= Solution
For the supplied equations, $\lambda_+\to1$ and $a_2/a_1=\alpha/(\alpha-\lambda_+)\sim-\alpha$. Thus the higher-<frequency> <normal mode> becomes an upper-mass <oscillation> at $\sqrt{g/l}$ with a vanishing, opposite-sign lower horizontal <displacement>. At $\alpha=0$ the supplied lower equation is $\ddot x_2=0$ and its other mode has zero <frequency>.
This limiting picture belongs to the given matrix. For the physical unequal-mass <double pendulum>, both <normal mode> <angular frequencies> tend to $\sqrt{g/l}$ as the lower mass tends to zero; at exactly zero mass its independent equation needs a limiting test-body interpretation. The source's extra $\alpha$ changes precisely this conclusion.
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