maximum number of zero elements (entries) in $A^{-1}?$ Let $A $ be an ${n \times n}$ symmetric invertible matrix with real positive elements. Then, find  the maximum number of zero elements in $A^{-1}$ ?
 A: Consider, for different positive numbers $a,b$, the matrix
$$M=\begin{bmatrix}
a & a & a\\
a & a & b\\
a&b&b
\end{bmatrix}$$
If $C_{ij}$ denotes the cofactor of the entry in the $i$th row and the $j$th column, then $$C_{12}=C_{21}=C_{33}=0$$
Consequently, $A^{-1}$ has at-least three zeros.
More generally, if $$M=\begin{bmatrix}
a&a&\cdots&a\\
a&a&\cdots&b\\
\vdots\\
a&b&\cdots&b
\end{bmatrix}$$
then the inverse of $A$ has at-least $n^2-2n$ zeros.
Finally, we prove that there cannot be more than $n^2-2n$ zeros in $A^{-1}$. To prove that, observe that if there are more than $n^2-2n$ zeros, then there are at-least $n^2-2n+1=(n-1)^2$ zeros. So, there are at-most $n^2-(n-1)^2=2n-1$ non-zero entries in $A^{-1}$. Using the pigeonhole principle, there must be a row in $A^{-1}$ that has at-most $1$ non-zero entry. That is, there is a row that has either no non-zero entry or exactly one non-zero entry. In the former case, matrix $A^{-1}$ is singular, a clear contradiction. In the latter case, without loss of generality, let the only non-zero entry in a row (if any) be the $ij$th entry in $A^{-1}$. The cofactor of any other entry in the same column in $A^{-1}$ is zero. Since a zero cofactor in $A^{-1}$ means a zero entry in $A$, This means that an entry in $A$ is zero, a contradiction. Thus, the maximum number of zeros in $A^{-1}$ is $n^2-2n$.
For $n=3$, it is $3^2-2\times 3=9-6=3$.
