How to simplify a fraction that has matrix

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Recently, I was studying about optimum beamforming in array signal processing. In it, the maximum output SINR is $$SINR=\frac{W^{H}R_sW}{W^{H}R_{i+n}W}$$ $$and\quad W=R_{i+n}^{-1}a \quad R_s=\alpha^2aa^H$$where $W \in \mathbb{C}^{N \times 1} \ R_s, R_{i+n} \in \mathbb{C}^{N \times N} \ a\in \mathbb{C}^{N \times 1}$ and $R_{i+n} \ $ is invertible, $\alpha$ is a scalar, $H$ denotes the transpose conjugate.

Then, it can be simplified as $$SINR=\alpha^2a^{H}R_{i+n}^{-1}a$$

I can't figure out how this fraction is simplified. Especially the realtion between inversion and divison of the matrix.

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The authors aren't simplifying, they're solving a Rayleigh Quotient problem.

The extremal values of the Rayleigh Quotient $$\eqalign{ \rho = \frac{w^HAw}{w^HBw} \\ }$$ occurs at the eigenvalues of the matrix $R=B^{-1}A$

Applying this result to the current problem $$\eqalign{ A &= \alpha^2 aa^H \\ B &= R_{i+n} \\ R &= \alpha^2B^{-1}aa^H = \alpha^2wa^H \\ }$$ results in an eigenvalue equation which is easily solved $$\eqalign{ Rw &= \rho w\\ \alpha^2wa^Hw &= \rho w \quad\implies\quad\rho = \alpha^2a^Hw \\ }$$ Since $\operatorname{rank}(A)=1$, this is the only non-zero eigenvalue and is therefore the maximum.