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.

**Answer**

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=trace(BA−1). Then g≠−1 and

(A+B)−1=A−1−11+gA−1BA−1.

From this lemma, we can take a general A+B that is invertible and write it as A+B=A+B1+B2+⋯+Br, where Bi each have rank 1 and such that each A+B1+⋯+Bk is invertible (such a decomposition always exists if A+B is invertible and rank(B)=r). Then you get:

**Theorem.** Let A and A+B be nonsingular matrices, and let B have rank r>0. Let B=B1+⋯+Br, where each Bi has rank 1, and each Ck+1=A+B1+⋯+Bk is nonsingular. Setting C1=A, then

C−1k+1=C−1k−gkC−1kBkC−1k

where gk=11+trace(C−1kBk). In particular,

(A+B)−1=C−1r−grC−1rBrC−1r.

(If the rank of B is 0, then B=0, so (A+B)−1=A−1).

**Attribution***Source : Link , Question Author : Tomek Tarczynski , Answer Author : Michael Hardy*