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I am preparing for a computer 3D graphics test and have a sample question which I am unable to solve.

The question is as follows:

For the following 3D transfromation matrix M, find its inverse. Note that M is a composite matrix built from fundamental geometric affine transformations only. Show the initial transformation sequence of M, invert it, and write down the final inverted matrix of M.

$M =\begin{pmatrix}0&0&1&5\\0&3&0&3\\-1&0&0&2\\0&0&0&1\end{pmatrix} $

I only know basic linear algebra and I don't think it is the purpose to just invert the matrix but to use the information in the question to solve this.

Can anyone help?

Thanks

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    The use of these 4x4 matr$i$ces in 3D graphics is somewhat specialized from ordinary matrix operations. Please see [my explanation for a previous Question](http://math.stackexchange.com/a/89937/3111) about the way it is used.2012-06-01

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Here $4\times4$ matrix $M$ represents an affine transformation in 3D. It does so by conveniently combining a $3\times3$ matrix $P$ and a translation $v$ in a way that allows the affine transformation $Pu + v$ to be computed by a single matrix multiplication:

$M \begin{pmatrix} u \\ 1 \end{pmatrix} = \begin{pmatrix} Pu + v \\ 1 \end{pmatrix} $

where $M = \begin{pmatrix} P & v \\ 0 & 1 \end{pmatrix}$.

It follows that "undoing" the affine transformation can be accomplished by multiplying by $M^{-1}$:

$M^{-1} = \begin{pmatrix} P^{-1} & -P^{-1}v \\ 0 & 1 \end{pmatrix} $

Given that $M = \begin{pmatrix} 0 & 0 & 1 & 5 \\ 0 & 3 & 0 & 3 \\ -1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix}$, one computes by any of a variety of ways:

$M^{-1} = \begin{pmatrix} 0 & 0 & -1 & 2 \\ 0 & ^1/_3 & 0 & -1 \\ 1 & 0 & 0 & -5 \\ 0 & 0 & 0 & 1 \end{pmatrix}$

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$\begin{pmatrix}0&0&1&5\\0&3&0&3\\-1&0&0&2\\0&0&0&1\end{pmatrix}$ $=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&2\\0&3&0&3\\0&0&1&5\\0&0&0&1\end{pmatrix}$ $=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix}$

$ =\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix} $

$ =\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&1\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&5\\0&0&0&1\end{pmatrix} $

The inverse of each factor individually is easy, so you can just compute those, then multiply in the reverse order to find the inverse of the matrix. (Remember for matrices $(AB)^{-1}=B^{-1}A^{-1}$, that is what I mean by reverse the factors.) Is this what you call fundamental affine transformations?

The first factor is permutes the rows, the second rescales them, then the last three are operations of the same type which produce a linear combination of of two rows.

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    @hardmath Thanks! I'm always interested in learning these practical approaches.2012-06-01
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I'll take a stab at a second method too, to see if that's what your teacher had in mind.

The idea is to use this post and/or this post, because:

$M=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&2\\0&3&0&3\\0&0&1&5\\0&0&0&1\end{pmatrix}$

$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}1&0&0&-2\\0&1&0&1\\0&0&1&5\\0&0&0&1\end{pmatrix}$

$=\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\left[\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}+\begin{pmatrix}0&0&0&-2\\0&0&0&1\\0&0&0&5\\0&0&0&0\end{pmatrix}\right]$

You can see that the brackets have a unit plus a nilpotent $B$ ($B^2=0$). I think the link above might be able to help you invert that right half quickly.

Yes, of course here $(I+B)^{-1}=I-B$. So we know: $M^{-1}=\begin{pmatrix}1&0&0&2\\0&1&0&-1\\0&0&1&-5\\0&0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0&0\\0&1/3&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\begin{pmatrix}0&0&1&0\\0&1&0&0\\1&0&0&0\\0&0&0&1\end{pmatrix}$ which isn't so hard to compute. Multiply the left pair first, then with the final matrix on the right to get this in only a minute:

\begin{pmatrix}0&0&-1&2\\0&1/3&0&-1\\1&0&0&-5\\0&0&0&1\end{pmatrix}

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    @hardmath Yup, looks like that crept in... will correct it now. That day the TeX was rendering *realllly* slowly and every time I changed something it took 15-20 seconds to come up. So I lost patience...2012-06-03