Question

Suppose that system of equations $x=cy+bz$, $y=az+cx$, $z=bx+ay$ has non-trivial solution. Then $a^2+b^2+c^2+2abc=$

• A. $2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$\left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$
Expanding the determinant, we get
$-1\left(1-a^2\right)-c\left(-c-ab\right)+b\left(ca+b\right)=0<br/>\Rightarrow a^2+b^2+c^2+2abc=1$.