Question

Answer the following question:

If A = ${\displaystyle \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}$, prove that Aⁿ = ${\displaystyle \left[\begin{array}{cc}\cos \text{n}\theta & \sin \text{n}\theta \\ -\sin \text{n}\theta & \cos \text{n}\theta \end{array}\right]}$, for all n ∈ N

Answer 1

A = ${\displaystyle \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}$

Let P(n) ≡ Aⁿ = ${\displaystyle \left[\begin{array}{cc}\cos \text{n}\theta & \sin \text{n}\theta \\ -\sin \text{n}\theta & \cos \text{n}\theta \end{array}\right]}$, for all n ∈ N.

Put n = 1

∴ R.H.S. = ${\displaystyle \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}$ = A = L.H.S.

∴ P(n) is true for n = 1.

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

∴ A^K = ${\displaystyle \left[\begin{array}{cc}\cos \text{k}\theta & \sin \text{k}\theta \\ -\sin \text{k}\theta & \cos \text{k}\theta \end{array}\right]}$   ...(i)

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

i.e., to prove that

A^k+1 = ${\displaystyle \left[\begin{array}{cc}\cos (\text{k}+1)\theta & \sin (\text{k}+1)\theta \\ -\sin (\text{k}+1)\theta & \cos (\text{k}+1)\theta \end{array}\right]}$

R.H.S. = ${\displaystyle \left[\begin{array}{cc}\cos (\text{k}+1)\theta & \sin (\text{k}+1)\theta \\ -\sin (\text{k}+1)\theta & \cos (\text{k}+1)\theta \end{array}\right]}$

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

= ${\displaystyle \left[\begin{array}{cc}\cos \text{k}\theta & \sin \text{k}\theta \\ -\sin \text{k}\theta & \cos \text{k}\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]}$  ...[From (i)]

= ${\displaystyle \left[\begin{array}{cc}\cos \text{k}\theta \cos \theta -\sin \text{k}\theta \sin \theta & \cos \text{k}\theta \sin \theta +\sin \text{k}\theta \cos \theta \\ -\sin \text{k}\theta \cos \theta -\cos \text{k}\theta \sin \theta & -\sin \text{k}\theta \sin \theta +\cos \text{k}\theta \cos \theta \end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}\cos \text{k}\theta \cos \theta -\sin \text{k}\theta \sin \theta & \sin \text{k}\theta \cos \theta +\cos \text{k}\theta \sin \theta \\ -(\sin \text{k}\theta \cos \theta +\cos \text{k}\theta \sin \theta )& \cos \text{k}\theta \cos \theta -\sin \text{k}\theta \sin \theta \end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}\cos (\text{k}\theta +\theta )& \sin (\text{k}\theta +\theta )\\ -\sin (\text{k}\theta +\theta )& \cos (\text{k}\theta +\theta )\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{cc}\cos (\text{k}+1)\theta & \sin (\text{k}+1)\theta \\ -\sin (\text{k}+1)\theta & \cos (\text{k}+1)\theta \end{array}\right]}$

= 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 = ${\displaystyle \left[\begin{array}{cc}\cos \text{n}\theta & \sin \text{n}\theta \\ -\sin \text{n}\theta & \cos \text{n}\theta \end{array}\right]}$ for all n ∈ N.