Question

Answer the following question:

If A = `[(costheta, sintheta),(-sintheta, costheta)]`, prove that Aⁿ = `[(cos"n"theta, sin"n"theta),(-sin"n"theta, cos"n"theta)]`, for all n ∈ N

Answer 1

A = `[(costheta, sintheta),(-sintheta, costheta)]`

Let P(n) ≡ Aⁿ = `[(cos"n"theta, sin"n"theta),(-sin"n"theta, cos"n"theta)]`, for all n ∈ N.

Put n = 1

∴ R.H.S. = `[(costheta, sintheta),(-sintheta, costheta)]` = A = L.H.S.

∴ P(n) is true for n = 1.

Let us consider that P(n) is true for n = k.

∴ A^K = `[(cos"k"theta, sin"k"theta),(-sin"k"theta, cos"k"theta)]`   ...(i)

We have to prove that P(n) is true for
n = k + 1,

i.e., to prove that

A^k+1 = `[(cos("k" + 1)theta , sin("k" + 1)theta),(-sin("k" + 1)theta, cos("k" + 1)theta)]`

R.H.S. = `[(cos("k" + 1)theta , sin("k" + 1)theta),(-sin("k" + 1)theta, cos("k" + 1)theta)]`

L.H.S. = A^k+1 = A^k.A

= `[(cos"k"theta, sin"k"theta),(-sin"k"theta, cos"k"theta)] [(costheta, sintheta),(-sintheta, costheta)]`  ...[From (i)]

= `[(cos"k"theta costheta - sin"k"theta sintheta, cos"k"theta sintheta + sin"k"theta costheta),(-sin"k"theta costheta - cos"k"theta sintheta, -sin"k"theta sintheta + cos"k"theta costheta)]`

= `[(cos"k"theta costheta - sin"k"theta sintheta, sin"k" theta costheta + cos"k"theta sintheta),(-(sin"k"theta costheta + cos"k"theta sintheta), cos"k"theta costheta - sin"k"theta sintheta)]`

= `[(cos("k"theta + theta), sin("k"theta + theta)),(-sin("k"theta + theta), cos("k"theta + theta))]`

= `[(cos("k" + 1)theta, sin("k" + 1)theta),(-sin("k" + 1)theta, cos("k" + 1)theta)]`

= R.H.S. 

∴ P(n) is true for n = k + 1.

From all steps above, by the principle of Mathematical induction, P(n) is true for all n ∈ N.

∴ Aⁿ = `[(cos"n"theta, sin"n"theta),(-sin"n"theta, cos"n"theta)]` for all n ∈ N.