Question 58 of 58intermediate🔧 ApplyMCQ1 mark

Let A = [1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1]\begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}, where 0 ≤θ≤2π\le \theta \le 2 \pi. Then

Exercise: Miscellaneous Exercises on Chapter 4 | Q: 9 | (Page 25)
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Explanation

To find the determinant of matrix A, we expand along the first row using the cofactor expansion method. The determinant simplifies to 2 + 2sin²θ. Since sin²θ ranges from 0 to 1 for all θ in the given interval, the determinant ranges from 2 to 4, making option (D) the correct answer.

Solution Steps

  1. Step 1: Expand det(A) along the first row: det(A) = 1·|1 sinθ; -sinθ 1| - sinθ·|-sinθ sinθ; -1 1| + 1·|-sinθ 1; -1 -sinθ|

  2. Step 2: Calculate each 2×22 \times 2 minor: First minor = 1 + sin²θ, Second minor = 0, Third minor = 1 + sin²θ

  3. Step 3: Substitute values: det(A) = 1·(1 + sin²θ) - sinθ·(0) + 1·(1 + sin²θ) = 2 + 2sin²θ

  4. Step 4: Find the range: Since 0 ≤\leq sin²θ ≤\leq 1, we get 2 ≤\leq det(A) ≤\leq 4, so det(A) ∈ [2, 4]

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