Evaluate : Δ = |(cosα cosβ, cosα sinβ,-sinα)(-sinβ, cosβ,0)(sinα cosβ,sinα sinβ,cosα)|

Question

 Evaluate : Δ = \(\begin{vmatrix} cos\alpha\,cos\beta& cos\,\alpha\,sin\,\beta& -sin\,\alpha \\[0.3em] -sin\,\beta& cos\,\beta & 0 \\[0.3em] sin\,\alpha\,cos\,\beta & -sin\,\alpha\,sin\,\beta &cos\,\alpha \end{vmatrix}\)

Answer 1

 Δ = \(\begin{vmatrix} cos\alpha\,cos\beta& cos\,\alpha\,sin\,\beta& -sin\,\alpha \\[0.3em] -sin\,\beta& cos\,\beta & 0 \\[0.3em] sin\,\alpha\,cos\,\beta & -sin\,\alpha\,sin\,\beta &cos\,\alpha \end{vmatrix}\)

Expanding along the second row,

⇒ |A| = sinβ (cosα×cosα sinβ + sinα × sinα sinβ) + cosβ (cosα cosβ × cosα + sinα×sinα cosβ) – 0 

|A| = sin²β(cos²α + sin²α) + cos²β (cos²α + sin²α)

|A| = sin²β(1) + cos²β (1) 

|A| = sin²β + cos²β 

|A| = 1