Question

A = `[(cos theta, - sin theta),(-sin theta, -cos theta)]` then find A⁻¹ 

Answer 1

|A| = `|(cos theta, - sin theta),(-sin theta, - cos theta)|`

= – cos²θ – sin²θ

= –1 ≠ 0

∴ A^–1 exists.

A₁₁ = (–1)¹⁺¹ M₁₁ = M₁₁ = – cos θ

A₁₂ = (–1)¹⁺² M₁₂ = – M₁₂ = sin θ

A₂₁ = (–1)²⁺¹ M₂₁ = – M₂₁ = sin θ

A₂₂ = (–1)²⁺² M₂₂ = M₂₂ = cos θ

∴ adj (A) = `[(- cos theta, sin theta),(sin theta, cos theta)]^"T"`

= `[(- cos theta, sin theta),(sin theta, cos theta)]`

A⁻¹ = `1/|"A"|` adj (A)

= `[(cos theta, -sin theta),(-sin theta, -cos theta)]`