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Please have a look at the image below. There is a transformation matrix and graphical representation of 2 coordinate systems. I understand what the numbers in the matrix represent, but I can't figure out how we could write this in terms of equations.

Link to image

I've been watching MIT open course videos on matrices, and from what I've learned the numbers in the matrices are just coefficients of some equations. For example, if we had a system of equations like this:

$x_1=3 y_1+2 y_2$

$x_2=2 y_1+4 y_2$

Then we would have a matrix like this:

\begin{pmatrix}3& 2\\ 2& 4\end{pmatrix}

So I'm trying to reverse engineer this matrix into an equation but can't figure it out.

lioness99a
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user1477107
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  • Are you trying to find an equivalent system of equations to this transformation matrix, or are you trying to formulate a system of equations with which to compute the entries in the matrix? – amd Feb 14 '20 at 22:10
  • I'm trying to find an equivalent system of equations to this transformation matrix – user1477107 Feb 15 '20 at 06:34

1 Answers1

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The equation for this matrix comes from transformation rule: $$ \begin{pmatrix}x_1\\y_1\\z_1\\1\end{pmatrix} = \begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\ R_{21}&R_{22}&R_{23}&t_2\\ R_{31}&R_{32}&R_{33}&t_3\\ 0&0&0&1\end{pmatrix} \begin{pmatrix}x\\y\\z\\1\end{pmatrix} $$ $$ x_1=R_{11}x+R_{12}y+R_{13}z+t_1,\\ y_1=R_{21}x+R_{22}y+R_{23}z+t_2,\\ z_1=R_{31}x+R_{32}y+R_{33}z+t_3, $$

Transformation matrix is usually presented as follows: $$ A_1 = \begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\ R_{21}&R_{22}&R_{23}&t_2\\ R_{31}&R_{32}&R_{33}&t_3\\ 0&0&0&1\end{pmatrix} = \begin{pmatrix} \mathbf R&\mathbf t \\ \mathbf 0& 1 \end{pmatrix}, $$ where $\mathbf R$ is $3\times3$ ($n\times n$ in general) rotation matrix and $\mathbf t$ is a $3\times1$ translation vector.

If we expand the transformation rule $$ \begin{pmatrix}\mathbf r_1\\1\end{pmatrix} = \begin{pmatrix}x_1\\y_1\\z_1\\1\end{pmatrix} = \begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\ R_{21}&R_{22}&R_{23}&t_2\\ R_{31}&R_{32}&R_{33}&t_3\\ 0&0&0&1\end{pmatrix} \begin{pmatrix}x\\y\\z\\1\end{pmatrix} = \begin{pmatrix} \mathbf R&\mathbf t \\ \mathbf 0& 1 \end{pmatrix}\begin{pmatrix}\mathbf r\\1\end{pmatrix} $$ we get an alternative way to present an affine transformation is like: $$ \mathbf r_1=\mathbf R\mathbf r + \mathbf t $$

So now we need to first deal with rotation (we forget about translation as if $\mathbf t=0$ or coordinate origins coincide). We see that if we take the vector $\mathbf i=(1,0,0)^\intercal$, then rotated vector $\mathbf i_1=\mathbf R\mathbf r=(R_{11}, R_{21}, R_{31})^\intercal$. From the picture we know, that $\mathbf i_1$ is looking down, so $\mathbf i_1=(0,0,-1)^\intercal$ and that help us to find the first column in matrix $\mathbf R$. We repeat the procedure for vectors $\mathbf j$ and $\mathbf k$ and find the row of matrix $\mathbf R$.

To find a translation vector is even simpler. The vector $\mathbf r=\mathbf 0$ will go to $\mathbf r_1=\mathbf t$, so $\mathbf t$ is just the position of new origin in old coordinate system. From the picture we see that $\mathbf t=(1,2,2)^\intercal$.

By arranging everything to the transformation matrix, we get the answer.

Vasily Mitch
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  • using the equation you wrote, wouldn't we get that x1=-y+1, which isn't correct? – user1477107 Feb 14 '20 at 14:59
  • Why do you think it isn't correct? Look at the end of the vector $j_1$. If unit vectors on the picture have the length $0.5$, then for $j_1$: $x_1=-y+1=-0.5+1=0.5$. Which is kinda of true: it ends in the middle of the box, just like vector $i$. – Vasily Mitch Feb 14 '20 at 15:25