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

Options

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):}]`

Answer

B

Explanation

Given `A=[{:( cos theta , - sin theta ),( sin theta , cos theta ):}]`
and `AB=BA=I`
`implies B=A^(-1)I=A^(-1)`
`=(1)/( cos^(2) theta + sin^(2) theta)[{:( cos theta , sin theta ),(- sin theta , cos theta ):}]`
`implies B=[{:( cos theta , sin theta),( - sin theta , cos theta ):}]`