Question 51 of 58intermediate🔧 ApplyNumerical4 marks

Evaluate ∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α∣\begin{vmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{vmatrix}.

Correct Answer

1

Exercise: Miscellaneous Exercises on Chapter 4 | Q: 2 | (Page 24)
For More Understanding

Explanation

This is Question 2 from Miscellaneous Exercises on Chapter 4. The determinant can be evaluated by expanding along any row or column. Expanding along the third column is efficient since it contains a zero element, reducing calculations. The key insight is recognizing that terms simplify using the trigonometric identity sin²θ + cos²θ = 1 at multiple stages.

Solution Steps

  1. Step 1: Let Δ = the given determinant. Expanding along C₃ (third column):

  2. Δ = (-sin α) × [determinant of submatrix from R₂,R₃ and C₁,C₂] - 0 + (cos α) × [determinant of submatrix from R₁,R₂ and C₁,C₂]

  3. Step 2: First submatrix: | -sin β cos β | | sin α cos β sin α sin β | = (-sin β)(sin α sin β) - (cos β)(sin α cos β) = -sin α(sin²β + cos²β) = -sin α

  4. Step 3: Second submatrix: | cos α cos β cos α sin β | | -sin β cos β | = (cos α cos β)(cos β) - (cos α sin β)(-sin β) = cos α(cos²β + sin²β) = cos α

  5. Step 4: Substituting back: Δ = (-sin α)(-sin α) + (cos α)(cos α) = sin²α + cos²α = 1

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