# Properties and formulas of determinant

Swap rows and columns

Let $A^{(i\leftrightarrow j)}$ be a matrix given by
swapping the $i$-th and $j$-th 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

Case two columns are equal

Let
$\mathbf{a}_{i}$
and
$\mathbf{a}_{j}$
$(i \neq j)$
denote the $i$-th and $j$-th column vectors of an $n \times n$ matrix $A$,
respectively,
and denote $A$ as

If
$\mathbf{a}_{i}$ is equal to $\mathbf{a}_{j}$,
the

determinant of $A$ is zero:

**Proof**
Let $A^{(i\leftrightarrow j)}$
be a matrix given by swapping the $i$-th and $j$-th columns of $A$.

Its determinant differs from the original determinant $|A|$ only in sign:

$$
\tag{2.1}
$$
(See

the above.)
If
$
\mathbf{a}_{i} = \mathbf{a}_{j}
$
,
we have

In this case,
swapping the $i$-th and $j$-th column vectors does not change the matrix..

Eq. $(2.1)$ and this give

We obtain

Case two rows are equal

Let
$\mathbf{a}_{i}$
and
$\mathbf{a}_{j}$
$(i \neq j)$
denote the $i$-th and $j$-th row vectors of an $n \times n$ matrix $A$,
respectively,
and denote $A$ as

If
$\mathbf{a}_{i}$ is equal to $\mathbf{a}_{j}$,
the

determinant of $A$ is zero:

**Proof**
Let $A^{(i\hspace{1mm}\updownarrow\hspace{1mm} j)}$ be a matrix given by
swapping the $i$-th and $j$-th rows of $A$.

Its determinant differs from the original determinant
only in sign
(

see the above).

$$
\tag{3.1}
$$
If
$\mathbf{a}_{i} = \mathbf{a}_{j}$,

In this case,
swapping the $i$-th and $j$-th rows does not change the matrix.
Therefore

Eq. $(3.1)$ and this give

We obtain

Case if a row is scalar-multiplied

Let $A_{ij}$
be the $i$-th row
and $j$-th column element of an $n \times n$ matrix $A$:

$$
\tag{4.1}
$$
Let $A'$ be the matrix obtained by multiplying $ A_{ij} $ $(j=1,\cdots,n)$ by a scalar
$C$:

$$
\tag{4.2}
$$
The

determinant of $A'$
is $C$ times the determinant of $A$, that is,

**Proof**
Let $A'_{ij}$
be the $i$-th row
and $j$-th column element of $A$.
The determinant of $A'$ is

(See

definition of determinant).
By $(4.1)$ and $(4.2)$,
we have

Therefore we obtain

Case if a column is scalar-multiplied

Let $A_{ij}$
be the $i$-th row
and $j$-th column element of an $n \times n$ matrix $A$:

$$
\tag{5.1}
$$
Let $A'$ be the matrix obtained by multiplying $ A_{ij} $ $(i=1,\cdots,n)$ by a scalar
$C$:

$$
\tag{5.2}
$$
The

determinant of $A'$
is $C$ times the determinant of $A$, that is,

Case if a scalar is multiplied

Let $A$ be an $n \times n$ matrix.
The determinant of $A$ multiplied by a scaler $\alpha$
is $\alpha^n$ times the determinant of $A$.

**Proof**
By

definition of determinant,
the determinant of $A$ is

where $A_{ij}$ is the $i$-th row and $j$-th column element of $A$,
$\sigma$ is the permutation,
$S_{n}$ is
the set of all the permutations,
and
$\mathrm{sgn}(\sigma)$ is the sign of permutation.

In the same way,
the determinant of
$\alpha A$ is

Therefore, we obtain

Case if a row vector is sum

Let $A$ be an $n \times n$ matrix,
and $\mathbf{a}_{i}$ be the $i$-th row vector of $|A|$.

$$
\tag{7.1}
$$
Let $B$ and $C$ be $n \times n$ matrices
given by replacing $\mathbf{a}_{i}$ with $\mathbf{b}_{i}$ and $\mathbf{c}_{i}$,
respectively.

$$
\tag{7.2}
$$
When $\mathbf{a}_{i}$ is represented as the sum ,

$$
\tag{7.3}
$$
$|A| = |B| + |C|$ holds, that is,

**Proof**
From $(7.1)$ and $(7.2)$,
the determinants of $|A|$, $|B|$, and $|C|$ are

Let
$A_{ij}$,
$B_{ij}$,
and
$C_{ij}$
be the $i$-th row and $j$-th column element of $A$, $B$, and $C$, respectively.
By

definition,
the determinants are

From $(7.3)$, we have

We see that
the

determinant of $|A|$ can be written as

We have