Say true or false:
If $x = c y + b z, y = a z + c x, z = b x + a y$ , where $x, y, z$ are not all zero, then $a^{2} + b^{2} + c^{2} + 2 a b c + 1 = 0$

True

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False

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Solution

The correct option is B False
$∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b c & - 1 & a b & a & - 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow (1 - a^{2}) + c (- c - a b) - b (a c + b) = 0$
$\Rightarrow 1 - a^{2} - c^{2} - a b c - b^{2} - a b c = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c - 1 = 0$
Hence, option B.

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