# Determinant properties and formulas

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: $$\tag{1.1}$$ 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: $$\tag{1.2}$$
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 $$\tag{1.3}$$ 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 $$\tag{1.4}$$ Here, the last line only changes the order of multiplication.
Let $\xi$ be a permutation that swaps $\sigma(i)$ and $\sigma(j)$: $$\tag{1.5}$$ For example, if $\sigma$ is and if $\xi$ swaps $\sigma(1)$ and $\sigma(3)$, then Using $(1.5)$, $(1.4)$ can be expressed as $$\tag{1.6}$$ 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 $(1.3)$ and this give holds for any permutation $\sigma$. Using this, $(1.6)$ can be written as $$\tag{1.7}$$ 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 $(1.7)$ 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: $$\tag{2.1}$$
Proof
Let $A^{(i\leftrightarrow j)}$ be a matrix given by swapping the $i$-th and $j$-th columns of $A$. By $(1.1)$, we have $$\tag{2.2}$$ If $\mathbf{a}_{i} = \mathbf{a}_{j}$ , In this case, swapping the $i$-th and $j$-th column vectors does not change the matrix.. Eq. $(2.2)$ 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: $$\tag{3.1}$$
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). $$\tag{3.2}$$ 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. $(3.2)$ 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, $$\tag{4.3}$$
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, $$\tag{5.3}$$
Proof
Since the determinant of transpose of $A$ is equal to the determinant of $A$ , we have By (4.3), we obtain 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 $A$. $$\tag{6.1}$$
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, $$\tag{7.4}$$
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 Case if a column vector is sum
Let $A$ be an $n \times n$ matrix, and $\mathbf{a}_{i}$ be the $i$-th column vector of $A$. $$\tag{8.1}$$ Let $B$ and $C$ be $n \times n$ matrices given by replacing $\mathbf{a}_{i}$ with $\mathbf{b}_{i}$ and $\mathbf{c}_{i}$, respectively. $$\tag{8.2}$$ When $\mathbf{a}_{i}$ is represented as the sum , $$\tag{8.3}$$ $|A| = |B| + |C|$ holds, that is, $$\tag{8.4}$$
Proof
By $(8.3)$, we have Since the determinant of the transposed matrix is equal to the determinant of the original matrix, we have Here, $\mathbf{b}_{i}^{T} + \mathbf{c}_{i}^{T}$ is the $i$-th row vector of $A^{T}$. By $(7.4)$, we have For the determinant of each term on the right side, we use again the property that the determinant of the transposed matrix is equal to the determinant of the original matrix, and obtain Let $A$ be an $n \times n$ matrix, and $\mathbf{a}_{i}$ be the $i$-th row vector of $A$. Let $A'$ be an $n \times n$ matrix given by adding the scalar multiple of $j$-th row vector of $A$ to $A$: , where $\alpha$ is the scalar. Then holds.
Proof
By (7.4), we have Applying $(4.3)$ and $(3.1)$ to the second term of the righthand side, we have Therefore we obtain Case if the 1st column is 1000...
Let $A$ be an $n \times n$ matrix whose elements after the second element in the first column are all $0$: $$\tag{10.1}$$ The determinant of $A$ is written as $$\tag{10.2}$$
Proof
The determinant of $A$ is $$\tag{10.3}$$ where $A_{ij}$ $(i,j=1,2,\cdots,n)$ 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 (See "Determinant definition").
Let us write $(10.3)$ by dividing it into the summation for $\sigma(1) = 1$ and the summation for $\sigma(1)\neq 1$. $$\tag{10.4}$$ Since the permutation $\sigma$ is a one-to-one mapping from $\{1,2,⋯,n \}$ to $\{1,2,⋯,n \}$, if $\sigma(1) \neq 1$, any one of the integers from $2$ to $n$ satisfies $\sigma(k) = 1$. For such $k$ , we have $$\tag{10.5}$$ By $(10.1)$, We have and This and $(10.4)$ give $$\tag{10.6}$$ Let $B$ an $(n-1)\times(n-1)$ matrix defined as and $B_{ij}$ be the $i$-th row and $j$-th column element of $B$. We have Using this, $(10.6)$ can be written as Let $\sigma'$ be a map defined as $$\tag{10.7}$$ We have Under the condition $\sigma(1)=1$, $\sigma$ is a one-to-one map from $\{2,3,\cdots,n \}$ to $\{2,3,\cdots,n \}$. By $(10.7)$, $\sigma'$ is a one-to-one map from $\{1,2,\cdots,n-1 \}$ to $\{1,2,\cdots,n-1 \}$. So, $\sigma'$ is a permutation of $\{1,2,\cdots,n-1 \}$ (see "Determinant definition"). We see that where $S_{n-1}$ is the set of all the permutations. By $(10.7)$, $\sigma'$ is an even permutation, if $\sigma$ is an even permutation. And $\sigma'$ is an odd permutation, if $\sigma$ is an odd permutation. Hence the permutation sign of $\sigma'$ is equal to the permutation sign of $\sigma$: $\mathrm{sgn}(\sigma) = \mathrm{sgn}(\sigma').$ We have The summation of the right-hand side is the determinant of $B$. We obtain 