I am trying to solve the problem as mentioned in the title. We can either take a general metric space $(M,\mathrm{d})$ and take $n$ points $\{x_1,\dots,x_n\}\subseteq M$ and define the matrix $$ A_{i,j}=\mathrm{d}(x_i,x_j) $$ or just consider a real valued matrix $B\in\mathrm{Mat}(\mathbb R^d,\mathbb R^d)$ such that \begin{align} B_{i,j}&\geq 0 \quad\forall i,j \\ B_{i,i}&=0 \quad \forall i \\ B_{i,j}&=B_{j,i} \quad \forall i,j \\ B_{i,j}&\leq B_{i,k}+B_{k,j} \quad \forall i,j,k. \\ \end{align}
It is trivial to solve this problem for 3 points, since the determinant of such a matrix would result in twice the product of the 3 elements in the upper diagonal, but already for 4 points it seems how positivity of the entries isn't enough and the other hypotheses should be used as well.
Using induction had me stuck as well since given the matrix $$A_d=\left( \begin{array}{ll} A_{d-1}&M_d\\ M_d^T& 0 \end{array} \right) $$ I would need to have $$ M_d A_{d-1}^{-1} M_d\neq0 $$ to get an invertible matrix, but I cannot evaluate it explicitly.
I have found some other questions on M.SE like here where OP never gets their question answered and people tag an article and another question where there is no explicit answer to the question.
I have tried searching for results in spectral graph theory to no avail, and most of the time the interest lies in the eigenvalues of the Laplacian matrix and not the distance or weight matrix.
Apparently it is trivial that the matrix has only one positive eigenvalue called Perron root due to its non-negativity but I couldn't find a reference. The fact that there is only one eigenvector with positive components on the other hand should be trivial but I didn't delve too much into it.
There are positive results for trees though. Given a tree with weights, then the distance matrix defined by shortest paths, is always invertible. This feels close to what I am searching but it is a very restrictive situation I believe.
So I am still with the question: Given a weight or distance matrix of a given graph or metric space, is it invertible?
