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A=[(cos theta, -sin theta),(sin theta, c...

`A=[(cos theta, -sin theta),(sin theta, cos theta)]` and `AB=BA=l`, then `B` is equal to

A

`[(-cos theta, sin theta),(sin theta, cos theta)]`

B

`[(cos theta, sin theta),(- sin theta, cos theta)]`

C

`[(-sin theta, cos theta),(cos theta, sin theta)]`

D

`[(sin theta, - cos theta),(- cos theta, sin theta)]`

Text Solution

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The correct Answer is:
B
Given `A=[(cos theta, -sin theta),(sin theta, cos theta)]` and `AB=BA=l`
`impliesB=A^(-1)l=A^(-1)=1/(cos^(2)theta+sin^(2)theta)[(cos theta, sin theta),(-sin theta, cos theta)]`
`:.B=[(cos theta, sin theta),(- sin theta, cos theta)]`
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