# Proving a matrix identity involving inverses

Assuming that all matrix inverses involved below exist, show that $$(A-B)^{-1}=A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}.$$ In particular $$\left(I+A\right)^{-1}=I-\left(A^{-1}+I\right)^{-1}$$ and $$\left\vert\left(I+A\right)^{-1}+\left(I+A^{-1}\right)^{-1}\right\vert=1.$$

Here's what I tried to do but I can't simplify that underlined part at all:

So I thought I should maybe try finding some sort of infinite geometric series in whose summation the $$(AB^{-1} - I)$$ term appears but that doesn't seem to help either and I'm not certain if that's mathematically justified either. What should I do next? Also, I've just started studying linear algebra so I would really appreciate an answer using only elementary properties of matrices.

• It may not be that straightforward. If I may, my advice is to first prove a formula for the inverse of X+Y whch may not depend on the inverse of the sum of inverses, apply the formula to compute $(X^{ -1}+Y^{-1})^{-1}$ and then tie those two equations together. Commented Dec 31, 2023 at 6:41
• I expect that someone will point out that you should use LaTeX for the question and not an image. This makes the question searchable. Commented Dec 31, 2023 at 6:55
• Yeah sorry about that, I'm new to all this and I don't know how to use latex at all Commented Dec 31, 2023 at 7:19

$$(A-B)\big(A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}\big) \\= I + (B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} ~(\text{as}~ AA^{-1}=I) \\= I + BB^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} ~(\text{as}~ BB^{-1}=I) \\= I + B(B^{-1} - A^{-1})(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} \\= I + BA^{-1} - BA^{-1} = I$$
That implies $$(A-B)^{-1} = \big(A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}\big)$$