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Question
If A = `[(cosx, sinx),(-sinx, cosx)]`, then A1 A–1 is
Options
`[(cos2x, - sin2x),(sin2x, cos2x)]`
`[(cos2x, sin2x),(-sin2x, -cos2x)]`
`[(1, - sin 2x),(sin 2x, 1)]`
`[(1, - sin 2x),(-sin 2x, 1)]`
MCQ
Solution
`[(cos2x, - sin2x),(sin2x, cos2x)]`
Explanation:
`|A| = cos^2x + sin^2x` = 1
∴ `A^(-1) = [(cosx, - sinx),(sinx, cosx)]` and `A^T = [(cosx, -sinx),(sin x, cos x)]`
Now `A^T. A^(-1) = [(cosx, -sinx),(sinx, cosx)][(cosx, -si nx),(sinx, cosx)]`
= `[(cos^2x - sin^2x, -2sinx. cosx),(2sinx.cosx, cos^2x - sin^2x)]`
= `[(cos2x, -sin2x),(sin 2x, cos 2x)]`
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