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Question
A = `[(1, tanx),(-tanx, 1)]`, show that AT A–1 = `[(cos 2x, - sin 2x),(sin 2x, cos 2x)]`
Solution
A = `[(1, tanx),(-tanx, 1)]`
|A| = 1 + tan2x
= sec2x ≠ 0.A–1 exists.
adj A = `[(1, - tanx),(tanx, 1)]`
A–1 = `1/|"A"|` adj A
= `1/(sec^2x) [(1, - tanx),(tanx, 1)]`
AT = `[(1, - tanx),(tanx, 1)]`
AT A–1 = `1/(sec^2x) [(1, - tanx),(tanx, 1)][(1, -tanx),(tanx, 1)]`
= `1/(sec^2x) [(1 - tan^2x, - tanx - tanx),(tanx + tanx, - tan^2x + 1)]`
= `1/(sec^2x) [(1 - tan^2x, -2tanx), (2tanx, 1 - tan^2x)]`
= `cos^2x [(1 - (sin^2)/(cos^2), -2sinx/cosx),(2 sinx/cosx, 1 - (sin^2x)/(cos^2x))]`
= `[(cos^2x - sin^2x, -2 sinx cosx),(2sinx cosx, cos^2x - sin^2x)]`
∵ cos 2A = cos2A – sin2A
sin 2A = 2 sin A cos A
AT A–1 = `[(cos2x, - sin2x),(sin2x, cos2x)]`
Hence proved
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