Let A=$(a_{ij})_n$ be a square matrix & B=$(b_{ij})_n$ be another square matrix obtained after interchanging two rows of A, say $p^{th}$ and $q^{th}$. Then |B|=-|A|.
Proof Given:- We have,
$b_{pj}$ = $a_{qj}$
$b_{qj} = $$a_{pj}$ and
$a_{ij}$ = $b_{ij}$ $\forall$ j,i($\neq$p,q)
Consider the transposition $\tau$=(p q). Note that $S_n$={$\sigma\circ\tau$ | $\sigma$ $\epsilon$ $S_n$}
Hence,
$|B|=\sum_{\sigma \epsilon S_n} = \operatorname{sign}(\sigma\circ\tau$) $b_{1,\sigma \circ \tau(1)} b_{2,\sigma \circ \tau(2)}b_{p,\sigma \circ \tau(p)}\dots b_{q,\sigma \circ\tau(q)} b_{n,\sigma \circ \tau(n)}$
$|B|=\sum_{\sigma \epsilon S_n} = \operatorname{sign}(\sigma) \operatorname{sign}( \tau ) b_{1\sigma(1)}$$b_{2\sigma(2)}$....$b_{p\sigma(q)}$.....$b_{q\sigma(p)}$....$b_{n\sigma(n)}$
|B| = sign($\tau$) $\sum_{\sigma \epsilon S_n} sign(\sigma) a_{1\sigma(1)} a_{2\sigma(2)}....a_{q\sigma(q)}.....a_{p\sigma(p)}....a_{n\sigma(n)}$
|B| = -|A|
My doubts are:-
Upon interchanging rows, the order will not change and hence, won't change the sign($\sigma$). Then why sign($\sigma$) is being replaced by sign($\sigma$ o $\tau$)?
How transposition (p q), whose elements represent the row numbers which have been interchanged, been used? Why not some other transposition?
How come $S_n$={$\sigma \circ\tau$ | $\sigma$ $\epsilon$ $S_n$}? How to explain this equality?
See link: Proof of Part 1
Thanks in advance!!