I have to find one affine transformation that maps the point P=(1,1,1) to P'=(-1,-1,-1), the point P=(-1,-1,-1)' to P=(1,1,1) and the point Q=(0,0,0) to Q'=(2,2,2). I started with a sketch and think that it is not possible to map both points with one affine transformation, but I must somehow prove that. So I take the formula: x' = a + Ax and started to fill in what we know about. We know that a = (2,2,2) to be able to map Q and we are looking for a matrix that can also transform P to P'. So I filled in P as x, and P' as x'.
$\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix} + \begin{pmatrix}a11+a12+a13\\a21+a22+a23\\a32+a32+a33\end{pmatrix}*\begin{pmatrix}1\\1\\1\end{pmatrix}$
Then I get 3 equations with 9 variables so I'm not sure how to solve it:
I: a11 + a12 + a13 = -3
II: a21 + a22 + a23 = -3
III: a31 + a32 + a33 = -3
Am I on the right way? How can I solve this equations?
You’ve got another three equations that come from the constraint $P'\mapsto P$ that you can add to the system. You will very quickly find that it’s inconsistent.
However, you can reach the same conclusion without expanding into individual matrix elements. Decompose the transformation into a linear transformation matrix $M$ and a translation, as you’ve done. The given constraints are that $M(1,1,1)^T+(2,2,2)^T=(-1,-1,-1)^T$ and $M(-1,-1,-1)^T+(2,2,2)^T=(1,1,1)^T$. Rearranging the the first one gives $M(1,1,1)^T=(-3,-3,-3)^T$, while by linearity, $M(-1,-1,-1)^T=-M(1,1,1)^T$, so the second constraint becomes $M(1,1,1)^T=(1,1,1)^T$, which is clearly impossible.