The system of equations is the following:
$$
\Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j,
$$
where $\Gamma = (\Gamma_i)$ is a vector  of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.


So there is a [paper](https://doi.org/10/gf5dw5) (_Numerical and computational aspects of cosmo-based activity coefficient models_, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

**I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?**


I don't know it it helps, but here is how the matrix $A$ is calculated:
$$
A = B \circ D
$$
Where $B$ is symmetric, dense with **only** positive (<s>and negative entries</s>) and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So:
$$
A_{ij}>0
$$