If $A = [\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}]$ then show that for all the positive integers $n$ $A^{n} = [\begin{matrix} cos n θ & sin n θ - sin n θ & cos n θ \end{matrix}]$

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Solution

$A = (\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix})$

$A^{2} = (\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix}) (\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix})$

$= (\begin{matrix} {cos}^{2} θ - {sin}^{2} θ & sin θ cos θ + sin θ cos θ - cos θ sin θ - cos θ sin θ & {cos}^{2} θ - {sin}^{2} θ \end{matrix})$

$= (\begin{matrix} cos 2 θ & sin 2 θ - sin 2 θ & cos 2 θ \end{matrix})$

The result is true for $n = 1$ and $2$

Let us assume the result is true for $n = k$

i.e., $A^{k} = (\begin{matrix} cos k θ & sin k θ - sin k θ & cos k θ \end{matrix})$

We have to prove the result for $n = k + 1$

$A^{k + 1} = A^{k} \cdot A$

$= (\begin{matrix} cos k θ & sin k θ - sin k θ & cos k θ \end{matrix}) (\begin{matrix} cos θ & sin θ - sin θ & cos θ \end{matrix})$

$= (\begin{matrix} cos k θ cos θ - sin k θ sin θ & cos k θ sin θ + sin k θ cos θ - (sin k θ cos θ + cos k θ sin θ) & cos k θ cos θ - sin k θ sin θ \end{matrix})$

$= (\begin{matrix} cos (k + 1) θ & sin (k + 1) θ - sin (k + 1) θ & cos (k + 1) θ \end{matrix})$

The result is true for $n = k + 1$ also. Hence by the principle of Mathematical Induction, we can understand that the result is true for all $n \in N$ .

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