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Q.If A=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}, then prove that An=[cos⁡nθsin⁡nθ−sin⁡nθcos⁡nθ]A^n=\begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}, where n∈Nn\in N.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2020Subjective· 5mImportance★★★★★
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Use mathematical induction: the base n=1n=1 is the matrix itself, and multiplying the n=kn=k form by AA and applying cos⁡(kθ+θ)\cos(k\theta+\theta), sin⁡(kθ+θ)\sin(k\theta+\theta) gives the n=k+1n=k+1 form.

Concept. Prove a statement P(n)P(n) for all natural nn by (a) base case and (b) P(k)⇒P(k+1)P(k)\Rightarrow P(k+1).

Base case n=1n=1. A1=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]A^1=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} — the formula holds.

Inductive step. Assume Ak=[cos⁡kθsin⁡kθ−sin⁡kθcos⁡kθ]A^k=\begin{bmatrix}\cos k\theta&\sin k\theta\\-\sin k\theta&\cos k\theta\end{bmatrix}. Then

Ak+1=AkA=[cos⁡kθsin⁡kθ−sin⁡kθcos⁡kθ][cos⁡θsin⁡θ−sin⁡θcos⁡θ].A^{k+1}=A^k A=\begin{bmatrix}\cos k\theta&\sin k\theta\\-\sin k\theta&\cos k\theta\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}. …

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