Finding the matrix of a rotation.

Question

I have to find the corresponding matrix for the following rotation:

$R_ \frac{π}{4}$ (1,1)

My general matrix formula is:

$$\begin{pmatrix}1&0&h\\ 0&1&k\\ 0&0&1 \end{pmatrix}\begin{pmatrix}cosα&-sinα&0\\ sinα&cosα&0\\ 0&0&1 \end{pmatrix}\begin{pmatrix}1&0&-h\\ 0&1&-k\\ 0&0&1 \end{pmatrix} $$

I have said that $h$=1 and $k$=1 therefore my matrix is

$$\begin{pmatrix}1&0&1\\ 0&1&1\\ 0&0&1 \end{pmatrix}\begin{pmatrix}cosα&-sinα&0\\ sinα&cosα&0\\ 0&0&1 \end{pmatrix}\begin{pmatrix}1&0&-1\\ 0&1&-1\\ 0&0&1 \end{pmatrix}$$

$$=\begin{pmatrix}(1)(1)+(0)(0)+(1)(0) & (1)(0)+(0)(1)+(1)(0) & (1)(-1)+(0)(-1)+(1)(1)\\(0)(1)+(1)(0)+(0)(0) & (0)(0)+(1)(1)+(1)(0) & (0)(-1)+(1)(-1)+(1)(1) \\ (0)(1)+(0)(0)+(1)(0) & (0)(0)+(0)(1)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1) \end{pmatrix} \begin{pmatrix}cosα&-sinα&0\\ sinα&cosα&0\\ 0&0&1 \end{pmatrix}$$

$$=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}cosα&-sinα&0\\ sinα&cosα&0\\ 0&0&1 \end{pmatrix}$$

$$=\begin{pmatrix} (1)(cosα)+(0)(sinα)+(0)(0) & (1)(-sinα)+(0)(cosα)+(0)(0) & (1)(0)+(0)(0)+(0)(1)\\ (0)(cosα)+(1)(sinα)+(0)(0) & (0)(-sinα)+(1)(cosα)+(0)(0) & (0)(-1)+(1)(-1)+(0)(1)\\ (0)(cosα)+(0)(sinα)+(1)(0) & (0)(-sinα)+(0)(cosα)+(1)(0) & (0)(-1)+(0)(-1)+(1)(1) \end{pmatrix}$$

$$=\begin{pmatrix} cosα&-sinα& 0\\sinα&cosα&-1 \\ 0 & 0 & 1 \end{pmatrix}$$

Does this matrix look right? My notes has something different for the third column.

Answer 1

You seem to have correctly set up your three matrices for rigid motions in the (coordinatized) Euclidean plane. We can write them $ABA^{-1}$. Then you made your first mistake, it would seem, by multiplying the outer two to each other, to get the correct answer (given your erroneous setup) for $AA^{-1}B$ That you did this right can be seen in the fact that the lefthand matrix after the second equals sign is the identity matrix. Then you seem to have made a mistake in multiplying the identity matrix times your rotation matrix $B$.

In any event, you made a serious mistake in multiplying the two outer matrices together.

Answer 2

Let $R_{\theta}(a,b)\colon \mathbb{R}^2\to\mathbb{R}^2$ be the affine transformation which rotates points by an angle $\theta$ anti-clockwise about the point $(a,b)\in\mathbb{R}^2$. This can be described by a matrix $M\in \mathbb{R}^{3\times 3}$ where $M(x,y,1)^T=(R_{\theta}(a,b)(x,y),1)^T$. In this case the matrix is given by the product of the translation matrix which maps $(a,b)$ to the origin, then the rotation in the $(x-y)$-plane by the corresponding angle and then the translation which maps the origin to $(a,b)$. All together this is the product of matrices given in your question (but with $\theta$ replacing $\alpha$ and $(a,b)=(1,1)$).

Your problem was that you didn't multiply the matrices correctly. You should get a matrix of the form $$M=\begin{pmatrix}\cos\theta&-\sin\theta&f(\theta,a,b)\\\sin\theta&\cos\theta&g(\theta,a,b)\\0&0&1\end{pmatrix}$$

Where I will leave you to find $f$ and $g$. You should also then substitute the relevant values in for $\theta,a$ and $b$.