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Prove that, for any $\theta$, $\lambda$, $\mu$

$Q = \pmatrix{\cos{\lambda}\cos{\mu} - \cos{\theta}\sin{\lambda}\sin{\mu}&\sin{\lambda}\cos{\mu} + \cos{\theta}\cos{\lambda}\sin{\mu}&\sin{\theta}\sin{\mu}\\-\cos{\lambda}\sin{\mu} - \cos{\theta}\sin{\lambda}\cos{\mu}&-\sin{\lambda}\sin{\mu} + \cos{\theta}\cos{\lambda}\cos{\mu}&\sin{\theta}\cos{\mu}\\ \sin{\theta}\sin{\lambda}&-\sin{\theta}\cos{\lambda}&\cos{\theta}}$

is a proper orthogonal matrix and write down a formula for $Q^{-1}$

How will I be able to do this problem? I know that in order to be a proper orthogonal matrix it must have $\det(Q) = 1$ and it must form orthonormal columns, but how can I show that here?

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en.wikipedia.or/wiki/Euler_angles This is associated with Euler angles and represents rotations about some combination of axes. That's the key to proof –  Ganesh Oct 22 '12 at 0:51
    
@Ganesh the book didn't mention anything about euler angles. –  diimension Oct 22 '12 at 0:55

1 Answer 1

up vote 2 down vote accepted

Just calculate $Q Q^T$ and simplify it to $I$, taking into account the trigonometric identities $\cos^2 \alpha + \sin^2 \alpha = 1$. It's a bit tedious, but has to work if your matrix is correct.

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That was what I was afraid of. I was hoping there was an easier way. Toda Raba!! –  diimension Oct 22 '12 at 1:02
    
Actually the $(1,3)$ element of your matrix is wrong: $\sin \lambda\; \sin \mu$ should be $\sin\theta\; \sin \mu$. –  Robert Israel Oct 22 '12 at 1:05
    
Yup, you are right. Thanks for catching that. I edited it. –  diimension Oct 22 '12 at 1:07

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