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

Question

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

Accepted Answer

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

Solutions · 1