Question

If the system of equations $(a-\lambda )x+by+cz=0bx+(c-\lambda )y+az=0cx+ay+(b-\lambda )z=0$
have non-trivial solution, then there will be three values of $\lambda$ whose product is the circulant.
$\left|\begin{array}{ccc}a& b& c\\ b& C& a\\ c& a& b\end{array}\right|$.

Answer 1

For non-trivial solution

$△=0\therefore \left|\begin{array}{ccc}a-\lambda & b& c\\ b& c-\lambda & a\\ c& a& b-\lambda \end{array}\right|=0$
Above is a cubic in $\lambda$ which will give three values of $\lambda$. If $p{\lambda }^3+q{\lambda }^2+r\lambda +s=0$ and the roots be ${\lambda }_1,{\lambda }_2,{\lambda }_3=-\frac{s}{p}$ where s is the constant term in the given equation and p is the coefficient of ${\lambda }^3$. Clearly the coefficient of ${\lambda }^3$ is $-1$ = p.
The constant term is obtained by putting $\lambda =0$ and it is
$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$ = D = circulant
${\lambda }_1,{\lambda }_2,{\lambda }_3=\frac{D}{-1}$ = D = circulant.