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In homogeneous coordinates, a rotation matrix around the origin can be described as

$R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix}$

with the angle $\theta$ and the rotation being counter-clockwise.

A translation amongst $x$ and $y$ can be defined as:

$T(x,y) = \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix}$

As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. The formula of this operations can be described in a simple multiplication of

$T(x,y) * R * T(-x,-y) \qquad (I)$

I find this to be counter-intuitive. In my understanding, it should be

$T(-x,-y) * R * T(x,y) \qquad (II)$

The two formulations are definitely not equal. The first equation yields

$E1 = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & -x\cdot\cos(\theta)+y\cdot\sin(\theta)+x\\\sin(\theta) & \cos(\theta) & -x\cdot\sin(\theta)-y\cdot\cos(\theta)+y \\ 0&0&1\end{bmatrix}$

The second one:

$E2 = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & x\cdot\cos(\theta)-y\cdot\sin(\theta)-x\\\sin(\theta) & \cos(\theta) & x\cdot\sin(\theta)+y\cdot\cos(\theta)-y \\ 0&0&1\end{bmatrix}$

So, which one is correct?

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Your first formula is correct. Remember, the point to which this is applied appears on the RIGHT: $$ T(x,y) * R * T(-x,-y) (P) $$ So to evaluate the expression above, we first translate $P$ by $(-x, -y)$, then rotate the result, then translate back. Let's see what happens when $P$ is the point $(x, y, 1)$. That amounts to evaluating the following product: \begin{align} f((x, y)) &= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix} \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix} \begin{bmatrix}1&0&-x\\ 0& 1&-y\\0&0&1\end{bmatrix} \begin{bmatrix}x\\ y\\1\end{bmatrix}\\ &= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix} \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix} \begin{bmatrix}0\\ 0\\1\end{bmatrix}\\ &= \begin{bmatrix}1&0&x\\ 0& 1&y\\0&0&1\end{bmatrix} \begin{bmatrix}0\\ 0\\1\end{bmatrix}\\ &= \begin{bmatrix}x\\ y\\1\end{bmatrix}\\ \end{align} as expected: the point $(x, y)$ remains fixed by this composite transformation.

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  • $\begingroup$ what would happen if you first translated by the negative Matrix (in other words flipped the translate order)? $\endgroup$ – Startec Jan 17 at 9:00
  • $\begingroup$ If you flipped the translate order, you'd find yourself rotation by the angle $\theta$ around the point $(-x, -y)$ instead of $(x, y)$. $\endgroup$ – John Hughes Jan 17 at 11:40
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The point is, that you're shifting the coordinate system, not the point.
So you don't actually shift the point to the origin, you shift the origin to the point, and then back.

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  • $\begingroup$ Regardless of whether you think of the math as "shifting the coordinate system" or "shifting the point", the first operation you apply, as John Hughes correctly explains, is T(-x, -y). If that transform is applied to the point, the result is (0, 0). IMHO its simpler to get this math correct, if you think of this operation as "shifting the point to the origin". Hmm, or maybe you are working with post-multiplications rather than pre-multiplications. So its a question of POV. $\endgroup$ – ToolmakerSteve Mar 22 '18 at 14:09
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These matrices are left-side multiplicated with vector positions, so the order of multiplication is from right to left - on the right side is the first operation, on the left side - the last one.

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