Approximating the inverse of a perturbed matrix Consider a matrix $A$ which we subject to a small perturbation $\partial A$.    If $\partial A$ is small, then we have $(A + \partial A)^{-1} \approx A^{-1} - A^{-1} \partial A A^{-1}$
I came across this approximation in some notes and I am trying to understand where it comes from.  This answer seems related, but I am having trouble translating the results from the cited paper into the provided equation.  
 A: The usual argument is that, if you perturb $A$ by a small $X$ and get $(A+X)^{-1}=A^{-1}+Y+O(\|X\|^2)$, where $Y$ is the first-order (i.e. linear) change in $A^{-1}$, then by comparing the first-order terms on both sides of $\left(A^{-1}+Y+O\left(\|X\|^2\right)\right)(A+X)=I$, you get $YA+A^{-1}X=0$. Hence $Y=-A^{-1}XA^{-1}$ and 
$$(A+X)^{-1}=A^{-1}+Y+O\left(\|X\|^2\right)\approx A^{-1}+Y=A^{-1}-A^{-1}XA^{-1}.
$$
Edit. The above is a rigorous argument provided that $A\mapsto A^{-1}$ is differentiable in the first place, but this is indeed the case because $A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)$ is a rational function in the entries of $A$.
A: In fact, you can understand the problem by the solution, given by mjqxxxx, of the related problem you give. The primary problem is to find a way to estimate the term in the right hand. Why not consider the determinant of the term.  The determinant of the product equals the product of the corresponding determinant. Then if we consider epsilon as being infinitesimal, then the higher term is approaching zero. $\square$ 
