Question

Find the inverse of A = `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]`

Answer 1

|A| = `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]`

= sec θ(sec θ – 0) – tan θ(tan θ – 0) + 0

= sec²θ – tan²θ

= 1 ≠ 0

∴ A⁻¹ exists.

Consider AA⁻¹ = I

∴ `[(sec theta, tan theta, 0),(tan theta, sec theta, 0),(0, 0, 1)]` A⁻¹ = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R₁ → (sec θ)R₁ – (tan θ)R₂, we get

`[(sec^2theta - tan^2theta, sectheta tantheta - sectheta tantheta, 0),(tantheta, sectheta, 0),(0, 0, 1)]` A⁻¹ = `[(sectheta, -tantheta, 0),(0, 1, 0),(0, 0, 1)]`

∴ `[(1, 0, 0),(tantheta, sectheta, 0),(0, 0, 1)]` A⁻¹ = `[(sectheta, -tantheta, 0),(0, 1, 0),(0, 0, 1)]`

Applying R₂ → R₂ – tanθ R₁, we get

`[(1, 0, 0),(0, sectheta, 0),(0, 0, 1)]` A⁻¹ = `[(sectheta, -tantheta, 0),(-sectantheta, 1 + tan^2theta, 0),(0, 0, 1)]`

∴ `[(1, 0, 0),(0, sectheta, 0),(0, 0, 1)]` A⁻¹ = `[(sectheta, -tantheta, 0),(-sectantheta, sec^2theta, 0),(0, 0, 1)]`

Applying R₂ → `(1/sectheta)` R₂, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` A⁻¹ = `[(sectheta, -tantheta, 0),(-tantheta, sectheta, 0),(0, 0, 1)]`

∴ A⁻¹ = `[(sectheta, -tantheta, 0),(-tantheta, sectheta, 0),(0, 0, 1)]`