Let $L$ be the Laplacian of a directed weighted graph on $n$ nodes, e.g., for $n=4$: $$ L = \left(\begin{array}{cccc} w_{1,1}+w_{1,2}+w_{1,3}+w_{1,4} & -w_{1,2} & -w_{1,3} & -w_{1,4}\\ -w_{2,1} & w_{2,1}+w_{2,2}+w_{2,3}+w_{2,4} & -w_{2,3} & -w_{2,4}\\ -w_{3,1} & -w_{3,2} & w_{3,1}+w_{3,2}+w_{3,3}+w_{3,4} & -w_{3,4}\\ -w_{4,1} & -w_{4,2} & -w_{4,3} & w_{4,1}+w_{4,2}+w_{4,3}+w_{4,4} \end{array}\right) $$ I am interested in a closed-form in terms of the $w_{i,j}$ of the inverse of its $(n-1)\times (n-1)$ leading submatrix $M = L_{1:n-1, 1:n-1}$.
Kirchoff's matrix-tree theorem states that $\det M$ can be written as a sum of products of $w_{ij}$ with a certain combinatorial interpretation as trees, and we all know that the inverse matrix can be written as $M^{-1} = \frac{1}{\det M} \operatorname{adj}(M)$. Heree $\operatorname{adj}(M)$ is the adjugate matrix, whose entries can be expressed as $(n-2)\times(n-2)$ determinants times a sign. It seems that also these numerators are sums of products of the $w_{ij}$, for instance for $n=4$:
$$ \operatorname{adj}(M) = \left(\begin{array}{ccc} \left(w_{2,1}+w_{2,2}+w_{2,3}+w_{2,4}\right)\,\left(w_{3,1}+w_{3,2}+w_{3,3}+w_{3,4}\right)-w_{2,3}\,w_{3,2} & w_{1,2}\,w_{3,1}+w_{1,2}\,w_{3,2}+w_{1,2}\,w_{3,3}+w_{1,3}\,w_{3,2}+w_{1,2}\,w_{3,4} & w_{1,3}\,w_{2,1}+w_{1,2}\,w_{2,3}+w_{1,3}\,w_{2,2}+w_{1,3}\,w_{2,3}+w_{1,3}\,w_{2,4}\\ w_{2,1}\,w_{3,1}+w_{2,1}\,w_{3,2}+w_{2,1}\,w_{3,3}+w_{2,3}\,w_{3,1}+w_{2,1}\,w_{3,4} & \left(w_{1,1}+w_{1,2}+w_{1,3}+w_{1,4}\right)\,\left(w_{3,1}+w_{3,2}+w_{3,3}+w_{3,4}\right)-w_{1,3}\,w_{3,1} & w_{1,1}\,w_{2,3}+w_{1,3}\,w_{2,1}+w_{1,2}\,w_{2,3}+w_{1,3}\,w_{2,3}+w_{1,4}\,w_{2,3}\\ w_{2,1}\,w_{3,1}+w_{2,1}\,w_{3,2}+w_{2,2}\,w_{3,1}+w_{2,3}\,w_{3,1}+w_{2,4}\,w_{3,1} & w_{1,1}\,w_{3,2}+w_{1,2}\,w_{3,1}+w_{1,2}\,w_{3,2}+w_{1,3}\,w_{3,2}+w_{1,4}\,w_{3,2} & \left(w_{1,1}+w_{1,2}+w_{1,3}+w_{1,4}\right)\,\left(w_{2,1}+w_{2,2}+w_{2,3}+w_{2,4}\right)-w_{1,2}\,w_{2,1} \end{array}\right) $$ I have written it in this form for simplicity, but if one expands the parentheses each diagonal entry contains a sum of 15 products, all with the plus sign. These diagonal entries can be given a combinatorial interpretation via the matrix tree theorem, since $(\operatorname{adj} M)_{ii}$ is the determinant of the $(n-2)\times(n-2)$ matrix obtained by removing row and columns $i$ and $n$ from $L$.
Is there a combinatorial interpretation also for the off-diagonal entries $\operatorname{adj}(M)_{ij}$, $i\neq j$? I imagine with sufficient work one can recurse one more level, but maybe there is already a known result stating this interpretation, since it is a similar setting to the matrix-tree theorem. My searches for "matrix tree theorem + inverse" have returned no useful result.
