# Non-square transfer matrix from square scattering matrix

I'm trying to construct the transfer matrix for an arbitrary system with 1 input and 2 outputs, like a splitter (shown below) or 1x2 multi-mode interferometer. (Image borrowed from http://www.fiberstore.com/images/ckfinder/images/tutorial/1x2_Splitter.jpg)

Currently, I have the entire scattering matrix for my device:

$$\begin{pmatrix} A_\text{in}^{(-)}\\ A_\text{out1}^{(+)}\\ A_\text{out2}^{(+)} \end{pmatrix} = \begin{pmatrix} S_{00} & S_{01} & S_{02}\\ S_{10} & S_{11} & S_{12}\\ S_{20} & S_{21} & S_{22}\\ \end{pmatrix} \begin{pmatrix} A_\text{in}^{(+)}\\ A_\text{out1}^{(-)}\\ A_\text{out2}^{(-)} \end{pmatrix}$$

where the $$(+)$$ and $$(-)$$ correspond to forward and backward propagating signals, respectively. I am looking for a transfer-matrix $$\hat{T}$$ such that $$\begin{pmatrix} A_\text{out1}^{(+)}\\ A_\text{out1}^{(-)}\\ A_\text{out2}^{(+)}\\ A_\text{out2}^{(-)}\\ \end{pmatrix} = \begin{pmatrix} T_{00} & T_{01}\\ T_{10} & T_{11}\\ T_{20} & T_{21}\\ T_{30} & T_{31}\\ \end{pmatrix} \begin{pmatrix} A_\text{in}^{(+)}\\ A_\text{in}^{(-)}\\ \end{pmatrix}$$

Likewise, I am interested in the inverse transfer matrix such that $$\begin{pmatrix} A_\text{in}^{(+)}\\ A_\text{in}^{(-)}\\ \end{pmatrix} = \begin{pmatrix} T_{00} & T_{01} & T_{02} & T_{03}\\ T_{10} & T_{11} & T_{12} & T_{13}\\ \end{pmatrix} \begin{pmatrix} A_\text{out1}^{(+)}\\ A_\text{out1}^{(-)}\\ A_\text{out2}^{(+)}\\ A_\text{out2}^{(-)}\\ \end{pmatrix}$$ which I figured I could just get from the pseudo-inverse of the previous result. Would this work? In both cases I am having trouble finding $$\hat{T}$$ because I am moving from a scattering matrix with 3 variables and 3 equations to either (1) an overdetermined system with 2 variables and 4 equations or (2) an underdetermined system with 4 variables and 2 equations.

It seems that generating a transfer-matrix for such a splitter should be quite straightforward, especially since I already have the scattering matrix data. Am I missing something here?