# How does multiplying by trigonometric functions in a matrix transform the matrix?

I found this comic:

But I can't understand the humor because I can't understand how trig functions affect matrix multiplication. Can someone please explain?

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(+1) for making me laugh... apparently by accident! – The Chaz 2.0 Jun 17 '11 at 3:23

The matrix \left[\begin{align} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{align} \right] when it acts on a vector it rotates the vector by $\theta$ in clockwise direction. Hence when $\theta = 90^{\circ}$, it rotates the vector \left[\begin{align} a_1 \\ a_2 \end{align} \right] from vertical to horizontal clockwise.

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$\theta=-90$... – lhf Jun 3 '11 at 19:12
@lhf: Thats why I have written clockwise direction. – user17762 Jun 3 '11 at 19:14
Oops, missed that clockwise. – lhf Jun 3 '11 at 19:20

Well, the rotation matrix they write there is

$$\left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$$

If you multiply a vector $\left( \begin{array}{c} a_{1} \\ a_{2} \\ \end{array} \right)$ by this matrix then you end up with $\left( \begin{array}{c} a_{2} \\ -a_{1} \\ \end{array} \right)$ If you draw a picture in the $xy$ plane connecting each of $(a_{1},a_{2})$ and $(a_{2},-a_{1})$ with the origin, it will be clear that the latter is a 90 degree rotation of the former.

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That's a $-90$-degree rotation matrix.

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Wow! A highly improbable case of two events occurring simultaneously! Your answer and mine show the same "answered 2 secs ago"! – user17762 Jun 3 '11 at 19:11

Because...

Sorry couldn't resist when it was bumped.

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