TITLE: Solving a simple system of equations
QUESTION [5 upvotes]: Given the simultaneous equations $$A\cos{(\sqrt{\lambda}\pi)} + B\sin{(\sqrt{\lambda}\pi)} = 0$$ $$A\cos{(2\sqrt{\lambda}\pi)}+B\sin{(2\sqrt{\lambda}\pi)} = 0$$ We want to show this has not trivial solutions (ie. solutions when $A\not=0$ and $B\not= 0$). In my notes I have that this gives non-trivial solutions when $$\sin{(2\sqrt{\lambda}\pi)}\cos{(\sqrt{\lambda}\pi)} - \cos{(2\sqrt{\lambda}\pi)}\sin{(\sqrt{\lambda}\pi)}  = 0$$ but can't quite see why. Can someone explain, thanks.

REPLY [6 votes]: If the system
$$Au+Bv=0$$
$$Aw+Bz=0$$
has a non-trivial solution, then
$$u=-\frac{Bv}A=\frac{Avw}{Az}=\frac{vw}z$$
Therefore,
$$uz-vw=0$$
Remark: The case $z=0$ must be considered in a different but easy way.