Question

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

Answer 1

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)]`

A^T = `[(1, - tanx),(tanx, 1)]`

A^T 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

A^T A^–1 = `[(cos2x, - sin2x),(sin2x, cos2x)]`

Hence proved