If $A = (\begin{matrix} cos θ & i sin θ i sin θ & cos θ \end{matrix})$ , where $i = \sqrt{- 1}$ , then by the principle of Mathematics Induction prove that $A^{n} = (\begin{matrix} cosn θ & i sinn θ i sinn θ & cosn θ \end{matrix})$

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Solution

$\begin{matrix} A = (\begin{matrix} cos θ & i sin θ i sin θ & cos θ \end{matrix}) i = \sqrt{- 1} A^{n} = (\begin{matrix} cosn θ & i sinn θ i sinn θ & cosn θ \end{matrix}) A^{n} = {cosn}^{2} θ - i^{2} {sin}^{2} n θ = 1 \end{matrix}$
Or else solve by putting
$\begin{matrix} n = 1, 2, 3, 4........ A^{2} = (\begin{matrix} cos 2 θ & i sin 2 θ i sin 2 θ & cos 2 θ \end{matrix}) A^{2} = cos 2^{2} θ - i^{2} {sin}^{2} 2 θ = 1 \end{matrix}$
Hence Prooved

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If A=(cosθisinθisinθcosθ), where i=√−1, then by the principle of Mathematics Induction prove that An=(cosnθisinnθisinnθcosnθ)

A=(cosθisinθisinθcosθ)i=√−1An=(cosnθisinnθisinnθcosnθ)An=cosn2θ−i2sin2nθ=1Or else solve by putting n=1,2,3,4........A2=(cos2θisin2θisin2θcos2θ)A2=cos22θ−i2sin22θ=1 Hence Prooved

If $A = (\begin{matrix} cos θ & i sin θ i sin θ & cos θ \end{matrix})$ , where $i = \sqrt{- 1}$ , then by the principle of Mathematics Induction prove that $A^{n} = (\begin{matrix} cosn θ & i sinn θ i sinn θ & cosn θ \end{matrix})$