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Q.
Suppose that system of equations $x = cy + bz$, $y = az + cx$, $z = bx +ay$ has non-trivial solution. Then $a^{2}+b^{2}+c^{2}+2abc=$
Determinants
Solution:
$\left|\begin{matrix}-1&c&b\\ c&-1&a\\ b&a&-1\end{matrix}\right|=0$
Expanding the determinant, we get
$-1\left(1-a^{2}\right)-c\left(-c-ab\right)+b\left(ca+b\right)=0
\Rightarrow \quad a^{2}+b^{2}+c^{2}+2abc=1$.