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
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.