TITLE: Provig that the number $z_n$ of zero element of $A^{-1} $ satisfies $z_n\le n^2-2n$
QUESTION [5 upvotes]: Let A be a $n\times n$ matrix $n\ge 2$ symmetric invertible with real positive entries. 

Then show that the number $z_n$ of zero element of $A^{-1} $ satisfies
  $$z_n\le n^2-2n$$
  How many zero elements are in the inverse of the $n\times n$ matrix A given by 
  $$A=\begin{pmatrix} 1&1&1&1&\cdots&1
\\1&2&2&2&\cdots&2
\\1&2&1&1&\cdots&1\\
\cdots &\cdots&\cdots&\cdots&\cdots&\cdots
\\1&2&1&2&\cdots&\cdots\end{pmatrix}$$

The try to use the multiplication formula for matrices which given by as follow
$$(AB)_{ij} = \sum_{k=1}^{n}a_{ik}b_{kj}$$
Then taking $B= A^{-1} = (b)_{i,j}$ we have the system,
$$ (I_n)_{ij}=(AB)_{ij} = \sum_{k=1}^{n}a_{ik}b_{kj} =\delta_{ij}$$
where $\delta_{ij}$ is the Kronecker symbol. 

From here I don't how to get further does any one ha an idea? 

for the second question the number of zeros elements seems to be $n^2-2n$ which prove that the inequality $z_n\le n^2-2n$ is sharp. see here

REPLY [2 votes]: Suppose to the contrary that $z_n \geq n^2 - 2n + 1$. Suppose $A^{-1} = (b_{ij})$. Then by pigeon hole principle there is a column with at least $n-1$ zeros, without loss of generality let it be the first column. This means that there is an $m$ such that $b_{i1} = 0$ for all $i \neq m$. Then the $m1$-minor of $A^{-1}$ has a zero column hence has zero determinant, which implies $A$ has a zero entry, contradiction.