# Matrix of an affine transformation

Let $$p_0,p_1,p_2,p_3$$ the vertices of a tetrahedron in $$R^3$$ and $$f$$ the affine transformation such that $$f(p_0)=p_1 \hspace{3ex} f(p_1)=p_2 \hspace{3ex} f(p_2)=p_3 \hspace{3ex} f(p_3)=p_0$$

Find the matrix of $$f$$ with $$p_0$$ as origin and $$B=\{\overline{p_0p_1},\hspace{2ex}\overline{p_1p_2},\hspace{2ex}\overline{p_2p_3}\}$$ as basis.

My work

$$\vec{f}(\overline{p_0p_1})=\overline{f(p_0)f(p_1)}=\overline{p_1p_2}$$

$$\vec{f}(\overline{p_1p_2})=\overline{f(p_1)f(p_2)}=\overline{p_2p_3}$$

$$\vec{f}(\overline{p_2p_3})=\overline{f(p_3)f(p_0)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$$

Then $$M_{B,B}(\vec{f})=\left(\begin{array}{ccc} 0&0&-1\\ 1&0&-1\\ 0&1&-1\\ \end{array}\right)$$

So the matrix is $$\left(\begin{array}{cccc} 1&0&0&0\\ ?&0&0&-1\\ ?&1&0&-1\\ ?&0&1&-1\\ \end{array}\right)$$

Question: Is my work correct? How can I find the $$f(p_0)$$coordinates?

• Each column is the image of a basis vector. You’re given that $f(p_0)=p_1$ and $\overline{p_0p_1}=p_1-p_0$, so...
– amd
Commented May 30, 2019 at 0:38

$$\vec{f}(\overline{p_2p_3})=\overline{f(p_3)f(p_0)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$$
$$\vec{f}(\overline{p_2p_3})=\overline{f(p_\color{red}2)f(p_\color{red}3)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$$
You ask "how do I find the $$f(p_0)$$ coordinates?"
I'm not sure exactly how your text/teacher is expressing affine xforms as matrices, so I don't know how to fill in the first column, but you do know that $$f(p_0) = p_1$$, so you can write $$f(p_0) = p_1 = p_0 + (p_1 - p_0) = p_0 + b_1$$ where $$b_1$$ is the first element of your basis. That makes me suspect that your matrix should be something like $$\left(\begin{array}{cccc} 1&0&0&0\\ \color{red}1&0&0&-1\\ \color{red}0&1&0&-1\\ \color{red}0&0&1&-1\\ \end{array}\right)$$