I have the equation of motion of a density matrix as
$$\dot{\rho} = -i[H, \rho] - \dfrac{1}{2}\left(\Gamma \rho + \rho \Gamma \right) + \lambda,$$
where
$$\rho = \begin{bmatrix} \rho_{aa} & \rho_{ab} \\ \rho_{ba} & \rho_{bb} \end{bmatrix}, \ \ \ \ \ H = \begin{bmatrix} W_{a} & V \\ V & W_{b} \end{bmatrix}, \\ \Gamma = \begin{bmatrix} \gamma_{a} & 0 \\ 0 & \gamma_{b} \end{bmatrix}, \ \ \ \ \ \lambda = \begin{bmatrix} \lambda_{a} & 0 \\ 0 & \lambda_{b} \end{bmatrix}$$
I also have that
$$V(t) = -A(t) \wp u(z)/\hbar$$
I am then told that the off-diagonal term in the density-matrix equation may be written as
$$\dot{\rho}_{ab} = -(i \omega + \gamma)\rho_{ab} - i(\rho_{aa} - \rho_{bb})A(t) \wp u(z)/\hbar,$$
where $\gamma = \dfrac{1}{2} (\gamma_a + \gamma_b)$.
I've calculated that
$$\dfrac{1}{2} \left(\Gamma \rho + \rho \Gamma \right) = \dfrac{1}{2} \left( \begin{bmatrix} \gamma_a \rho_{aa} & \gamma_a \rho_{ab} \\ \gamma_b \rho_{ba} & \gamma_b \rho_{bb} \end{bmatrix} + \begin{bmatrix} \gamma_a \rho_{aa} & \gamma_a \rho_{ab} \\ \gamma_b \rho_{ba} & \gamma_b \rho_{bb} \end{bmatrix} \right) = \begin{bmatrix} \gamma_a \rho_{aa} & \frac{1}{2} \rho_{ab}(\gamma_a + \gamma_b) \\ \frac{1}{2} \rho_{ba}(\gamma_b + \gamma_a) & \gamma_b \rho_{bb} \end{bmatrix}$$
But it still isn't clear to me how we get
$$\dot{\rho}_{ab} = -(i \omega + \gamma)\rho_{ab} - i(\rho_{aa} - \rho_{bb})A(t) \wp u(z)/\hbar$$
from
$$\dot{\rho} = -i[H, \rho] - \dfrac{1}{2}\left(\Gamma \rho + \rho \Gamma \right) + \lambda$$
In particular, I really don't understand what $[H, \rho]$ is supposed to be. It looks like a block/partition matrix, but, from what I've seen, block matrices typically don't have a comma; and, even then, I still can't figure out how $[H, \rho]$ would fit into $\dot{\rho}_{ab} = -(i \omega + \gamma)\rho_{ab} - i(\rho_{aa} - \rho_{bb})A(t) \wp u(z)/\hbar$.
EDIT
Following user Paul's comment that $[a, b] = ab - ba$, I get
$$[H, \rho] = H \rho - \rho H \\ = \begin{bmatrix} \rho_{aa} W_a + \rho_{ba} V & \rho_{ab} W_a + \rho_{bb} V \\ \rho_{aa}V + \rho_{ba} W_b & \rho_{ab}V + \rho_{bb}W_b \end{bmatrix} - \begin{bmatrix} \rho_{aa} W_a + \rho_{ab} V & \rho_{aa} V + \rho_{ab} W_b \\ \rho_{ba}W_a + \rho_{bb} V & \rho_{ba}V + \rho_{bb} W_b \end{bmatrix} \\ = \begin{bmatrix} \rho_{ba} V - \rho_{ab} V & \rho_{ab} W_a + \rho_{bb} V - (\rho_{aa} V + \rho_{ab} W_b) \\ \rho_{aa}V + \rho_{ba} W_b - (\rho_{ba} W_a + \rho_{bb} V) & \rho_{ab}V - \rho_{ba}V \end{bmatrix}$$
EDIT2
Ok, I think I've figured it out. It seems that $\dot{\rho}_{ab}$ means that I need to use only those equations that are off-diagonal and contain subscript $ab$ terms. This means we use $\rho_{ab} W_a + \rho_{bb} V - (\rho_{aa} V + \rho_{ab} W_b)$ in the final matrix of the above edit. We then indeed get
$$\begin{align} \dot{\rho}_{ab} &= -[i(W_a - W_b) + \frac{1}{2}(\gamma_a + \gamma_b)]\rho_{ab} - i(\rho_{aa} - \rho_{bb}) A(t) \wp u(z)/\hbar \\ &= -(i \omega + \gamma)\rho_{ab} - i(\rho_{aa} - \rho_{bb})A(t) \wp u(z)/\hbar \end{align}$$
(the $\lambda$ matrix term has off-diagonals $0$).