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Find the inverse of the rotation matrix where $\theta$ is a fixed angle. Then use your result to solve the system $x=a \cos \theta-b\sin\theta$, $y=a\sin\theta+b\cos\theta$ for $a$ and $b$ in terms of $x$ and $y$.

I'm just really not sure on where to begin with this. Do I need to use a determinant because it is a 2x2 matrix?

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The system is

$$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} \cos \theta & - \sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} \begin{pmatrix} a \\ b \end{pmatrix} $$

If the 2×2 matrix is a rotation, when you invert it you will get the inverse rotation. So you either do it the long way (with 2×2 matrix inversion) or the short way of negating $\theta$.

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  • $\begingroup$ Okay, I've successfully found the inverse of my matrix, but now what am I supposed to do with the system of x and y? $\endgroup$ – jerry2144 Sep 16 '14 at 20:16
  • $\begingroup$ @jerry2144 Now that you have the inverse matrix, just multiply both sides by it to solve for a and b in terms of x and y. $\endgroup$ – user84413 Sep 16 '14 at 21:24
  • $\begingroup$ Did I mention the awkward but correct way of transposing the matrix to get the inverse. $\endgroup$ – John Alexiou Sep 17 '14 at 18:10
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Hint: What is the opposite operation of rotating a vector by $\theta$ in the anti-clockwise direction?

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