<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] Investigation with Maple indicates that for $n>1$, the inverse of $A$ has the following form:
(1) the main diagonal starts with $2$, finishes with $\pm1$, has zeros in between;
(2) the next diagonal each way has $1$ and $-1$ alternating;
(3) everything more than "one step away" from the main diagonal is $0$.
For example
$$\pmatrix{1&1&1&1\cr 1&2&2&2\cr 1&2&1&1\cr 1&2&1&2\cr}^{-1}
  =\pmatrix{2&-1&0&0\cr -1&0&1&0\cr 0&1&0&-1\cr 0&0&-1&1\cr}$$
and
$$\pmatrix{1&1&1&1&1\cr 1&2&2&2&2\cr 1&2&1&1&1\cr 1&2&1&2&2\cr 1&2&1&2&1\cr}^{-1}
  =\pmatrix{2&-1&0&0&0\cr -1&0&1&0&0\cr 0&1&0&-1&0\cr 0&0&-1&0&1\cr 0&0&0&1&-1\cr}\ .$$
I can't see a simple proof that this is true for all $n$, but if it is then the number of zeros is $n^2-2n$, provided $n>1$.
</s>