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Let a, b, c be distinct rational numbers. The value of the 3x3 determinant whose every row is a cyclic permutation of (a²+b²+c², ab+bc+ca, ab+bc+ca) — specifically the matrix with all diagonal entries equal to (a²+b²+c²) and all off-diagonal entries equal to (ab+bc+ca) — is always:
- Zero
- Rational and non-negative
- Rational and negative
- Irrational and positive
Correct answer: Rational and non-negative
Solution
Let p = a²+b²+c², q = ab+bc+ca. The matrix M = (p-q)I + qJ (J = all-ones matrix). Det(M) = (p-q)² * (p+2q). Now p-q = (1/2)[(a-b)²+(b-c)²+(c-a)²] > 0 (since a,b,c distinct). And p+2q = (a+b+c)² >= 0. All quantities are rational (a,b,c rational). So Det = (p-q)²*(a+b+c)² >= 0 and rational.
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