<p>Evaluate <br>\[\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}\]</p>
Step-by-Step Solution
Key Concept: Recognize this as an orthogonal transformation matrix (rotation in 3D). For orthogonal matrices, the determinant is always ±1, and the product of rows (or columns) with themselves gives 1 while different rows are orthogonal.
<p><strong>Step 1:</strong> Recognize the matrix structure. The rows are:</p><p>R₁ = (cos α cos β, cos α sin β, -sin α)</p><p>R₂ = (-sin β, cos β, 0)</p><p>R₃ = (sin α cos β, sin α sin β, cos α)</p><p><strong>Step 2:</strong> Verify orthogonality by checking R₁·R₁ = cos²α(cos²β + sin²β) + sin²α = cos²α + sin²α = 1</p><p><strong>Step 3:</strong> Check R₁·R₂ = -cos α cos β sin β + cos α sin β cos β + 0 = 0</p><p><strong>Step 4:</strong> Check R₁·R₃ = cos α sin α cos²β + cos α sin α sin²β - sin α cos α = 0</p><p><strong>Step 5:</strong> Similarly verify R₂·R₂ = sin²β + cos²β = 1 and all cross products = 0</p><p><strong>Step 6:</strong> Since all rows are orthonormal, the matrix is orthogonal. For orthogonal matrices: det(Q)·det(Qᵀ) = det(I) = 1, so [det(Q)]² = 1, giving det(Q) = ±1</p><p><strong>Step 7:</strong> Since the determinant of the product of three orthonormal row vectors equals 1 (proper rotation):</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1