Let $A$ be an $n\times n$, symmetric, invertible matrix with real positive elements. Show that $z_n\leq n^2-2n$, where $z_n$ is the number of zero elements in $A^{-1}$.
Let us prove this by contradiction. Let us suppose that $z_n\geq n^2-2n+1$. Then by the Pigeonhole principle, there is at least one column with $\lceil{\frac{n^2-2n+1}{n}\rceil}$ elements, which turns out to be $n-1$. Hence, there is at least one column, say the $i^{th}$ column, with all elements except one as $0$. Let the $j^{th}$ row in the $i^{th}$ column be non-zero. Then the cofactor of any $a_{jk}$, where $k\neq i$, is a matrix with one column full of $0's$. Hence, $(A^{-1})^{-1}=A$ should also have a $0$ element, which is a contradiction.
Does this proof work? I have not used the fact that $A$ is symmetric anywhere.