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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
Solution
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.
∴ AK = `[(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
Ak+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. = Ak+1 = Ak.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.
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