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I need to prove that if $A$, $B$ and $(A + B)$ are invertible then $(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$

I'm a bit lost with this one,
I can't find a way to make any assumptions about $(A^{-1} + B^{-1})$,
Neither by using $A^{-1}$, $B^{-1}$ and $(A + B)^{-1}$.

If someone could clue me in I'll be grateful.


UPDATE:

Thanks for all of your input - it really helped,
I got a solution, I'd like to know if my way of proving it is valid:

$(A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$


multiplying both sides by $A^{-1}$ and $B^{-1}$

$A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1} = (A+B)^{-1}$


now multiplying both sides by $(A + B)$, I get to

$(A+B)(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1}) = I$

Lets call $(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})$ -> $C$

Now I determine that $C$ is $(A+B)^{-1}$,
because $(A+B)C$ equals the Identity Matrix.


So lastly to verify that I place $C$ in the original equation:

$(A^{-1} + B^{-1})^{-1} = A(A^{-1}(A^{-1} + B^{-1})^{-1}B^{-1})B$

and from this I get:

$(A^{-1} + B^{-1})^{-1} = (A^{-1} + B^{-1})^{-1}$

  • 0
    updated my solution2012-11-20

5 Answers 5