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Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $$A^{-1} + B^{-1} =(A+B)^{-1}$$ then prove that $\operatorname{det}(A)=\operatorname{det}(B)$. Also show that this result is not valid for complex matrices.

I'm not getting any way out to solve this. Can anybody solve the problem?
Thanks for assistance in advance.

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By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$\det(U)=1$$or equivalently$$\det(A)=\det(B)$$


Counter example on $\Bbb C$

Let $A=I_2$ and $b=kI_2$ with $k={-1+i\sqrt 3\over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1\over k}I\\(A+B)^{-1}={1\over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$\det(B)=k^2=-1-k={-1-i\sqrt 3\over 2}\ne 1=\det(A)$$


Comment

Even on $\Bbb C$, from $U^3=I$ we can conclude that$$\left|\det(A)\right|=\left|\det(B)\right|$$

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  • $\begingroup$ Over $\mathbb{R}$, $U^2+U+I=0$ implies that $n$ is even. Over $\mathbb{C}$, that does no imply that $\det(U)=1$. $\endgroup$ – loup blanc Jan 5 at 18:44
  • $\begingroup$ You are right. The statement doesn't hold generally over $\Bbb C$ but the question has mentioned with real entries $\endgroup$ – Mostafa Ayaz Jan 5 at 18:48
  • $\begingroup$ The OP asked a second question about the complex case. It's true that he could work (that would not hurt him) in order to clarify this point. $\endgroup$ – loup blanc Jan 5 at 19:21
  • $\begingroup$ Thank you for pointing that out. I will add a counter example on $\Bbb C$ $\endgroup$ – Mostafa Ayaz Jan 5 at 19:34

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