Tamil Nadu Board of Secondary EducationHSC Arts Class 12

A = [1tanx-tanx1], show that AT A–1 = [cos2x -sin2xsin2xcos2x] - Mathematics

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Sum

A = `[(1, tanx),(-tanx, 1)]`, show that AT A–1 = `[(cos 2x,  - sin 2x),(sin 2x, cos 2x)]`

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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

Concept: Inverse of a Non-singular Square Matrix
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Chapter 1: Applications of Matrices and Determinants - Exercise 1.1 [Page 16]

APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 12th Mathematics Volume 1 and 2 Answers Guide
Chapter 1 Applications of Matrices and Determinants
Exercise 1.1 | Q 11 | Page 16

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