TITLE: Inverse of the sum of matrices
QUESTION [181 upvotes]: I have two square matrices: $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that help with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{-1}$ is also known.

REPLY [2 votes]: I know the question has been answered multiple times with great answers, but with my answer you don't need to memorize any lemmas or formulas.
Suppose $(A+B)x=y$, then $x=(A+B)^{-1}y$. This is all we need to get. The steps are:
(1) Start with $(A+B)x=y$.
(2) Then $Ax=y-Bx$, so $x=A^{-1}y -A^{-1}Bx$.
(3) Multiply $x$ in step (2) by $B$ to get 
 $$Bx=BA^{-1}y -BA^{-1}Bx$$
 which is equivalent to
 $$(I+BA^{-1})Bx=BA^{-1}y $$
 or,
 $$Bx=(I+BA^{-1})^{-1}BA^{-1}y $$
(3) Substitute this $Bx$ into the $x$ in step (2) to get
 $$x=A^{-1}y -A^{-1}(I+BA^{-1})^{-1}BA^{-1}y $$
(4) Now factorizing the $y$ gives you the required result.
 $$x=(A^{-1} -A^{-1}(I+BA^{-1})^{-1}BA^{-1})y $$
(5)The assumptions we have used are $A$ and $I+BA^{-1}$ are nonsingular.
(6) We can factorize the $A^{-1}$ to get:
$$(A+B)^{-1}=A^{-1}(I -(I+BA^{-1})^{-1}BA^{-1})$$