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.


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 $$\det(B)=k^2=-1-k={-1-i\sqrt 3\over 2}\ne 1=\det(A)$$


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

  • $\begingroup$ 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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