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$$T(x,y)=(2x+y+5,3x+2y+2)$$ Check if $T$ is affine and compute $T^{-1}$.

I'm not sure if I'm using the right methods here or if I am missing something, but I went about proving that $T(A)+T(B)=T(A+B)$ where $A=(x_1,y_1)$ and $B=(x_2, y_2)$. I then went on to show $c[T(A)]=T(cA)$ which I believe shows that $T$ is affine. Not too sure if this is the correct way to go about it and not sure where to go form here to get $T^{-1}$? Do I multiply by a $2×2$ matrix $A$ and leave it equal to the identity matrix?

Any assistance would be greatly appreciated!

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  • $\begingroup$ I don't understand what the matrix is in this case. All I see is badly formatted equations ... $\endgroup$ – Matti P. Feb 28 at 13:29
  • $\begingroup$ Sorry I didn't know how to format a matrix in the question the matrix is as displayed better now I feel. Do you understand it now? $\endgroup$ – 11276 Feb 28 at 13:41
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    $\begingroup$ The point is : there is no matrix associated to an affine transformation. Matrices represent linear transformations... So there is no matrix in your task, just an application you have to invert. $\endgroup$ – TheSilverDoe Feb 28 at 13:57
  • $\begingroup$ @TheSilverDoe That’s not entirely true. If you use homogeneous coordinates, then $T$ can be represented by a matrix. $\endgroup$ – amd Feb 28 at 19:41
  • $\begingroup$ Well, yes, thank you for the precision. Anyway it is not as usual as for a linear transformation. $\endgroup$ – TheSilverDoe Feb 28 at 21:20
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$$T\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}2x+y+5\\ 3x+2y+2\end{pmatrix}=\begin{pmatrix}2x+y\\ 3x+2y\end{pmatrix}+\begin{pmatrix}5\\ 2\end{pmatrix}=\begin{pmatrix}2&1\\ 3&2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}5\\ 2\end{pmatrix}$$ showing that $T$ is affine. To calculate the inverse, we have to invert $$A=\begin{pmatrix}2 & 1\\ 3&2\end{pmatrix}$$ first. $$A^{-1}=\begin{pmatrix}2 & -1 \\-3& 2\end{pmatrix}$$ and we get the inverse affine transformation $T^{-1}$ by $$T(x,y)=A^{-1}\left(\begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}5\\ 2\end{pmatrix}\right)=A^{-1}\begin{pmatrix}x\\ y\end{pmatrix}-A^{-1}\begin{pmatrix}5\\2\end{pmatrix}\\=\begin{pmatrix}2 & -1\\ -3& 2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}8\\ -11\end{pmatrix}.$$

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