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Consider the determinant equation: det([[a1 + b1*x, a1*x + b1, c1], [a2 + b2*x, a2*x + b2, c2], [a3 + b3*x, a3*x + b3, c3]]) = 0. Which of the following are possible conditions that guarantee this holds for all choices of ai, bi, ci?

1. x = 1 for all ai, bi where 1 <= i <= 3
2. x = -1 for all ai, bi where 1 <= i <= 3
3. det([[a1, b1, c1], [a2, b2, c2], [a3, b3, c3]]) = 0
4. x = +-2 for all ai, bi where 1 <= i <= 3

Correct answer: x = -1 for all ai, bi where 1 <= i <= 3

Solution

By expanding: det = (1 - x²) * det([[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]]). So the expression is zero when x = 1, x = -1, or when the base determinant is zero. However x = 1 makes ALL columns proportional in a trivial way that reduces the determinant to 0 regardless; and x = -1 similarly collapses it. Both x=1 and x=-1 work universally.

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