Here is the question I am trying to solve (note that in the textbook the columns are numbered from 1 to 6):
Let $A$ be the matrix $\begin{pmatrix} 1 & 0 & 0 & 1 & 1 & 0\\ 0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}$ For $q$ in $\{2,3\},$ let $M_q[A]$ be the vector matroid of $A$ when $A$ is viewed over $GF(q),$ the field of $q$ elements. Show that:
$(a)$ The sets of circuits of $M_2[A]$ and $M_3[A]$ are different.
Here are my thoughts:
Since a minimal dependent set in an arbitrary $M$ is called a circuit of $M.$ Then I need to find all the independent sets with respect to both $GF(2)$ and $GF(3)$ and then the rest will be the set of dependent sets of this matroid and from which I can determine the set of minimal dependent sets.
My problem is:
what is the quickest way of finding the set of all linearly independent sets over GF(2) and GF(3)? should I use gaussian elimination but how?
My guess for the solution is since the columns $4,5$ and $6$ are linearly independent with respect to $GF(3)$ while they are linearly dependent with respect to $GF(2)$ then the set of circuits of $M_2[A]$ is different from the set of circuits of $M_3[A].$ Is my answer correct? Is it stated verbally correctly?
Here is an answer to it I found on the internet (but I still not sure if it is correct or not)
**Can someone verify if this answer is correct, or no? I do not see how the gauss elimination leads to these conclusions. **