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Question
If P(x) = `[(cosx, sinx),(-sinx, cosx)]`, then show that P(x) . (y) = P(x + y) = P(y) . P(x)
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Solution
We have, P(x) = `[(cosx, six),(-sinx, cosx)]`
∴ P(y) = `[(cosy, siny),(-siny, cosy)]`
Now,
P(x) . P(y) = `[(cosx, sinx),(-sinx, cosx)] [(cosy, siny),(-siny, cosy)]`
= `[(cosx * cosy - sinx * siny, cosx * siny + sinx * cosy),(-sinx * cosy - cosx * siny, -sinx * siny + cosx * cosy)]`
= `[(cos(x + y), sin(x + y)),(-sin(x + y), cos(x + y))]`
= P(x + y) ......(i)
Also,
P(y) . P(x) = `[(cosy, siny),(-siny, cos y)] [(cosx, sinx),(-sinx, cosx)]`
= `[(cosy * cosx - siny * sinx, cosy * sinx + siny * cosx),(-siny * cosx - sinx * cosy, -siny * sinx + cosy * cosx)]`
= `[(cos(x + y), sin(x + y)),(-sin(x + y), cos(x + y))]` .....(ii)
Thus, from (i) and (ii), we get
P(x) . (y) = P(x + y) = P(y) . P(x)
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