Inverse of the sum of matrices 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.
In general, $A+B$ need not be invertible, even when $A$ and $B$ are. But one might ask whether you can have a formula under the additional assumption that $A+B$ is invertible.

As noted by Adrián Barquero, there is a paper by Ken Miller published in the Mathematics Magazine in 1981 that addresses this. 

He proves the following:

Lemma. If $A$ and $A+B$ are invertible, and $B$ has rank $1$, then let $g=\mathrm{trace}(BA^{-1})$. Then $g\neq -1$ and 
$$(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}.$$

From this lemma, we can take a general $A+B$ that is invertible and write it as $A+B = A + B_1+B_2+\cdots+B_r$, where $B_i$ each have rank $1$ and such that each $A+B_1+\cdots+B_k$ is invertible (such a decomposition always exists if $A+B$ is invertible and $\mathrm{rank}(B)=r$). Then you get:

Theorem. Let $A$ and $A+B$ be nonsingular matrices, and let $B$ have rank $r\gt 0$. Let $B=B_1+\cdots+B_r$, where each $B_i$ has rank $1$, and each $C_{k+1} = A+B_1+\cdots+B_k$ is nonsingular. Setting $C_1 = A$, then
$$C_{k+1}^{-1} = C_{k}^{-1} - g_kC_k^{-1}B_kC_k^{-1}$$
where $g_k = \frac{1}{1 + \mathrm{trace}(C_k^{-1}B_k)}$. In particular,
$$(A+B)^{-1} = C_r^{-1} - g_rC_r^{-1}B_rC_r^{-1}.$$

(If the rank of $B$ is $0$, then $B=0$, so $(A+B)^{-1}=A^{-1}$).
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:
$$(A+B)^{-1}=A^{-1}(I -(I+BA^{-1})^{-1}BA^{-1})$$