Inverse of $A^{-1}+B^{-1}$ If $A, B$ and $A + B$ are all $n × n$ invertible matrices. Prove that $A^{−1} + B^{−1}$ is invertible and the inverse is $A(A + B) ^{−1}B$.
I am afraid I am really stuck on this one, and I haven't really tried much because I don't know what to try.
Thanks for all the help guys, I understand now.
 A: \begin{equation}
(A^{-1} + B^{-1}) A (A+B)^{-1} B = A^{-1}A(A+B)^{-1}B + B^{-1} A (A+B)^{-1} B
\end{equation} by the distributive law.  Now here comes the trick:  Write $A = (A+B) - B$ to get
$$
A^{-1}A(A+B)^{-1}B + B^{-1} A (A+B)^{-1} B = (A+B)^{-1} B + B^{-1}(A+B)(A+B)^{-1} B -B^{-1} B (A+B)^{-1} B \\ = (A+B)^{-1}B + I - A+B)^{-1} B = I
$$
A: Here's half of the solution.
$$A(A+B)^{-1}B=((A+B)A^{-1})^{-1}B=(I+BA^{-1})^{-1}B$$
(Note that inversion reverses the order of products.) What happens if you do the same thing with $B$? We then get
$$(B^{-1}(I+BA^{-1}))^{-1}=(B^{-1}+A^{-1})^{-1}$$
We didn't even have to multiply anything out!
A: Here's a concise proof:
$$
\begin{align}
A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\
&= B^{-1}AA^{-1} + B^{-1}BA^{-1}\\
&= B^{-1}(A + B)A^{-1} \\
\end{align}
$$
So $A^{-1} + B^{-1}$ is the product of invertible matrices and thus is invertible, with inverse equal to
$$
\begin{align}
(A^{-1} + B^{-1})^{-1} &= (B^{-1}(A + B)A^{-1})^{-1} \\
&= A(A + B)^{-1}B \\
\end{align}
$$
