Interchange of rows changes the sign of the determinant

Question

Let $A$ be a square matrix. The determinant of $A$ can be defined as $$\sum_{b\in S_n}a_{1b_1}a_{2b_2}...a_{nb_n}$$

Not breaking this down further, apart from the fact that the first term of this sum is always the product of the terms in the main diagonal of $A$. We take one term from each row and choose the column according to the permutation $b = {1,2,3,...,n}$. If two rows are now interchanged, so does the permutation change its parity from even to odd. Therefore the sign of the determinant is changed as well.

For example, if we swap rows 2 and 3 in a $3 \times 3 $ matrix the first terms $$a_{11}a_{22}a_{33}$$ $$a_{11}a_{32}a_{23}$$ which can be rewritten as $$a_{11}a_{23}a_{32}$$

Which corresponds to an odd permutation. I think my idea is correct, but I'm not sure if its appropriately explained here - thus the example.

I don't understand your definition of the determinant. Should that say $$ \sum_{b \in S_n} a_{1b_1} a_{2b_2} \cdots a_{n b_n}? $$ — Ben Grossmann, Dec 03 '19 at 10:37
Yes you are correct, the b values correspond to the columns of A. — In the blind, Dec 03 '19 at 10:40
Yes, the sign is determined per the parity of permutation, (+) if even, ( -1 ) if odd. — In the blind, Dec 03 '19 at 13:26

Ben Grossmann · Answer 1 · 2019-12-03T11:27:42.620

I find your explanation and example incomprehensible, so I can't tell you whether your ideas are correct. However, since you're looking for a proof that swapping rows changes the sign of the determinant that uses your definition, here is one such proof.

Suppose we permute the rows of $A$. That is, we would like to compute $\det M$ where $m_{ij} = A_{\tau(i),j}$ for some $\tau \in S_n$. We can state that $$ \det(M) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)a_{\tau(1),\sigma(1)} \cdots a_{\tau(n),\sigma(n)}. $$ We can rearrange each product in the sum so that the row-indices appear in increasing order. Note that in the term corresponding to a given choice of $\sigma$, $\tau(i)$ is the row-index corresponding to the column-index $\sigma(i)$. Thus, $i = \tau(\tau^{-1}(i))$ is the row-index corresponding to the column-index $\sigma(\tau^{-1}(i))$. So, we can rewrite the sum as $$ \det(M) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)a_{i,\sigma(\tau^{-1}(1))} \cdots a_{n,\sigma(\tau^{-1}(n))}. $$ We can now reindex this sum. Note that $\{\sigma \circ \tau^{-1} | \sigma \in S_n\} = S_n$. So, if we take $\pi := \sigma \circ \tau^{-1}$, then we can rewrite this sum as $$ \det(M) = \sum_{\pi \in S_n} \operatorname{sgn}(\sigma)a_{i,\pi(1)} \cdots a_{n,\pi(n)} \\= \sum_{\pi \in S_n} \operatorname{sgn}(\pi \circ \tau)a_{i,\pi(1)} \cdots a_{n,\pi(n)} \\= \operatorname{sgn}(\tau) \sum_{\pi \in S_n} \operatorname{sgn}(\pi)a_{i,\pi(1)} \cdots a_{n,\pi(n)} \\ = \operatorname{sgn}(\tau) \det(A). $$ So, we can indeed include that if $\tau$ is a transposition (i.e. a switch of two rows), then $\operatorname{sgn}(\tau) = -1$ and $\det(M) = -\det(A)$ as desired.

For your $3 \times 3$ example: suppose we swap rows $2$ and $3$. Then the sums I write in my proof can be expanded as $$ \det(M) = a_{11}a_{32}a_{23} - a_{11}a_{33}a_{22} + \cdots\\ = a_{11}a_{23}a_{32} - a_{11}a_{22}a_{33} + \cdots\\ = -a_{11}a_{22}a_{33} + a_{11}a_{23}a_{32} \\ = -[a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} + \cdots]. $$ I suspect that this is the kind of thing you were trying to convey with your example.

Interchange of rows changes the sign of the determinant

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