TITLE: New base of matroid from old
QUESTION [7 upvotes]: Let $M$ be a matroid of rank 3 and $E_1, E_2, E_3$ 3 basis of $M.$ Let $e_{i,j}$ be the $i$-th element of base $E_j$.

Is it true that you can always find a permutation $s: \{1,2,3\} \to \{1,2,3\}$ such that $e_{s(1),1}, e_{s(2),2}, e_{s(3),3}$ is also a base? 

I could not find a counter example.
I am also interested in arbitrary rank (it is obviously false for a matroid of rank 2).

REPLY [4 votes]: Let $M$ be the cycle matroid of the graph shown below.

Consider this grid:
1 2 3
2 3 4
1 2 5

Each row gives a basis for $M$.  However, no transversal gives a basis.