We have two square matrices $A, B \in C^{n \times n}$. We need to compare the values of $||A^{-1}||_{F}$ and $||B^{-1}||_{F}$.
The brute force approach will be to compute the inverses $A^{-1}, B^{-1}$ and then compare their norms which has a computational complexity of $O(n^{3})$. But can we somehow compare $||A^{-1}||_{F}$ and $||B^{-1}||_{F}$ without actually computing the inverses in order to reduce the complexity?
We have that $$ A, B \in \mathbb{C}^{n \times n}$$
Then we know the following
$$ ||A||_{2} = ||A||_{F}= \sqrt{\sum_{i=1}^{M}\sum_{j=1}^{N}|a_{ij}|^{2}} = \sqrt{tr(A^{T}A)} = \sqrt{\sum_{i=1}^{R}\lambda_{i}} = \sqrt{\sum_{i=1}^{R}\sigma_{i}^{2}} $$
we also know that for any matrix $$ A \in \mathbb{C}^{n \times n}$$ it admits an SVD which is given by $$ A = U \Sigma V^{*} $$ then we know that the inverse of A is given by $$ A^{-1} = V\Sigma^{-1} U^{*} $$
where the inverse of the diagonal matrix of singular values is simply their reciprocal. From the previous information what can you conclude about the inverse norm.
Edited: Technically you'd have to compute some decomposition here which wouldn't work for your time complexity. There are faster matrix decompositions. However, if you just construct the matrix like you said and have some arbitrary singular values given by $$ a_{1}, a_{2}$$ supposing that the matrix is 2 x 2 then the inverse matrix has singular values given by $$ \frac{1}{a_{1}}, \frac{1}{a_{2}}$$
Then we have $$ || A^{-1}|| = \sum_{i=1}^{2} \tilde{\sigma}_{i}^{2} = (\frac{1}{a_{1}})^{2} + (\frac{1}{a_{2}})^{2} = \frac{1}{a_{1}^{2}} + \frac{1}{a_{2}^{2}}$$