$M^{-1}$ has at most $n^2-2n$ coefficients equal to zero Let $M\in GL_n(\mathbb{R})$ such that all its coefficients are non zero.
How can one show that $M^{-1}$ has at most $n^2-2n$ coefficients equal to zero ?
I have no idea how to tackle that problem, I've tried drawing some contradiction if $M^{-1}$ had $n^2+1-2n$ zero coefficients but couldn't find any.

Notations :
$GL_n(\mathbb{K})$ : Set of invertible matrices in $\mathcal{M}_n(\mathbb{K})$

Hint: Show that if there are at least $n^2-2n+1$ zeros, there will be a column of $M^{-1}$ with exactly 1 nonzero entry. Convince yourself that without loss of generality this might as well be the first column. Then convince yourself that $M M^{-1}$ will have the first column a multiple of one of the columns of $A$. Finally, see that this means $M M^{-1}$ cannot be the identity matrix since all the entries of $M$ are nonzero.
