$M^{-1}$ 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 $\mathcal{M}_n(\mathbb{K})$
 A: Equivalently you need to prove that $M^{-1}$ has at least $2n$ nonzero entries. Let's in fact prove that every row of $M^{-1}$ had at least two nonzero entries. Suppose that the $i$th row of $M^{-1}$ has less than two nonzero entries; of course it cannot have a zero row (otherwise it would be singular), so it has exactly one nonzero entry. But then, in the equality $M^{-1}M = I$, we reach a contradiction, because in the $i$th row you would get:
$$M^{-1} M = \begin{pmatrix}
\dots &&&&&& \dots \\
0 & \dots & 0 & a_{i,j} & 0 & \dots & 0 \\
\dots &&&&&& \dots
\end{pmatrix}
\times
\begin{pmatrix}
\vdots & b_{j,1} & \vdots \\
\vdots & \vdots & \vdots \\
\vdots & b_{j,n} & \vdots
\end{pmatrix}
=
\begin{pmatrix}
\dots && \dots \\
a_{i,j}b_{j,1} & \dots & a_{i,j}b_{j,n} \\
\dots && \dots
\end{pmatrix}$$
And since all the entries of $M$ are nonzero, the $i$th row of $I$ would only have nonzero entries, which is absurd.
A: 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.
