Sum inverse implies equal determinant

Question

Let,$A,B$ are two $n\times n,(n\ge 2)$ non-singular matrices with real entries $(a)$ If $A^{-1}+B^{-1}=(A+B)^{-1}$ , then show that $\text{det(A)}=\text{det(B)}.$ I am getting $$B^{-1}A+A^{-1}B=-I_n.$$ Then how to proceed

Answer 1

The answer by @TsemoAristide has an obvious gap. Since my comment under the answer was not replied, I'm just posting it here so that people can see it. In the answer, when taking the determinant of the right-hand side, the inverse on $I+AB^{-1}$ was omitted. What should be correct is $$\det(A^{-1})\det(I+AB^{-1})=\det(B^{-1})\det(I+AB^{-1})^{-1},$$ which does not immediately imply $\det(A^{-1}) = \det(B^{-1})$.

I give an answer below, the complexity of which exceeds what I have expected. So any comments to simplify are welcome.

As the op has derived, we have (in slightly different way but basically the same): $$AB^{-1} + BA^{-1} + I = \mathbf{0}.$$ Setting $C = AB^{-1}$ one realizes $C^{-1} = BA^{-1}$, so the above equation becomes $$C + C^{-1} + I = \mathbf{0},\quad\text{or}\hspace{1.5ex}C^2 + C + I = \mathbf{0}.$$ Let $\lambda$ be any eigenvalue of $C$, then $\lambda$ satifies $$\lambda^2 + \lambda + 1 = 0,$$ so $\lambda$ is either of the two conjugate roots $\omega$, $\bar\omega$ of the quadratic, whose product $\omega\bar\omega=1$. Since $C$ is real matrix its eigenvalues appear in conjugate pairs (counting multiplicity), so it must be that $$\det C = \prod\lambda_i = (\omega\bar\omega)^\frac{n}{2} = 1,$$ i.e., $$\det\left(AB^{-1}\right) = 1 \implies \det(A) = \det(B).$$ (By the proof we also showed that $n$ must be even.)

Edit: also by the proof, the result holds if $$A^{-1} + B^{-1} = r(A+B)^{-1},\text{ for some }0<r<4.$$

Answer 2

$A^{-1}+B^{-1}=(A+B)^{-1}$ implies that $A^{-1}(I+AB^{-1})=((AB^{-1}+I)B)^{-1}=B^{-1}(AB^{-1}+I)^{-1}$ we deduce that $det(A^{-1})det(I+AB^{-1})=det(B^{-1})det(I+AB^{-1})$ and $det(A^{-1})=det(B^{-1})$ and this implies that $det(A)=det(B)$.