Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x=cy+bz, y=az+cx, and z=bx+ay. Then {a^2} + {b^2} + {c^2} + 2abc is equal to :

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Answer: D. 1 The given equations are \begin{matrix} { - x + cy + bz = 0} \ {cx - y + az = 0} \ {bx + ay - z = 0} \ \end{matrix} As x,y,z are not all zero 	herefore The above system should not have unique (zero) solution \Rightarrow \Delta = 0 \Rightarrow \left| {\begin{matrix} { - 1} & c & b \ c & { - 1} & a \ b & a & { - 1} \ \end{matrix} } ight| = 0 \Rightarrow - 1\left( {1 - {a^2}} ight) - c\left( { - c - ab} ight) + b\left( {ac + b} ight) = 0 \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0 \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1

Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to :

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