Let $a . b, c$ be any real numbers, suppose that there are real number $x, y, z$ not all zero such that $x = c y + b z, y = a z + c x, z = b x + a y,$ then $a^{2} + b^{2} + c^{2} + 2 a b c$ equal to

-1

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Solution

The correct option is A 1
Given system of equations can be written as $A X = 0$
where $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - c & - b - c & 1 & - a - b & - a & 1 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦, O = ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦$
For non-trivial solution,
$| A | = 0$
$∣ ∣ ∣ ∣ \begin{matrix} 1 & - c & - b - c & 1 & - a - b & - a & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 1 - a^{2} - b^{2} - c^{2} - 2 a b c = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b c = 1$

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