Question

If A = `[(cosθ, -sinθ, 0),(sinθ, cosθ, 0),(0, 0, 1)]`, find A^–1

Answer 1

A = `[(cosθ, -sinθ, 0),(sinθ, cosθ, 0),(0, 0, 1)]`

∴ |A| = `|(cosθ, -sinθ, 0),(sinθ, cosθ, 0),(0, 0, 1)|`

= cos θ (cos θ) + sin θ (sin θ)

= cos²θ + sin²θ

= 1 ≠ 0

`\implies` A^–1 exists.

We write: AA^–1 = I

∴ `[(cosθ, -sinθ, 0),(sinθ, cosθ, 0),(0, 0, 1)]A^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

`R_1 → R_1 cos θ + R_2 sin θ`

`[(1, 0, 0),(sinθ, cosθ, 0),(0, 0, 1)]A^-1 = [(cosθ, sinθ, 0),(0, 1, 0),(0, 0, 1)]`

`R_2 → R_2 - (sinθ)R_1`

`[(1, 0, 0),(0, cosθ, 0),(0, 0, 1)]A^-1 = [(cosθ, sinθ, 0),(-sinθcosθ, 1 - sin^2θ, 0),(0, 0, 1)]`

`R_2 → 1/cosθ R_2`

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]A^-1 = [(cosθ, sinθ, 0),(-sinθ, cosθ, 0),(0, 0, 1)]`

∴ A^–1 = `[(cosθ, sinθ, 0),(-sinθ, cosθ, 0),(0, 0, 1)]`