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Question
If A = `[[ cos 2θ sin 2θ],[ -sin 2θ cos 2θ]]`, find A2.
Solution
Given : A= `[[ cos 2θ sin 2θ],[ -sin 2θ cos 2θ]]`
Now,
`A^2=A A`
`⇒A^2= [[ cos 2θ sin 2θ],[ -sin 2θ cos 2θ]]` `[[ cos 2θ sin 2θ],[ -sin 2θ cos 2θ]]`
`⇒A^2=[[cos^2(2θ)-sin^2(2θ) cos(2θ) sin2θ+cos(2θ)sin2θ],[-cos(2θ)sin2θ-sin2θcos2θ -sin^2(2θ)+cos^2(2θ)]]`
`⇒A^2=[[cos(2xx2θ) 2sin2θcos2θ],[-2sin2θcos(2θ) cos(2xx2θ)]]` `[∵cos^2θ-sin^2θ=cos^2(2θ)]`
`⇒A^2=[[cos4θ sin(2xx2θ)],[-sin(2xx2θ) cos4θ]]` `[∵ sin2θ = 2sinθcosθ]`
`⇒A^2=[[cos 4θ sin4θ],[-sin4θ cos4θ]]`
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