Matrix inverse of $\left(A-I\right)$ given $A^{-1}$ I am wondering if the inverse of $$B = A-I$$ can be written in terms of $A^{-1}$ and/or $A$.  I am able to accurately compute $A$ and $A^{-1}$, which are very large matrices. Is it possible to calculate $B^{-1}$ without directly computing any inverses?
For example, if $A = 2I$, then $B^{-1}=\frac{1}{2}A$.
 A: It is not a complete answer, but if $\sum\limits_{k \geq 0} A^k$ converges then $$-(A-I) \sum\limits_{k \geq 0} A^k= \operatorname{Id}$$hence $$(A-I)^{-1}=- \sum\limits_{k \geq 0} A^k$$
A: In analogy with $\sum_{i=1}^\infty a^{-i}=(a-1)^{-1}$ ($|a|>1$), one could study the convergence of the series
$(A-I)^{-1}=A^{-1}+A^{-2}+\cdots$. 
Equivalently, if $C=(A-I)^{-1}$ then $C=A^{-1}(C+I)$ which could suggest an iterative calculation $C_{n+1}=A^{-1}(C_{n}+I)$ . This is esentially (if we start with $C=0$) the same as the series above.
A: If $A$ is nilpotent that is if there exists $k\in N$ such that $A^k=0$ then we have  
$(\sum_{i=0}^{k-1}A^i)(I-A)=(I-A)(\sum_{i=0}^{k-1}A^i)=I$ so the inverse is $-(\sum_{i=0}^{k-1}A^i)$
If $A^2=-A$ then $(I-A)(I+A/2)=(I+A/2)(I-A)=I$
And more importantly $B=I-A$ may not be invertible at all if $1$ is an eigen value of A, and if $1$ is not an eigen value then it is always invertible.
A: Why don't you just found $A-I$ and then take it's inverse? I think it's way more easy. For exemple : $$A=\;\; \begin{pmatrix}
a_1 & a_2\\ 
a_3 & a_4
\end{pmatrix}$$
$$ A-I= \;\; \begin{pmatrix}
a_1-1 & a_2\\ 
a_3 & a_4-1
\end{pmatrix}$$
Therefore we get : 
$$(A-I)^{-1}=\frac{1}{(a_1-1)(a_4-1)-a_2a_3}\;\; \begin{pmatrix}
a_4-1 & -a_2\\ 
-a_3 & 1_1-1
\end{pmatrix}$$
