Consider the matrix differential equation $$X'(t) = \beta M(t)X(t)$$
where $M$ is a non-inversible matrix and $M(t)$ does not commute with $$\exp \left(\int_a^t M(s)ds \right)$$ and, hence, there is no simple solution. I now $X_0(t)$ the solution of $M(t)X(t)=0$ that can be considered as the approximate solution when $\beta \gg 1$.
How can I re-use $X_0(t)$ to get a second more precise approximation of $X(t)$ for big $\beta$ ?
The formal solution to your problem is $$ X(t) = T \exp\left(\beta \int_0^t M(s) ds\right) X(0) $$ where $T$ is the time-ordering operator.
(I do not see any simple way how to construct a series expansion in $1/\beta$ as already the first order would contain an equation as this. )