Tamil Nadu Board of Secondary EducationHSC Arts Class 12

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

Sum

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

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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