# 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