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Why is the determinant of an elementary permutation matrix equal to $-1$?

I am brand new to determinants and I've tried expanding it and using cofactor expansion, but it's messy and complicated. I would prefer if someone could show me using expansion, but alternative methods are welcome. Thanks.

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Not every permutation matrix has determinant $-1$, but the elementary matrices which are permutation matrices (corresponding to interchanges of two rows) have determinant $-1$. The easy way to see this is that (1) the identity matrix has determinant $1$, and (2) interchanging two rows or columns of a matrix multiplies its determinant by $-1$.

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  • $\begingroup$ Ok, I guess what I'm asking is why does interchanging two rows or columns of a matrix multiply its determinant by -1? $\endgroup$ – user85362 Aug 22 '13 at 1:44
  • $\begingroup$ @AbhishekMallela The determinant is defined to be an alternate multilinear map and alternate means changing the sign if we interchage two vector (i.e. coulumns) and we prove that $\det A=\det A^T$ so we have the same proprety for the rows. $\endgroup$ – user63181 Aug 22 '13 at 1:51
  • $\begingroup$ I'm sorry I don't understand. I would like to see by cofactor expansion. $\endgroup$ – user85362 Aug 22 '13 at 2:05
  • $\begingroup$ For the $2 \times 2$ case it's obvious. For $n > 2$, expand along a row that is not being interchanged, note that the interchange switches two rows of each cofactor, and use mathematical induction. $\endgroup$ – Robert Israel Aug 22 '13 at 4:09
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Recall that if we interchange two rows (or two coulumns) of a matrix $A$ then its determinant change the sign.

The permutation matrix is obtained from the identity matrix (with determinant $1$) by interchanging their rows so its determinant is $\pm 1$.

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The elementary permutation matrix that swaps rows $i$ and $j$ is

$$\mathrm E_{ij} := \mathrm I - \mathrm{e}_i \mathrm{e}_i^\top - \mathrm{e}_j \mathrm{e}_j^\top + \mathrm{e}_i \mathrm{e}_j^\top + \mathrm{e}_j \mathrm{e}_i^\top = \mathrm I - \left( \mathrm{e}_i - \mathrm{e}_j \right) \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top$$

Using Sylvester's determinant identity,

$$\det \left( \mathrm E_{ij} \right) = \det \left( \mathrm I - \left( \mathrm{e}_i - \mathrm{e}_j \right) \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top \right) = 1 - \left( \mathrm{e}_i - \mathrm{e}_j \right)^\top \left( \mathrm{e}_i - \mathrm{e}_j \right) = 1 - \underbrace{\| \mathrm{e}_i - \mathrm{e}_j \|_2^2}_{=2} = \color{blue}{-1}$$

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