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If A = [cosαsinα-sinαcosα], show that A2=[cos2αsin2α-sin2αcos2α]

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प्रश्न

If A = `[(cosalpha, sinalpha),(-sinalpha, cosalpha)]`, show that `"A"^2=[(cos2alpha, sin2alpha),(-sin2alpha, cos2alpha)]`

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उत्तर

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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पाठ 4: Determinants and Matrices - Exercise 4.6 [पृष्ठ ९५]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 4 Determinants and Matrices
Exercise 4.6 | Q 23 | पृष्ठ ९५

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