Question

If matrix P = `[(0, -tan (θ//2)),(tanθ//2, 0)]`, then find (I – P) `[(cosθ, -sinθ),(sinθ, cosθ)]`

Options

• I + 2P

• 2I + P

• I + P

• None of these

Answer 1

I + P

Explanation:

I – P = `[(1, 0),(0, 1)] - [(0, -tan(θ//2)),(tan(θ//2), 0)]`

= `[(1, tan(θ//2)),(-tan(θ//2), 1)]`

∴ (I – P) `[(cosθ, -sinθ),(sinθ, cosθ)]`

= `[(1, tan(θ//2)),(-tanθ//2, 1)].[(cosθ, -sinθ),(sinθ, cosθ)]`

= `[(cosθ + tan(θ//2)sinθ, -sinθ + tan(θ//2)cosθ),(-tan(θ//2) cosθ + sinθ, tan(θ//2)sinθ + cosθ)]`

= `[(1 - 2sin^2(θ//2), -2sin(θ//2)cos(θ//2)),(+2sin^2(θ//2), + tan(θ//2)(2cos^2 (θ//2) - 1)),(-tan(θ//2)(2cos^2θ//2 -1), tan(θ//2)(2sin(θ//2) cos(θ//2))),(+2sin(θ//2)cos(θ//2), +(1 - 2sin^2(θ//2)))]`

= `[(1, -tan(θ//2)),(tan θ//2, 1)]`

= I + P