Matrix inverse of $A + \epsilon I$, where $A$ is invertible

Question

Let $A$ be a square invertible matrix, and $\epsilon$ a small positive quantity. To first-order in $\epsilon$, what is the inverse of $A + \epsilon I$, where $I$ is the identity matrix?

Answer 1

I found a way to do it right after posting.

$$(A + \varepsilon I)^{- 1} = A^{- 1} A (A + \varepsilon I)^{- 1} = A^{- 1} (I + \varepsilon A^{- 1})^{- 1} \approx A^{- 1} (I - \varepsilon A^{- 1})$$

Answer 2

You can write it as

$\left((I+\varepsilon A^{-1})A\right)^{-1}=A^{-1}\sum \limits_{n=0}^\infty A^{-n}\varepsilon^n= \sum \limits_{n=0}^\infty A^{-n-1}\varepsilon^n,$

where $A^0=I$. ((I+B) is invertible if $B$ has norm, as a linear operator, less than 1, so for $\varepsilon$ small this works).