= Solution
Maps act on column vectors, so the first element appears at the right of the product. Starting immediately before the focusing kick, one full cell has
$$
M_x=D(L_2)G D(L_1)F.
$$
Multiplying gives
$$
\boxed{M_x=\begin{pmatrix}
1-L_1/f-L_1L_2/f^2&L_1+L_2+L_1L_2/f\\
-L_1/f^2&1+L_1/f
\end{pmatrix}.}
$$
It has $\det M_x=1$, as required for this lossless linear <beam optics>. The other transverse plane experiences the opposite quadrupole signs, so
$$
\boxed{M_y=M_x\big|_{f\to-f}=\begin{pmatrix}
1+L_1/f-L_1L_2/f^2&L_1+L_2-L_1L_2/f\\
-L_1/f^2&1-L_1/f
\end{pmatrix}.}
$$
The two planes have different envelopes but the same trace and therefore the same stable phase-advance magnitude for this equal-strength doublet.
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