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If A=$\begin{bmatrix}cos\theta & -sin\theta\\sin\theta & cos\theta\end{bmatrix}$.Then show that $A^n=\begin{bmatrix}cos n \theta&-sin n\theta\\sin n\theta & cos n\theta\end{bmatrix}$where n is a positive integer.

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  • Here we will use the principle of mathematical induction.
Let P(n):$A^n=\begin{bmatrix}cos n\theta & -sin n\theta\\sin n\theta & cos n\theta\end{bmatrix}$
When n=1
LHS: $A=\begin{bmatrix}cos \theta & -sin \theta\\sin \theta & cos \theta\end{bmatrix}$
RHS: $\begin{bmatrix}cos \theta & -sin \theta\\sin \theta & cos \theta\end{bmatrix}$
P(1) is true--------(1)
Let P(m) be true $\Rightarrow A^m=\begin{bmatrix}cos m\theta & -sin m\theta\\sin m\theta & cos m\theta\end{bmatrix}$-----(2)
To prove P(m+1) is true
(i.e)$A^{m+1}=\begin{bmatrix}cos (m+1)\theta & -sin(m+1) \theta\\sin(m+1) \theta & cos(m+1) \theta\end{bmatrix}$------(3)
Multiply both side of (3) by matrix A,we get
$A^{m+1}=A\begin{bmatrix}cos m\theta & -sin m\theta\\sin m\theta & cos m\theta\end{bmatrix}$--------(4)
$\Rightarrow \begin{bmatrix}cos \theta & -sin \theta\\sin \theta & cos \theta\end{bmatrix}\begin{bmatrix}cos m\theta & -sin m\theta\\sin m\theta & cos m\theta\end{bmatrix}$
$\quad=\begin{bmatrix}cos\theta cos m\theta& -cos\theta sin m\theta-sin\theta cos m\theta\\sin\theta cos m\theta+cos\theta sinm\theta&-sin\theta sin m\theta+cos\theta cos m\theta\end{bmatrix}$
$\quad=\begin{bmatrix}cos (m+1)\theta & -sin(m+1) \theta\\sin(m+1) \theta & cos(m+1) \theta\end{bmatrix}$-----(4)
Hence P(m+1) is true whenever P(m) is true.
From (1) & (4) it is proved from the principle of mathematical induction that P(n) is true for all natural numbers n.
answered Apr 4, 2013 by sreemathi.v

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