# Isometry and its inverse

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.

• A linear function maps $0$ to $0$, so $f$ is not linear... – Andrew Mar 26 '17 at 11:07
• Ignoring the very obvious abuse of notation and language, think about what the inverse function (say $g$) maps $0$ to? – mdave16 Mar 26 '17 at 11:55
• I edited the notation, but realised I could edit the grammar too - woe is me – mdave16 Mar 26 '17 at 11:59

$\overline{b} = g(0) = f^{-1}(0)$, so solve $f(x) = 0$ so $Ax = (0,1,-1)$. The answer (which depends on the 3 constants in $A$) is the required $\overline{b}$.
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.
• 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. – mdave16 Mar 26 '17 at 18:54