Perturbation of linear system of equations: Is the solution still non-negative? Let $A = (a_{ij})_{i,j=1,\dots,n}$ be a matrix such that 


*

*$a_{ij} \ge 0$ for all $i,j = 1, \dots, n,$ and

*$A$ is positive definite.


Let $I$ be the identity matrix, and $\pmb{1}$ the vector containing only ones.
Suppose that the solution $x = (x_i)_{i=1,\dots,n}$ to the system of linear equations
$$
(A+I)x=\pmb{1}
$$
is non-negative, i.e. $x_i \ge 0$ for $i=1,\dots,n$.
Define $f(t) = \frac{e^t + 1}{e^t - 1}.$
Is the solution $x$ to the system of linear equations
$$
(A+f(t)I)x=\pmb{1}
$$
non-negative for every $t>0$?
 A: I found a proof under the additional assumption that $(A+f(t)I)^{-1}$ is a $Z$-matrix.
It works for arbitary vectors $x$, not just the vector containing only ones.
Let me write $M \ge 0$ ($M > 0$) if every entry of the matrix $M$ is non-negative (positive).
Pick $t > 0$. 
First of all, $A+I$ and $A+f(t)I$ are positive definite and therefore invertible.
Also, $f(t) > 1$.
Define $B := (A+f(t) I)^{-1} (A+ I).$
If we can show $B \ge 0,$ then $B^{-1}$ is a monotone matrix, and
$$
B^{-1} (A+f(t)I)^{-1} x = (A+I)^{-1} (A+f(t)I)(A+f(t)I)^{-1}x=(A+I)^{-1}x \ge 0
$$
implies $(A+f(t)I)^{-1}x \ge 0$ as desired.
Writing
$$
B = I - (f(t) - 1)(A+f(t)I)^{-1}
$$
shows that all off-diagonal entries of $B$ are non-negative:
All off-diagonal entries of the $Z$-matrix $(A+f(t)I)^{-1}$ are non-positive, and $f(t)-1 > 0.$
The matrices $(A + f(t)I) - (f(t)-1)I = A + I$ and $(f(t)-1)I$ are both positive definite.
Hence 
$$
\Big( \frac1{f(t)-1}I - (A+f(t)I)^{-1} \Big) = (f(t)-1)) B
$$
is also positive definite (see 1.).
A positive definite matrix has non-negative entries on its diagonal (see 5.).
Since $f(t) - 1 > 0,$ the entries on $B$'s main diagonal must also be non-negative.
