Question

If $x=cy+bz,y=az+cx,z=bx+ay$ where $x,y,z$ are not all zeros, then find the value of $a^2+b^2+c^2+2abc$

Answer 1

Given system of equations can be written as
$\left[\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
where $A=\left[\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
Since, x, y, z are not all zero, so for the system to be consistent
$\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$
$-1(1-a^2)-c(-c-ab)+b(ac+b)=0$
$\Rightarrow a^2+b^2+c^2+2abc=1$