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I got this affine map:

$$ f: R^3 \rightarrow R^3: \begin{pmatrix}x\\ y\\ z\\\end{pmatrix} \rightarrow A \cdot \begin{pmatrix}x\\ y\\ z\\\end{pmatrix} + \begin{pmatrix}0\\ -1\\ 1\\\end{pmatrix} $$

with $$ A = \begin{bmatrix} 1 & a_{12} & a_{22}\\ 0 & 1 & a_{21}\\ 0 & 0 & 1 \end{bmatrix}$$

Also given was this information about the inverse (which should also be an affine map?): $$ g=f^{-1}: R^3 \rightarrow R^3: \begin{pmatrix}x\\ y\\ z\\\end{pmatrix} \rightarrow B \cdot \begin{pmatrix}x\\ y\\ z\\\end{pmatrix} + \bar b $$

I have to find $\bar b$. Does anyone have an idea how to find it? I tried inputting some values but I can't seem to get there.

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    $\begingroup$ A linear function maps $0$ to $0$, so $f$ is not linear... $\endgroup$ – Andrew Mar 26 '17 at 11:07
  • $\begingroup$ Ignoring the very obvious abuse of notation and language, think about what the inverse function (say $g$) maps $0$ to? $\endgroup$ – mdave16 Mar 26 '17 at 11:55
  • $\begingroup$ I edited the notation, but realised I could edit the grammar too - woe is me $\endgroup$ – mdave16 Mar 26 '17 at 11:59
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$\overline{b} = g(0) = f^{-1}(0)$, so solve $f(x) = 0$ so $Ax = (0,1,-1)$. The answer (which depends on the 3 constants in $A$) is the required $\overline{b}$.

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So $f$ is an isometry (that is, it's bijective and can be seen as a translation (the $ + \bar a$) and rotation ($A \cdot \bar x$). As you can see geometrically, they must have inverses, because I can translate back and rotate again.

HINT: Let $g$ be the inverse. Then $g(\bar 0) = \bar b$. So $f(\bar b) = 0$ as they are inverses. Now that looks like an equation we can solve.

I'll answer completely in a while, but I'd rather you do the next step first.

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  • $\begingroup$ Sorry I'm new to this but I'm not really able to figure it out yet? I don't think that's an equation I can quite solve but I might just be wrong? $\endgroup$ – Addy E Mar 26 '17 at 16:36
  • $\begingroup$ call $\bar b = [b_1, b_2, b_3]^T$. Then evaluate $f(\bar b)$. You should get three linear equations. I think that's pretty soluble. $\endgroup$ – mdave16 Mar 26 '17 at 18:54

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