Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 3 / iv

For the slow mode choose $e_B=(1,\beta {)}^T$. The second row gives $c+s\beta ={\lambda }_B\beta$, so to the requested orderFor the fast mode choose $e_A=(\alpha ,1{)}^T$. The first row gives $({\lambda }_A-q^{2}s)\alpha =c{q}^2$, henceThus the conserved mode contains an order-one slaved nonconserved component, while the fast mode contains only an $O(q^2)$ conserved component.