Let $A^{(i\leftrightarrow j)}$ be a matrix given by swapping the $i$ and $j$ columns of a matrix $A$.
Its determinant differs from the original determinant only in sign:

Let $A^{(i\hspace{1mm} \updownarrow \hspace{1mm}j)}$ be a matrix given by swapping the $i$ and $j$ rows of a matrix $A$.
Its determinant differs from the original determinant only in sign:

**Proof**
●

$| A^{(i\hspace{1mm} \updownarrow \hspace{1mm}j)} | = -|A|$
Let $A_{kl}$ be the $k$-th row and $l$-th column element of an $n \times n$ matrix of $A$.
The determinant of $A$ is

(See

"Definition of determinant").
Here, the

sign of the permutation is

$$
\tag{1.1}
$$
Focusing on the $i$ and $j$ rows,
let us write $|A|$ as

Let $A^{(i\hspace{1mm} \updownarrow \hspace{1mm}j)}$ be a matrix given by swapping the $i$ and $j$ rows of $A$,
and $A^{(i \hspace{1mm}\updownarrow \hspace{1mm}j)}_{kl}$
be the $k$-th row and $l$-th column element of $A^{(i\hspace{1mm} \updownarrow \hspace{1mm}j)}$.
We have

The determinant of
$A^{(i \hspace{1mm}\updownarrow \hspace{1mm}j)}$ is

$$
\tag{1.2}
$$
Here,
the last line only changes the order of multiplication.

Let $\xi$ be a

permutation that swaps $\sigma(i)$ and $\sigma(j)$:

$$
\tag{1.3}
$$
For example,
if $\sigma$ is

and if $\xi$ swaps $\sigma(1)$ and $\sigma(3)$,
then

Using $(1.3)$,
$(1.2)$ can be expressed as

$$
\tag{1.3}
$$
The permutation $\xi$ is an odd permutation
because it swaps only once.
Therefore,
if $\sigma$ is an even permutation,
the composite permutation $\xi \circ \sigma(\cdot)$
is an odd permutation,
and if $\sigma$ is an odd permutation,
the composite permutation $\xi \circ \sigma(\cdot)$
is an even permutation.
We therefore see that

$(1.1)$ and this give

holds for any permutation $\sigma$.
Using this, $(1.3)$ can be written as

$$
\tag{1.4}
$$
The permutation $\sigma$ is a mapping that only changes the order of the set
$
\{1,2,\cdots,n\}.
$
For example, if $n=3$,
all the permutations $\sigma$ are

If $\xi$ swaps $\sigma(1)$ and $\sigma(3)$,
all the composite permutations $\xi \circ \sigma $ are

As seen this example,
the composite permutation $\xi \circ \sigma$ is also a mapping that only changes the order.
And
the set

is identical to the set

Therefore,
the summation
$\sum_{\sigma \in S_{n}}$ in $(1.3)$
can be replaced with
$\sum_{\xi \circ \sigma \in S_{n}}$:

Let $\tau$
be
$\xi \circ \sigma$. We have

The right hand side is equal to -$|A|$.
So we obtain

●

$| A^{(i \leftrightarrow j)} | = -|A|$
First,
applying the same discussion as above
to the

transposed matrix $A^{T}$ ,
we have

Generally,
a matrix transposed after swapping columns
is equal to a matrix swapping rows after transpose:

We obtain

Finally,
the matrix transpose property

$|A^{T}| = |A|$
gives