Suppose we have two matrices $A$ and $B$ (both have size $n\times n$), where both $A$ and $B$ are positive definite.
We want to determine if every element of $A^{-1}$ is greater than the corresponding element of $B^{-1}$. However, we cannot directly calculate $A^{-1}$ and $B^{-1}$.
Of course, the comparison result will depend on the specific forms of $A$ and $B$. But I want to know if one can tell if $A^{-1}$ is element-wise greater than $B^{-1}$ by just observing $A$ and $B$. Or in other words, what properties should $A$ and $B$ have so that $A^{-1}$ is element-wise greater than $B^{-1}$?
The best I could come up with is the following.
With the well-known representation of the inverse matrix in terms of the adjungate matrix, we find that
\begin{align} &&(A^{-1})_{ij} &> (B^{-1})_{ij} \\ \Longleftrightarrow \qquad &&\frac1{\det(A)}(\def\adj{\mathrm{adj}}\adj(A))_{ij} &>\frac1{\det(B)}(\adj(B))_{ij} \\ \Longleftrightarrow \qquad &&(-1)^{i+j}\frac{\det(A^{(ji)})}{\det(A)} &> (-1)^{i+j}\frac{\det(B^{(ji)})}{\det(B)} \end{align}
where $M^{(ij)}$ denotes the matrix $M$ after deletion of row $i$ and column $j$. So $A^{-1}$ is component-wise greater than $B^{-1}$ if and only if for all $i,j=1,...,n$
$$\frac{\det(A^{(ij)})}{\det(A)} \;\begin{matrix}\color{red}>\\\color{blue}<\end{matrix}\; \frac{\det(B^{(ij)})}{\det(B)}\qquad\text{if $i+j$ is}\; \begin{matrix}\color{red}{\text{even}}\\\color{blue}{\text{odd}}\end{matrix}.$$
Whether this is a helpful criterion for you will depend on the source of your matrices $A$ and $B$.