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Question
If A = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`, show that `"A"^2=[(cos2alpha, sin2alpha),(-sin2alpha, cos2alpha)]`
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Solution
A2 = A.A = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)] [(cosalpha, sinalpha),(-sinalpha, cosalpha)]`
`= [(cos^2alpha - sin^2alpha, cosalphasinalpha + cosalphasinalpha),(-cosalphasinalpha - cosalphasinalpha, -sin^2alpha + cos^2alpha)]`
= `[(cos^2alpha - sin^2alpha, 2sinalpha cosalpha),(-2sinalpha cosalpha, cos^2alpha - sin^2alpha)]`
`=[(cos2alpha, sin2alpha),(-sin2alpha, cos2alpha)]`
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