Question

If A = `[(0, -tan  θ/2),(tan  θ/2, 0)]` and (I₂ + A) (I₂ – A)^–1 = `[(a, -b),(b, a)]` then 13(a² + b²) is equal to ______. 

Options

• 11

• 12

• 13

• 14

Answer 1

If A = `[(0, -tan  θ/2),(tan  θ/2, 0)]` and (I₂ + A) (I₂ – A)^–1 = `[(a, -b),(b, a)]` then 13(a² + b²) is equal to 13. 

Explanation:

I₂ + A = `[(1, -tan  θ/2),(tan  θ/2, 1)]` I₂ – A = `[(1, tan  θ/2),(-tan  θ/2, 1)]`  ...(i)

`\implies` (I₂ – A)^–1 = `1/(1 + tan^2  θ/2) [(1, -tan  θ/2),(tan  θ/2, 1)] `  ...(ii)

Now, (I₂ + A) (I₂ – A)^–1 

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

= `[(cosθ, -sinθ),(sinθ, cosθ)]`

= `[(a, -b),(b, a)]`

On comparing

a = cosθ and b = sinθ, then 13(a² + b²) = 13.