Question

Given that the system of equations $x=cy+bz,y=az+cx,z=bx+ay$ has nonzero solutions and atleast one of the $a,b,c$ is a proper fraction.
$a^2+b^2+c^2$ is

• A. $>2$
• B. $>3$
• C. $<3$
• D. $<2$

Answer 1

The correct option is C $<3$

We know that foe some $x,y,z$ not all $0$
$\left[\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
Since we have non-zero solution, that mean
$\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0\Rightarrow a^2+b^2+c^2+2ab-1=0\Rightarrow c=-ab\pm \sqrt{(1-a^2)(1-b^2)}$
Suppose $\left|a\right|\le 1$ and since $c\in R$
thus $\sqrt{(1-a^2)(1-b^2)}\in R$$\Rightarrow (1-a^2)(1-b^2)\ge 0$
So $\left|b\right|\le 1$ as well
So there are angle $\alpha ,\beta$ such that
$a=\cos \alpha$ and $b=\cos \beta$
So $\left|c\right|=|-ab\pm ab\pm \sqrt{(1-a^2)(1-b^2)}|=|-\cos \alpha \cos \beta \pm \sin \alpha \sin \beta |$

$=|-\cos (\alpha +\beta )|\le 1$
$\Rightarrow a^2+b^2+c^2\le 3$ and $\left|abc\right|\le 1$