let the population of degenerated lower states be $n_i$ and the upper degenerated state be $m_j$. the excitation rate from state $n_i$ to $m_j$ is $e_{ji}$ and the de-excitation rate from state $m_j$ to $n_i$ is $d_{ij}$. the notation of subscript is that, the first letter is target, and the later is the source.

the rate equations are:

$\frac{d n_i}{dt} = - n_i \sum_j{e_{ji}}+\sum_j{d_{ij} m_j}$

$\frac{d m_j}{dt} = - m_j \sum_i{d_{ij}}+\sum_i{e_{ji} n_i}$

combine these equations into one matrix equation by setting a population vector for h lower states and k higher states:

$p=(n_1,n_2,...,n_h,m_1,m_2,...,m_k)$

$\frac{d p}{dt} = M.p$

where M is the matrix.

$M=\begin{pmatrix}Ln&D \\ E&Lm\end{pmatrix}$

$E = \begin{pmatrix} e_{11} & ... & e_{1h} \\ ... & ... & ... \\ e_{k1} & ... & e_{kh} \end{pmatrix}$

$D= \begin{pmatrix} d_{11} & ... & d_{1k} \\ ... & ... & ... \\ d_{h1} & ... & d_{hk} \end{pmatrix}$

$Ln= \begin{pmatrix} \sum_j{e_{j1}}& 0 & 0 \\ 0& ... & 0 \\ 0 & 0 & \sum_j{e_{jh}}\end{pmatrix}$

$Lm= \begin{pmatrix} \sum_i{d_{i1}}& 0 & 0 \\ 0& ... & 0 \\ 0 & 0 &\sum_i{d_{ik}}\end{pmatrix}$

so the Matrix M is a $(h+k) \times (h+k)$ matrix. the meaning of matrix E is the excitation matrix. D is de-excitation matrix. the matrix Ln is the matrix of the lost of lower states. the matrix Lm is the lost of the upper state. at equilibrium, the change will be zero. thus, the solution of population is:

$M.p = 0$

____________________example_____________________

3 upper states and 3 lower states with a circular polarization.

$M = \begin{pmatrix}1/2&0&0&1/2&1/3&0 \\ 0&1/2&0&1/2&1/3&1/2 \\ 0&0&0&0&1/3&1/2 \\ 0&0&0&1&0&0 \\ 1/2&0&0&0&1&0 \\ 0&1/2&0&0&0&1 \end{pmatrix}$

the solution is

$p = ( 0, 0, 1 ,0, 0, 0)$

which means, all population stay in $n_3$, the highest lower state.