If $A$, $B$, $A-B$ and $I+A$ are invertible $n×n$ matrices then prove the following i. $(A-B)^{-1} = A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}$
ii. $(I+A)^{-1} = I-(A^{-1} + I)^{-1}$
iii. $tr((I+A)^{-1}) + tr((A^{-1} + I)^{-1}) = n$
I'm stuck on these. So far I only thought about taking the second half of each equation and try to find the first half but I had no success.
 A: Hint: If you want to show that the inverse of $X$ is (a properly defined expression) $Y$, it is sufficient to compute $XY$ (and in general also $YX$, but you can drop that for $n\times n$ matrices as in this problem).
A: For i) and ii) these things tend to be a pile of algebraic manipulations based on two things:
1) Square matrices $M$ and $N$ are inverses to each other iff $MN = I$ iff $NM = I$.
2) If $M$ and $N$ are invertible matrices then $(MN)^{-1} = N^{-1}M^{-1}$.
So for example for the second one, it suffices to show that
$$(I + A)(I -(A^{-1} + I)^{-1}) = I$$
Multiplying this out, it's equivalent to
$$I + A - (A^{-1} + I)^{-1} - A (A^{-1} + I)^{-1} = I$$
Cancelling out the $I$ terms, this is the same as
$$A - (A^{-1} + I)^{-1} - A (A^{-1} + I)^{-1} = 0$$
Left multiplying by $A^{-1}$ this is the same as
$$I - A^{-1}(A^{-1} + I)^{-1} - (A^{-1} + I)^{-1} = 0$$
Combining the second and third terms this is the same as
$$I - (A^{-1} + I) (A^{-1} + I)^{-1} = 0$$
Equivalently, 
$$I - I = 0$$
And we're done. And we didn't even have to use 2) here.
