If A-cos-sin- 2 sin-2 sin-cos-sin- prove thatAn -cos n -sin n - 2 sin n -2 sin n -cos n -sin n - for all n - N

We shall prove the result by the principle of mathematical induction on n. Step 1: nbsp; If n = 1, by definition of integral power of a matrix, we have A1=c ...

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Standard XII

Mathematics

Linear Functions

If A=cosα+sin...

Question

If $A = [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}]$ , prove that

$A^{n} = [\begin{matrix} \cos n α + \sin n α & \sqrt{2} \sin n α \\ - \sqrt{2} \sin n α & \cos n α - \sin n α \end{matrix}]$ for all n ∈ N.

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Solution

We shall prove the result by the principle of mathematical induction on n.

Step 1: If n = 1, by definition of integral power of a matrix, we have

$A^{1} = [\begin{matrix} \cos 1 α + \sin 1 α & \sqrt{2} \sin 1 α \\ - \sqrt{2} \sin 1 α & \cos 1 α - \sin 1 α \end{matrix}] = [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}] = A$

So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,
$A^{m} = [\begin{matrix} \cos m α + \sin m α & \sqrt{2} \sin m α \\ - \sqrt{2} \sin m α & \cos m α - \sin m α \end{matrix}]$ ...(1)

Now we shall show that the result is true for $n = m + 1$ .
Here,
$A^{m + 1} = [\begin{matrix} \cos (m + 1) α + \sin (m + 1) α & \sqrt{2} \sin (m + 1) α \\ - \sqrt{2} \sin (m + 1) α & \cos (m + 1) α - \sin (m + 1) α \end{matrix}]$

By definition of integral power of matrix, we have
$A^{m + 1} = A^{m} . A \Rightarrow A^{m + 1} = [\begin{matrix} \cos m α + \sin m α & \sqrt{2} \sin m α \\ - \sqrt{2} \sin m α & \cos m α - \sin m α \end{matrix}] [\begin{matrix} \cos α + \sin α & \sqrt{2} \sin α \\ - \sqrt{2} \sin α & \cos α - \sin α \end{matrix}] [From eq . (1)] \Rightarrow A^{m + 1} = [\begin{matrix} (\cos m α + \sin m α) (\cos α + \sin α) - \sqrt{2} \sin m α (\sqrt{2} \sin α) & (\cos m α + \sin m α) (\sqrt{2} \sin α) + \sqrt{2} \sin m α (\cos α - \sin α) \\ - \sqrt{2} \sin m α (\cos α + \sin α) - (\cos m α - \sin m α) (\sqrt{2} \sin α) & - \sqrt{2} \sin m α (\sqrt{2} \sin α) + (\cos m α - \sin m α) (\cos α - \sin α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos m α \cos α + \sin m α \cos α + \cos m α \sin α + \sin m α \sin α - 2 \sin m α \sin α & \sqrt{2} \sin α \cos m α + \sqrt{2} \sin α \sin m α + \sqrt{2} \sin m α \cos α - \sqrt{2} \sin m a \sin α \\ - \sqrt{2} \sin m a \cos α - \sqrt{2} \sin m a \sin α - \sqrt{2} \sin α \cos m α + \sqrt{2} \sin α \sin m α & - 2 \sin α \sin m α + \cos m α \cos α - \sin m α \cos α - \cos m α \sin α + \sin m α \sin α \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m α - α) + \sin (m α + α) - \cos (m α - α) + \cos (m α + α) & \sqrt{2} \sin (m α + α) \\ - \sqrt{2} \sin (m α + α) & \cos (m α + α) - \sin (m α + α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m α + α) + \sin (m α + α) & \sqrt{2} \sin (m α + a) \\ - \sqrt{2} \sin (m α + α) & \cos (m α + α) - \sin (m α + α) \end{matrix}] \Rightarrow A^{m + 1} = [\begin{matrix} \cos (m + 1) α + \sin (m + 1) α & \sqrt{2} \sin (m + 1) α \\ - \sqrt{2} \sin (m + 1) α & \cos (m + 1) α - \sin (m + 1) α \end{matrix}]$

This show that when the result is true for n = m, it is also true for n = m +1.

Hence, by the principle of mathematical induction, the result is valid for all n $\in N$ .

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