= Solution
The <stationary balance equations> at level zero are $(\lambda+\alpha)\pi_{0C}=\beta\pi_{0W}$ and $\beta\pi_{0W}=\alpha\pi_{0C}+\mu\pi_{1W}$. At every $i\geq1$ they are
$$
(\lambda+\alpha)\pi_{iC}=\lambda\pi_{i-1,C}+\beta\pi_{iW},\qquad
(\mu+\beta)\pi_{iW}=\alpha\pi_{iC}+\mu\pi_{i+1,W}.
$$
The zero-level equations give $\pi_{0C}=\beta\pi_{0W}/(\lambda+\alpha)$ and $\pi_{1W}=\beta\lambda\pi_{0W}/[\mu(\lambda+\alpha)]$. Substituting the latter in the level-one $C$ equation gives
$$
(\pi_{1C},\pi_{1W})=\frac{\beta\pi_{0W}}{\lambda+\alpha}\left(\frac{\lambda(\mu+\beta)}{\mu(\lambda+\alpha)},\frac\lambda\mu\right).
$$
Solve the $W$ equation for $\pi_{i+1,W}$, then substitute it into the next $C$ equation. This yields the required row-vector recurrence with
$$
\boxed{B=\begin{pmatrix}\frac{\lambda\mu-\beta\alpha}{\mu(\lambda+\alpha)}&-\frac\alpha\mu\\\frac{\beta(\beta+\mu)}{\mu(\lambda+\alpha)}&\frac{\beta+\mu}\mu\end{pmatrix}.}
$$
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