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Is there a way to calculate the Covariance of the inverse of a random matrix from the Covariance of the original matrix? So say you have C1 = Cov(X), is there a direct way to calculate C2 = Cov(inv(X))?

EFG001
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No. You can't even calculate $E[1/X]$ from $E[X]$ in general for a single random variable, let alone the (co)variance of it.

Example: let $X$ be a Rademacher RV, so its variance is $1$ and the variance of $1/X$ is also $1$ (because $1/X=X$ for this particular distribution).

And, let $Y$ be a RV with possible values $2$, and $-2$, i.e. a slight variation of the Rademacher RV, and let $P(Y=2)=p$. So, the variance of $Y$ is $$\operatorname{var}(Y)=E[Y^2]-E[Y]^2=4-(2p-2(1-p))^2$$ And, this is equal to $1$ if $p=(2+\sqrt 3)/4$. Now, calculating $\operatorname{var}(1/Y)$: $$\operatorname{var}(1/Y)=E[1/Y^2]-E[1/Y]^2=1/4-(p/2-(1-p)/2)^2\neq 1$$ when calculated. So, both random variables have the same variance but not the same variance for their reciprocals, which means you can't actually find $\operatorname{var}(1/Y)$ from $\operatorname{var}(Y)$.

gunes
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    Although this is true, it would be nice to include a simple demonstration of this fact. The logical minimum would be to exhibit two variables $X$ and $Y$ with the same variance but for which $1/X$ and $1/Y$ have different variances. A stronger demonstration would exhibit a set of variables $X$ for which the variances of $1/X$ include every positive real number (along with positive infinity). – whuber Sep 26 '21 at 18:58
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    You're right. I started with the logical minimum, but also found the following as a good answer for a very similar question: https://stats.stackexchange.com/questions/41896/varx-is-known-how-to-calculate-var1-x – gunes Sep 26 '21 at 19:22