I have a Continuous Time Markov Chain with transition rate matrix given by the folowing recipe:
Let $\lambda_1,\lambda_2 \geq 0$ and $\lambda_N := \lambda_1 + \lambda_2$. We denote $e_n$ for the row unit vector of dimension $n$, $\alpha := \begin{pmatrix} 1 & 0 & \dots & 0 \end{pmatrix} \in \mathbb{R}^n$, $\beta := \begin{pmatrix} 0 & \dots & 0 & 1 \end{pmatrix} \in \mathbb{R}^n$ and $m := S_1 \cdot S_2$. The transition rate matrix is given by $$ Q = -\lambda_N \cdot I_m + \lambda_1 \cdot \begin{pmatrix} \begin{pmatrix} I_{S_2 - c_2} \otimes \begin{pmatrix} 0 & I_{S_1-1}\\ \alpha_{S_1-1} & 0 \end{pmatrix} & 0 \end{pmatrix} \\ \begin{pmatrix} 0 & I_{c_2} \end{pmatrix} \otimes \begin{pmatrix}0 & I_{S_1 - 1}\\ 0 & 0 \end{pmatrix} + (e_{S_1}^{\intercal} \otimes \beta_{c_2} ^{\intercal}) \cdot \alpha_m \end{pmatrix} + \lambda_2 \cdot \begin{pmatrix} 0 & I_{m-S_1}\\ I_{S_1-c_1} & 0\\ e_{c_1} \cdot \alpha_m \end{pmatrix} $$
After analyzing this matrix one can see that its structure is fairly simple as it only has 3 non-zero elements in each row.
The question is simple: I would like to find the steady state of this process (or a way to compute it, or an argument why you think this is virtually impossible to do).
I do know that it can by done by solving $\pi \cdot Q = 0 \mbox{ with } \sum \pi_i = 1$ but I don't really see how to easily solve it in this setting.