If A, B and (adj(A^–1) + adj(B^–1)) are non-singular matrices of same order, then the inverse of A(adj(A^–1) + adj(B^–1))^–1B, is equal to

$\begin{array}{l}{\left[\mathrm{A}{\left(\mathrm{adj}\left({\mathrm{A}}^{-1}\right)+\mathrm{adj}\left({\mathrm{B}}^{-1}\right)\right)}^{-1}\cdot \mathrm{B}\right]}^{-1}\\ {\mathrm{B}}^{-1}\cdot \left(\mathrm{adj}\left({\mathrm{A}}^{-1}\right)+\mathrm{adj}\left({\mathrm{B}}^{-1}\right)\right)\cdot {\mathrm{A}}^{-1}\\ {\mathrm{B}}^{-1}\mathrm{adj}\left({\mathrm{A}}^{-1}\right){\mathrm{A}}^{-1}+{\mathrm{B}}^{-1}\left(\mathrm{adj}\left({\mathrm{B}}^{-1}\right)\right)\cdot {\mathrm{A}}^{-1}\\ {\mathrm{B}}^{-1}\left|{\mathrm{A}}^{-1}\right|\mathrm{I}+\left|{\mathrm{B}}^{-1}\right|{\mathrm{IA}}^{-1}\\ \frac{{\mathrm{B}}^{-1}}{\left|\mathrm{A}\right|}+\frac{{\mathrm{A}}^{-1}}{\left|\mathrm{B}\right|}\\ \Rightarrow \frac{\mathrm{adjB}}{\left|\mathrm{B}\right|\left|\mathrm{A}\right|}+\frac{\mathrm{adjA}}{\left|\mathrm{A}\right|\left|\mathrm{B}\right|}\\ =\frac{1}{\left|\mathrm{A}\right|\left|\mathrm{B}\right|}(\mathrm{adjB}+\mathrm{adjA})\end{array}$