Problem of determinant when $A^{-1}+B^{-1}=(A+B)^{-1}$

Question

I have two $4\times 4$ real matrices $A$ and $B$, and it is known that $A^{-1}+B^{-1}=(A+B)^{-1}$ ($A$, $B$ and $A+B$ are invertible). How can I prove that $\det (A)=\det (B)$?

Answer 1

If multiply both sides of $A^{-1}+B^{-1}=(A+B)^{-1}$ by $A+B$ on the left , then we have
$$AB^{-1} + BA^{-1}= -I$$
Now put $AB^{-1}=C$ then
$$C^{-1} =-(C+I)$$ And multiply both sides by $C$
$$C+I=-C^2$$
Hence $$C^{-1}=C^2$$ $$(\det(C))^3=1$$ $$\det(C)=1$$ Thus $\det(A)=\det(B)$ .