Math miscellaneousHard
Question
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 a2 + b2 + c2 + 2abc is equal to
Options
A.2
B.-1
C.0
D.1
Solution
The system of equations x - cy - bz = 0, cx - y + az = 0 and bx + ay - z = 0 have non-trivial solution if
= 0 ⇒ 1(1 - a2) + c(-c - ab) - b(ca + b) = 0
⇒ a2 + b2 + c2 + 2abc = 1.
= 0 ⇒ 1(1 - a2) + c(-c - ab) - b(ca + b) = 0⇒ a2 + b2 + c2 + 2abc = 1.
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