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Question

For (i)A=[cosαsinαsinαcosα], verify that A'A=I.

For (ii)A=[sinαcosαcosαsinα],  verify that A'A=I.


Solution

Here, A=[cosαsinαsinαcosα]A=[cosαsinαsinαcosα]=[cosαsinαsinαcosα]AA=[cosαsinαsinαcosα][cosαsinαsinαcosα]

=[(cosα)(cosα)+(sinα)(sinα)(cosα)(sinα)+(sinα)(cosα)(sinα)(cosα)+(cosα)(sinα)sin2α+cos2α]=[cos2α+sin2αcosαsinαsinαcosαsinαcosαcosαsinαsin2α+cos2α]=[1001]=I.[sin2α+cos2α=1]

Here, A=[sinαcosαcosαsinα]A=[sinαcosαcosαsinα]=[sinαcosαcosαsinα]
AA=[sinαcosαcosαsinα][sinαcosαcosαsinα]=[(sinα)(sinα)+(cosα)(cosα)(sinα)(cosα)+(cosα)(sinα)(sinα)(cosα)+(sinα)(cosα)(cosα)(cosα)+(sinα)(sinα)]=[sin2α+cos2αsin α cos αcosα sin αsin α cos αsin α cos αcos2α+sin2 α]=[1001]=1.[sin2α+cos2α=1]
Hence, we have verified that A'A=I.


Mathematics

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