sum of all entries of the inverse matrix of a positive definite, symmetric matrix Let $A$ be a positive definite, symmetric $n\times n$ matrix such that each entry $a_{i,j}$ of $A$ is a number in $[0,1]$ and $\text{diag}[A]=(1,1,\cdots,1)$. Does the sum of all entries of $A^{-1}$ have a finite bound? That is,
$$
\text{sup}_{A}\sum_{1\leq i,j\leq n}\text{entry}_{i,j}[A^{-1}]<\infty?
$$
What is the exact value of the supremum?
 A: When $n=2$, if we denote the common value of the two off-diagonal entries of $A$ by $x\in(0,1)$, the sum of entries of $A^{-1}$ will be the decreasing function $\frac{2}{1+x}$ and hence it is bounded above by $2$.
When $n\ge3$, the sum is unbounded above. Let $t>0$ and
$$
A_t=\pmatrix{1&\frac12&\frac{\sqrt{3}}2\\ \frac12&1&\frac{\sqrt{3}}2-t\\ \frac{\sqrt{3}}2&\frac{\sqrt{3}}2-t&1}.
$$
Then $\det(A_t)=t\left(\frac{\sqrt{3}}2-t\right)$. By Sylvester's criterion, $A_t$ is positive definite whenever $t\in(0,\frac{\sqrt{3}}2)$. Also, straightforward calculation shows that the sum of all its entries of $A_t^{-1}$ is given by
$$
S_t=\frac1{\det(A_t)}\left(\frac74-\sqrt{3}+O(t)\right).
$$
(The big-oh term is $t-t^2$, but that is unimportant here. There is no need to calculate $A^{-1}$. One merely needs to use cofactor expansion to calculate the constant terms in the adjugate of $A_t$ and sum them up to obtain the constant term in $S_t$). Therefore $S_t\to+\infty$ when $t\to0^+$. It follows that when $n\ge3$ and $t\in(0,\frac{\sqrt{3}}2)$, the matrix defined by $A=A_t\oplus I_{n-3}$ is positive definite and entrywise between $0$ and $1$, but the sum of all entries in $A^{-1}$ diverges to $+\infty$ when $t\to0^+$.
