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+1 vote
20.2k views
in Matrices & determinants by (41.3k points)

If the system of equations x = cy + bz, y = az + cx and z = bx + ay has a non-zero solution and at least one of a, b, c is a proper fraction, prove that a2 + b2 + c2 < 3 and abc > -1.

1 Answer

+2 votes
by (41.5k points)
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Best answer

We are given that system of equations has non-trivial solution. So

⇒ 1( 1 - a2) + c(-c - ab) - b(ac + b) = 0

⇒ 1 - 2abc - a2 - b2 - c2 = 0

or a2 + b2 + c2 + 2abc = 1

⇒ a2 + b2c2 + 2abc = 1 - b2 - c2 + b2c2 = (1 - b2)(1 - c2)

⇒ (a + bc)2 = (1 - b2)(1 - c2)

Similarly, (b + ac)2 = (1 - a2)(1 - b2) and (c + ab)2 = (1 - a2)(1 - b2)

Hence, (1 - a2), (1 - b2) and (1 - c2) all have same sign. Since at least one of them is proper fraction, it implies all of them are positive.

So 1 - a2 > 0, 1 - b2 > 0, 1 - c2 > 0

⇒ a2 + b2 + c2 < 3

⇒ 1 - 2abc < 3

⇒ abc > -1

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