}-2\\4\end{bmatrix}[/tex]
