If P(x) = [cosxsinx-sinxcosx], then show that P(x) . (y) = P(x + y) = P(y) . P(x) - Mathematics

Question

If P(x) = $[\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$ , then show that P(x) . (y) = P(x + y) = P(y) . P(x)

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Solution

We have, P(x) = $[\begin{matrix} \cos x & s i x \\ - \sin x & \cos x \end{matrix}]$

∴ P(y) = $[\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}]$

Now,

P(x) . P(y) = $[\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}] [\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}]$

= $[\begin{matrix} \cos x \cdot \cos y - \sin x \cdot \sin y & \cos x \cdot \sin y + \sin x \cdot \cos y \\ - \sin x \cdot \cos y - \cos x \cdot \sin y & - \sin x \cdot \sin y + \cos x \cdot \cos y \end{matrix}]$

= $[\begin{matrix} \cos (x + y) & \sin (x + y) \\ - \sin (x + y) & \cos (x + y) \end{matrix}]$

= P(x + y) ......(i)

Also,

P(y) . P(x) = $[\begin{matrix} \cos y & \sin y \\ - \sin y & \cos y \end{matrix}] [\begin{matrix} \cos x & \sin x \\ - \sin x & \cos x \end{matrix}]$

= $[\begin{matrix} \cos y \cdot \cos x - \sin y \cdot \sin x & \cos y \cdot \sin x + \sin y \cdot \cos x \\ - \sin y \cdot \cos x - \sin x \cdot \cos y & - \sin y \cdot \sin x + \cos y \cdot \cos x \end{matrix}]$

= $[\begin{matrix} \cos (x + y) & \sin (x + y) \\ - \sin (x + y) & \cos (x + y) \end{matrix}]$ .....(ii)

Thus, from (i) and (ii), we get

P(x) . (y) = P(x + y) = P(y) . P(x)

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Chapter 3: Matrices - Exercise [Page 58]

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Chapter 3 Matrices
Exercise | Q 46 | Page 58

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