Given a symmetric real matrix $A$, is there any potentially easier way to evaluate $(A^{-1}+\lambda I)^{-1}$ than directly?
(I'm hoping the matrix inversions can be avoided somehow.)
Background: https://math.stackexchange.com/a/3664647/86969
2nd prize edit: How about if both $A$ and $A^{-1}$ are available?
Edit
Given that $A$ is symmetric, we have $A=R^{-1}DR$ where $D$ is a diagonal matrix and $R$ is an orthogonal matrix (e.g. rotation).
So FWIW the expression is equivalent to
$$ (A^{-1}+\lambda I)^{-1}= ((R^{-1}DR)^{-1}+\lambda I)^{-1} =(RD^{-1}R^{-1}+\lambda I)^{-1} =(R(D^{-1}+\lambda R^{-1}IR)R^{-1})^{-1} =R(D^{-1}+\lambda R^{-1}IR)^{-1}R^{-1} =R(D^{-1}+\lambda I)^{-1}R^{-1} $$
Note that for $\lambda=0$ we have $$(A^{-1}+\lambda I)^{-1}=A$$
Now, if $\lambda \neq 0$, a simple computation yields $$ (I+\lambda A)^{-1}=I- \lambda (A^{-1}+\lambda I)^{-1} $$
Assume by contradiction that there is a formula for calculating $(I+\lambda A)^{-1}$ without using iversion.
Then, for each symmertic matrix $B$, setting $A=\frac{1}{\lambda} (B-I)$ we can calculate $$B^{-1}=(I+\lambda A)^{-1}=I- \lambda (A^{-1}+\lambda I)^{-1}$$ without using inversion.