Form the composition of the given rotation and scaling

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Form both compositions of the rotation $$d:\mathbb{R}^2 \rightarrow \mathbb{R}^2 \mbox{ with } d:\begin{pmatrix} x\\ y \end{pmatrix} \mapsto \begin{pmatrix} x \cos \alpha - y \sin \alpha\\ x \sin \alpha + y \cos \alpha \end{pmatrix}$$ and the scaling $$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \mbox{ with } f:\begin{pmatrix} x\\ y \end{pmatrix} \mapsto \begin{pmatrix} 2x\\ 2y \end{pmatrix}$$ and respectively state the transformation matrix.

If I understood correctly, we need to concatenate the rotation with the scaling in order to get the composition?

Because $d$ and $f$ are linear transformations, $d \circ f$ is also linear.

Now we need to multiply the matrices:

$$\begin{pmatrix} x \cos \alpha - y \sin \alpha\\ x \sin \alpha + y \cos \alpha \end{pmatrix} \begin{pmatrix} 2x\\ 2y \end{pmatrix}$$

But the problem is that both matrices have the exact same amount of rows and columns so a multipication is not possible..

So I think there is a mistake or how do you do this correctly? : /

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You have to represent the linear transformations by matrices: the rotation with angle $\alpha$ corresponds to matrix $$R=\begin{pmatrix}\cos\alpha &-\sin\alpha\\\sin\alpha & \cos\alpha \end{pmatrix}$$ and scaling by $2$ to matrix $$2I=\begin{pmatrix}2&0 \\ 0&2\end{pmatrix}.$$ So the composition is represented by the product of the matrices (in any order, since the unit matrix commutes with all matrices): $$\begin{pmatrix}2\cos\alpha &-2\sin\alpha\\2\sin\alpha & 2\cos\alpha \end{pmatrix}.$$

Note: This is called a similitude with angle $\alpha$ and ratio $2$. It can be seen, in the Argand-Cauchy plane as the multiplication of a complex number by the number $\;2\,\mathrm e^{i\alpha}$.

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What you need to figure out is a matrix representation for both $d$ and $f$,

$$ d = \begin{pmatrix} \cos \alpha & -\sin \alpha\\ \sin \alpha & \cos \alpha \end{pmatrix} $$

and

$$ f = \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix} $$

The matrix representation of $d\circ f$ is just the product of this two matrices

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We have $(d \circ f)(\begin{pmatrix} x\\ y \end{pmatrix} )=d(f(\begin{pmatrix} x\\ y \end{pmatrix} ))=d(\begin{pmatrix} 2x\\ 2y \end{pmatrix} )=\begin{pmatrix} 2x \cos \alpha - 2y \sin \alpha\\ 2x \sin \alpha + 2y \cos \alpha \end{pmatrix}$.