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Let Delta = |1, sin(theta), 1; -sin(theta), 1, sin(theta); -1, -sin(theta), 1| be a 3x3 determinant. Find the interval in which Delta lies.

1. [2, 3]
2. [3, 4]
3. [2, 4]
4. (2, 4)

Correct answer: [2, 4]

Solution

Expanding along the first row: Delta = 1*(1 + sin² theta) - sin(theta)*(-sin(theta) + sin(theta)) + 1*(sin² theta + 1) = (1 + sin² theta) - 0 + (1 + sin² theta) = 2 + 2*sin²(theta). Since sin²(theta) is in [0, 1], Delta is in [2, 4]. Both bounds are attained, so Delta is in [2, 4].

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