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In learning how to rotate vertices about an arbitrary axis in 3D space, I came across the following matrices, which I need to calculate the inverse of to properly "undo" any rotation caused by them:

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How can I get the inverse of both of these matrices?
Thanks.

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2 Answers 2

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The submatrices of interest are rotation matrices, so they are of the form

$$\left[ \begin{array}{ccc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{array} \right]$$

Since $\sin -\theta = -\sin \theta$ and $\cos -\theta = \cos\theta$ and rotating by $-\theta$ is the inverse of rotating by $\theta$, it's easily seen and verified that the inverse is simply

$$\left[ \begin{array}{ccc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{array} \right]$$

So in both cases just negate the off-diagonal entries.

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  • $\begingroup$ Very observant. I would have never figured that out on my own. Thank you. $\endgroup$
    – HartleySan
    Apr 8, 2014 at 23:54
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These are actually quite nice block matrices, for instance the first matrix is already the identity in the bottom right $2\times 2$ matrix, So if we let $\underline{0}=\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0 \\ \end{array} \right]$, and let $\underline{A}=\left[ \begin{array}{ccc} u/\sqrt{u^2+v^2} & v/\sqrt{u^2+v^2} \\ -v/\sqrt{u^2+v^2} & u/\sqrt{u^2+v^2}\\ \end{array} \right]$,

Then $T_{xz}=\left[ \begin{array}{ccc} \underline{A} & \underline{0} \\ \underline{0} & \underline{I} \\ \end{array} \right]$, where $\underline{I}$ is the identity matrix.

Here $T_{xz}^{-1}=\left[ \begin{array}{ccc} \underline{A}^{-1} & \underline{0} \\ \underline{0} & \underline{I} \\ \end{array} \right]$

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  • $\begingroup$ Good observation. Thank you. $\endgroup$
    – HartleySan
    Apr 8, 2014 at 23:53

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