Inverse of $FR(X)R'F'$ Suppose I have three matrices $F$, $R$ and $X$ , where $F$ and $X$ are non-singular and $R$ is full row rank and whose respective dimensions are `N×N`, `N×K` and `K×K`. I have to solve \begin{gather} \\{FR(X)R'F'\\}^{-1} \end{gather} My guess is to use the formula for the inverse of a product of matrices, i.e. $(AB)^{-1} = B^{-1}A^{-1}$. Thus, I have proceeded as follows \begin{gather} (R'F')^{-1}(X)^{-1}(FR)^{-1} = (F')^{-1}(R')^{-1}X^{-1}R^{-1}F^{-1} \end{gather} but $R^{-1}$ and $F^{-1}$ are not conformable because it is a `N×K` times a `N×N` matrix. What is wrong with my reasoning?

Before answering your question, few comments:
a) It is not possible to invert a matrix that has not an equal number of rows and columns; so it is not possible to invert $R$ which not a square matrix.
b) Even if the matrix has an equal number of rows and columns it is not always possible to compute the inverse: it must also be full rank, or its determinant must but nonzero. If $rank(X)=K$ then $X$ has an inverse (and conversely).
c) It is well known that $rank(AB) \leq \min(rank(A),rank(B))$.
These rules imply that your matrix $M = FRXR'F'$ is $(N\times N)$ and has an inverse matrix if $rank(M)=N$, this requires that (i) $N\leq K$ (see condition c), that (ii) $rank(F)=N$ and (iii) $rank(R'XR)=N$. Under these conditions
$$(FRXR'F')^{-1}=(F')^{-1}(RXR')^{-1}(F)^{-1}.$$