# What transformation matrix can be used to move a triangle?

In image there is a triangle on a coordinate plane. Coordinates of a triangle are A(1, 2), B(1, 0) and C(3, 0)

If I represent all three coordinates in a 3x3 matrix, it will look like below:

$$\begin{bmatrix}1&1&3\\2&0&0\end{bmatrix}$$

Now I want to translate this triangle to new triangle whose coordinates will be A'(3,2), B'(3,0) and C'(5,0). And my new 3x3 matrix will look like below:

$$\begin{bmatrix}3&3&5\\2&0&0\end{bmatrix}$$

I know this transformation can be done using matrix multiplication. But I don't know with what transformation matrix I should multiply? And How to write transformation Matrix for this type of transformation?

Thank you

• What year/grade are you in? Do you know about homogeneous coordinates? Commented Jul 2, 2020 at 19:32

Take the original vectors $$\begin{bmatrix}1&1&3\\2&0&0\end{bmatrix}$$ and "augment" by adding a new "z" coordinate with the value $$1$$: $$\begin{bmatrix}1&1&3\\2&0&0\\1&1&1\end{bmatrix}$$. Now you can write the $$\begin{bmatrix} 2\\0\end{bmatrix}$$ translation as $$\begin{bmatrix}1&0&2\\0&1&0\\0&0&1 \end{bmatrix}$$ and the effect of the translation on the vectors is $$\begin{bmatrix}1&0&2\\0&1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}1&1&3\\2&0&0\\1&1&1\end{bmatrix}$$.

Note that after the matrix multiplication, the "z" coordinate remains $$1$$.

• The "right" solution depends on the theory that the person who asked the question knows and can use. My bad if I misinterpreted the question, I just meant there is no linear transformation from $\Bbb R^2$ to $\Bbb R^2$ that allows you to do that, I edited my answer. Commented Jul 2, 2020 at 19:29
• @Flutterblaxi so you were telling me this type of translation is not possible? Commented Jul 3, 2020 at 16:44
• @zipperblock I just said it's not possible using 2x2 matrices. Commented Jul 3, 2020 at 17:16
• @Flutterblaxi okay. Then it is possible above two dimension like 3x3, 4x4. Thank you for your answer Commented Jul 5, 2020 at 7:51
• @zipperblock No, it's not about dimensions. It is always possible in $\Bbb R^n$ by using (n+1)x(n+1) matrix multiplication (homogeneous coordinates) but it is not possible to use nxn matrix multiplication (linear transformations), because translations are not linear. So for example in a plane (isomorphic to $\Bbb R^2$) you can't use 2x2 matrices to compute translations and you'll need 3x3. Commented Jul 5, 2020 at 8:04

You can only compute rotations and symmetries by using 2x2 matrix multiplication (i.e. linear transformations from $$\Bbb R^2$$ to $$\Bbb R^2$$), not translations. To obtain the second triangle you'll just need to translate by $$\vec v=(2,0)^T$$.

• By augmenting the triangle vertex vectors you can write the translation as a matrix multiplication. Commented Jul 2, 2020 at 18:38
• What do you mean by "augmenting the (...) vectors"? Commented Jul 2, 2020 at 18:54
• see my answer below Commented Jul 2, 2020 at 19:17