= Solution
The block matrix is
$$
K=\begin{pmatrix}0&M\\M&0\end{pmatrix},
\qquad
K^{-1}=\begin{pmatrix}0&M^{-1}\\M^{-1}&0\end{pmatrix}.
$$
If both $q\in\mathcal L$ and $q'$ condense, choose their boundary excursions so that the two string operators cross once. Their commutator is their mutual full-braiding phase:
$$
W_qW_{q'}=exp(2\pi i q^TK^{-1}q')W_{q'}W_q.
$$
Both operators act as the identity on every ground state, so consistency requires
$$
q^TK^{-1}q'\in\mathbb Z
\qquad\hbox{for every }q=(q_1,q_2,0,0)^T\in\mathcal L.
$$
Writing $q'=(q'_{\rm top},q'_{\rm bot})$, this says
$$
M^{-1}q'_{\rm bot}\in\mathbb Z^2.
$$
Set $l_{\rm top}=M^{-1}q'_{\rm bot}$ and choose $l_{\rm bot}=0$. Then $Kl=(0,q'_{\rm bot})^T$, and
$$
u=q'-Kl=(q'_{\rm top},0)^T\in\mathcal L.
$$
Therefore every additionally condensable anyon has the form
$$
\boxed{q'=u+Kl,qquad u\in\mathcal L,quad l\in\mathbb Z^4.}
$$
The $Kl$ term is a local particle in the <anyon lattice of an Abelian Chern--Simons theory>, so the condensate is maximal modulo local excitations, as required for a <Lagrangian subgroup of Abelian anyons>.
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