= Solution
Take a reference point immediately before the focusing quadrupole and let matrices act on $(x,x')$ from the right. With $q=1/f$, define $Q_F=\bigl(\begin{smallmatrix}1&0\\-q&1\end{smallmatrix}\bigr)$, $Q_D=\bigl(\begin{smallmatrix}1&0\\q&1\end{smallmatrix}\bigr)$ and $D(L)=\bigl(\begin{smallmatrix}1&L\\0&1\end{smallmatrix}\bigr)$. One cell is
$$
\boxed{M=D(L_2)Q_DD(L_1)Q_F=
\begin{pmatrix}1-L_1q-L_1L_2q^2&L_1+L_2+L_1L_2q\\-L_1q^2&1+L_1q\end{pmatrix}.}
$$
Its <determinant> is one. A different starting point gives a conjugate matrix with the same trace and stability; the orthogonal transverse plane replaces $q$ by $-q$.
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