Bounding the inverse of a diagonally dominant matrix entry-wise I have a $d \times d$ matrix $A$ whose entries are bounded (C1): $I - \epsilon X \preceq A \preceq I + \epsilon X$, where $I$ is the identity matrix and $X = 11^\top - I$ is  the matrix with ones everywhere but the diagonal. Furthermore, $\epsilon < 1/d$, so $A$ is guaranteed to be diagonally dominant and thus non-singular.


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*Firstly, is the following true (C2): $(I + \epsilon X)^{-1} \preceq A^{-1} \preceq (I - \epsilon X)^{-1}$?

*If not, does there exist an $\epsilon'$ such that $c(I - \epsilon' X) \preceq A^{-1} \preceq c(I + \epsilon' X)$ for some constant, $c$?


Empirically, it seems like the first statement is true: I generated a few 1,000 matrices $A$ satisfying the bound (C1) and observed that (C2) holds always. Assuming that the first statement is true, I have been able to prove the second statement for an $\epsilon' \approx \epsilon/(1 - \epsilon d)$.
Note that $A \preceq B$ if is $B - A$ contains only positive entries; it is not the Loewner ordering.

Let $A\leq B$ be true iff $B-A$ contains only non-negative entries (with your definition $I-\epsilon X\leq A\leq I+\epsilon X$ is impossible). You have a big problem with the diagonals of the matrices.
Clearly 1. is false. 
 Take $d=2$ and $A=\begin{pmatrix}1&a\\b&1\end{pmatrix}$ where $a,b\in (-\epsilon,\epsilon)$. Then $(I+\epsilon X)^{-1}\leq A^{-1}\leq (I-\epsilon X)^{-1}$ implies $1/(1-\epsilon^2)\leq 1/(1-ab)\leq1(1-\epsilon^2)$ and then $ab=\epsilon^2$, that is contradictory.
Clearly 2. is false. Indeed $c(I-\alpha X)\leq A^{-1}\leq c(I+\alpha X)$ implies $c\leq 1/(1-ab)\leq c$, that is contradictory ($ab$ goes through $(-\epsilon^2,\epsilon^2)$).
