Let $A$ be the matrix $$A=\begin{pmatrix}1 & 0 & 0 & 1 & 1 & 0 \\ 0& 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1\end{pmatrix}$$ in the field $\mathbb{F}_q$. Let $M_q[A]$ be the matroid associated to $A$. Prove that if $q=2$ then $M_q[A]$ is graphic, but it's not if $q=3$.
My (primitive) attempt
Clearly the rank of $A$ is $3$. And for $q=2$ clearly a set of $3$ columns is a basis iff their sum isn't equal to $0$ or one of the three columns. This allows me to write quite easily the list of basis of $M_2[A]$ so I have the list of the maximal subtrees of the graph. It's not clear to me how to construct a graph given its family of maximal subtrees.
For $q=3$ I have no clue on how to proceed...