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This question asks why $A^{-1}+B^{-1}\ne(A+B)^{-1}$ when $A,B$ are $n\times n$ matrices. Of course, there is no particular reason to think that equality should hold, but it does suggest the question: Are the cases where equality holds easily classified? For concreteness, let's work with $2\times 2$ matrices. Using the fact that matrices of the form $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ are really complex numbers in disguise, one can, without too much trouble, find the matrices of this specific form which satisfy the desired equation: These are exactly the matrices corresponding to complex numbers $z,w$ with $z\ne0$ and $w=\left(-\frac12\pm\frac{\sqrt 3}2i\right)z$. In general, one can check that we must have $\mathrm{det}(A)=\mathrm{det}(B)$, as in this question, but beyond that I am stuck.

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    $\begingroup$ Since the determinant of a matrix corresponding to the complex number $z$ is $|z|^2$, your matrices would have determinants $|z|^2$ and $|\frac12+\frac34i|^2|z|^2$, but the multiplying term on the latter isn't $1$, so they can't have equal determinants. Did you maybe mean $w=(-\frac12\pm\frac{\sqrt{3}}2i)z$? $\endgroup$ – Steven Stadnicki Sep 1 '17 at 5:15
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    $\begingroup$ FWIW, "beg the question" in its traditional meaning doesn't mean what you think it means. The phrase you want is something like "raise the question." See the Usage Note at ahdictionary.com/word/… $\endgroup$ – symplectomorphic Sep 1 '17 at 5:16
  • $\begingroup$ @StevenStadnicki yes I did. Good catch. $\endgroup$ – A. Howells Sep 1 '17 at 5:30
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Multiplying on the right by $A+B$, you get $I+A^{-1}B + B^{-1}A+I=I$ or $A^{-1}B+B^{-1}A=-I$. Letting $U=A^{-1}B$ then this means that $U+U^{-1}=-I$, or $U^2+U+I=0$. This tells you that the eigenvalues of $U$ come from $\frac{-1}{2}\pm \frac{\sqrt{-3}}{2}$. If $A,B$ are real matrices, then $U$ must have both these values as eigenvalues, but if $A,B$ are complex matrices, then you could have a single eigenvalue, repeated.

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    $\begingroup$ Is this necessary condition also sufficient? $\endgroup$ – vadim123 Sep 1 '17 at 5:20
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    $\begingroup$ Yes, one can reverse the argument. If $A,B$ are invertible, and $U=A^{-1}B$ has the property that $U^2+U+I=0$ then $U+U^{-1}=-I$ and, leading to $(A^{-1}+B^{-1})(A+B)=I$. $\endgroup$ – Thomas Andrews Sep 1 '17 at 5:24

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