Question
If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, prove that $\mathrm{A}^n=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$, for all $n \in \mathbb{N}$.

Answer

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

Let $P(n)=A^{\mathbb{n}}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$, for all $n \in N$.

Step 1: Put $n=1$

$\therefore \quad$ R.H.S. $=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]=\mathbf{A}=$ L.H.S.

$\therefore \quad P(n)$ is true for $n=1$.

Step 2: Let us consider that $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{k}$.

$\therefore \quad A^k=\left[\begin{array}{cc}\cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta\end{array}\right]$

...(i)

Step 3: We have to prove that $P(n)$ is true for

$\mathrm{n}=\mathrm{k}+\mathrm{l}$

i.e., to prove that

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

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

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

$=$ R.H.S.

$\therefore \quad P(n)$ is true for $n=k+1$.

Step 4: From all steps above, by the principle of

Mathematical induction, $P(n)$ is true for all

$\mathrm{n} \in \mathrm{N}$.

$\therefore \quad A^n=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$ for all $n \in N$.

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