<s>[INST] How many zero elements are there in the inverse of the $n\times n$ matrix How many zero elements are there in the inverse of the $n\times n$ matrix
$A=\begin{bmatrix}
1&1&1&1&\cdots&1\\
1&2&2&2&\cdots&2\\
1&2&1&1&\cdots&1\\
1&2&1&2&\cdots&2\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&1&2&\cdots&\cdots
\end{bmatrix}$
My try: Denote by $a_{ij}$ and $b_{ij}$ the elements of $A$ and $A^{-1}$,respectively
then for $k\neq m$,we have
$$\sum_{i=0}^{n}a_{ki}b_{im}=0$$
Then I can't.Thank you very much
 [/INST] I am excavating this post because this post needs an affirmative answer.
After some try I feel that this can be (ironically) done fastest by brute force. Let $A^{-1} = (a_{ij})$ and we just solve column-wisely $AA^{-1} = I$.
For the first column, by subtracting equations derived from neighboring row of $A$, you will get:
$$
\begin{aligned}
a_{11} + a_{21} + a_{31} + a_{41} + \cdots &= 1\\
a_{21} + a_{31} + a_{41} + \cdots &= -1 \\
-a_{31}- a_{41} - \cdots &= 0 \\
a_{41} + \cdots &= 0\\
&\vdots
\end{aligned}
$$ 
You then have $a_{11} = 1$, $a_{21} = -1$ and the rest are 0.
From the second to the penultimate columns, $a_{ij}$ for $i < j$ can be derived directly from the fact that $A^{-1}$ is symmetric as $A$ is, for the rest, play the same "compare neighboring row game", you will get:
$$
\begin{aligned}
(-1)^ia_{ii} + (-1)^ia_{i+1,i} + (-1)^ia_{i+2,i} + \cdots &= 1\\
(-1)^{i+1}a_{i+1,i} + (-1)^{i+1}a_{i+2,i} + \cdots &= -1 \\
(-1)^{i+2}a_{i+2,i} \cdots &= 0 \\
&\vdots
\end{aligned}
$$ 
As the equations are easy enough, you will immediately see that $a_{i, i+1} = (-1)^i$ and $a_{i, i-1} = (-1)^{i-1}$.
For the last column, the only thing new is that $(-1)^na_{nn} = 1$, so $a_{nn} = (-1)^n$ and $a_{n, n-1} = (-1)^{n-1}$.
At the end you see that $A^{-1}$ has $2n$ non-zero entries, where exactly 2 of them are on each row and each column.
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