Properties and formulas of determinant

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

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

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

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

Proof
  Since the determinant of transpose of $A$ is equal to the determinant of $A$ , we have The determinant of a scalar multiplication of a row vector is equal to the scalar multiplication of the determinant of the original matrix. Therefore, we obtain

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

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