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Given a rectangular matrix $A$, what is the general form to rotate the matrix about the center term, e.g. such that

$$\pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2,1} & a_{2,2}}\longrightarrow\pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\ a_{0,0} & a_{1,0} & a_{2,0}} $$

and possibly the reverse case as well.

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XKCD may be helpful in solving this question. xkcd.com/184 –  akkkk Apr 17 '12 at 21:52
    
@Auke no its not –  gardian06 Apr 17 '12 at 21:54
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1 Answer

up vote 2 down vote accepted

$$\pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0}A^T = \pmatrix{a_{0,2} & a_{1,2} & a_{2,2} \\ a_{0,1} & a_{1,1} & a_{2,1} \\ a_{0,0} & a_{1,0} & a_{2,0}} =A_r $$

Edit: reverse is,

$$A_r^T \pmatrix{0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0} = \pmatrix{a_{0,0} & a_{0,1} & a_{0,2} \\ a_{1,0} & a_{1,1} & a_{1,2} \\ a_{2,0} & a_{2,1} & a_{2,2}} = A $$

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that somewhat answers the example (being that it is a square matrix), but what about the general case which will be a "rectangular" matrix where transpose has no meaning –  gardian06 Apr 17 '12 at 22:14
    
It works for non-square matrices, just make sure the transformation matrix is the right size. The transpose works on rectangular matrices (are you getting mixed up with the inverse?) –  Jeff E Apr 17 '12 at 22:16
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In short: transpose, and multiply by an appropriately-sized exchange matrix. –  J. M. Apr 18 '12 at 0:57
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