Evaluate : [(cosα cosβ , cosα sinβ , -sinα ),(-sin β , cosβ , 0),(sinα cosβ , sinα sinβ , cosα )]

Question 1

Question 2

Expanding along the first row, we get

Δ = cosα cosβ(cosβ cosα – 0) – cosα sinβ(–sinβ cosα) – sinα(–sinα sin2β – sinα cos2β)

= cos2α cos2β + cos2α sin2β + sin2α sin2β+ sin2α cos2β

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

= cos2α· 1 + sin2α· 1 = 1

kratos · Answer 1 · 2020-04-18T20:20:55+0000

Expanding along the first row, we get

Δ = cosα cosβ(cosβ cosα – 0) – cosα sinβ(–sinβ cosα) – sinα(–sinα sin2β – sinα cos2β)

= cos2α cos2β + cos2α sin2β + sin2α sin2β+ sin2α cos2β

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

= cos2α· 1 + sin2α· 1 = 1

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