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I'm looking for a transformation matrix (or set of transformation matrices) that transforms matrix $\mathbf A = \begin{pmatrix} a&b&i&j\\ c&d&k&l \\ e&f&m&n \\ g&h&o&p \end{pmatrix}$ into matrix $\mathbf B = \begin{pmatrix} -p&o&-h&g\\ -n&m&-f&e \\ -l&k&-d&c \\ -j&i&-b&a \end{pmatrix}$.

I tried multiplying $\mathbf A$ by the anti-diagonal identity matrix on the left and by $\begin{pmatrix} 1&0&0&0\\ 0&-1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \end{pmatrix}$ on the right.

But this provides me with the matrix $\begin{pmatrix} g&-h&o&-p\\ e&-f&m&-n \\ c&-d&k&-l \\ a&-b&i&-j \end{pmatrix}$, which different from matrix $\mathbf B$.

Does anyone have any clue?

Thanks in advance.

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1 Answer 1

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Now multiply that result by

$$\pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}\;:$$

$$\begin{pmatrix} g&-h&o&-p\\ e&-f&m&-n \\ c&-d&k&-l \\ a&-b&i&-j \end{pmatrix}\pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}=\begin{pmatrix}-p&o&-h&g\\-n&m&-f&e\\-l&k&-d&c\\-j&i&-b&a\end{pmatrix}\;,$$

which is what you want. You can do this in one step by multiplying by

$$\begin{pmatrix} 1&0&0&0\\ 0&-1&0&0 \\ 0&0&1&0 \\ 0&0&0&-1 \end{pmatrix}\pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}=\pmatrix{0&0&0&1\\0&0&-1&0\\0&1&0&0\\-1&0&0&0}\;.$$

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  • $\begingroup$ Great! Works just fine! Thanks a lot! I really need to exercise in the elementary matrices business... $\endgroup$
    – Meclassic
    Commented Feb 17, 2013 at 11:34
  • $\begingroup$ @jrojasqu: You’re welcome! Yes, it really is just an exercise in elementary matrices. $\endgroup$ Commented Feb 17, 2013 at 11:36

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