Deleting a line of $A \in \mathcal{M}_n(\mathbb{Z}/2\mathbb{Z})$ to get again a matrix with columns pairwise distinct. I have seen this exercise for a while but I don't have the slightest good idea, any hint is appreciated.

Problem. Let $ n \geqslant 2, A \in \mathcal{M}_n(\mathbb{Z}/2\mathbb{Z})$ 
  whose columns are pairwise distinct.
Show that there is a row $L$ of $A$ such that the matrix $B$, obtained from $A$ by removing $L$, also has pairwise distinct columns.

 A: Proof by contradiction. Denote by $X$ the set of columns of $A$.
Assume the statement is not true for $A$, i. e. for any row number $i$ there exist two distinct columns $x_i, y_i\in X$ such that they are equal in all rows except the $i$-th one. The vectors themselves must be distinct, thus they do differ  in $i$-th row and only in it. Let's denote this relation by $x_i \sim_i y_i$.
So there are exactly $n$ such relations (they do surely not coincide) and $m$ vectors participating in these relations, $m\le n$. This situation can be considered as an undirected graph: vectors are the vertices and relations are the edges of the graph.
And the number of edges $n$ is greater than or equal to the number of vertices $m$. Thus there is a nontrivial loop in this graph: $v_1 \sim_{i_1} v_2 \sim_{i_2} \ldots \sim_{i_{k-1}} v_{k-1} \sim_{i_k} v_1$, where $k>1$ is the order of the loop. Since all $i_1, \ldots, i_k$ are different (by construction), one can deduce that $v_1$ differs from $v_{k-1}$ in $(k-1)$ distinct rows. On the other hand, $v_{k-1} \sim_{i_k} v_1$, thus $v_1$ differs from $v_{k-1}$ in one row, also distinct from the previous $(k-1)$. This is a contradiction.
