Proof of Rules for the Determinant: Rows, Scalar Multiples and the Transpose

Each result cited is universally quantified over the data in its own statement. We work in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices.

Preliminaries. $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, hence a field by Ordered Fields §ordered-field, and sums and products over finite sets are as in Commutative Rings, Fields and Ordered Fields: Standard Notation §rings. $S_{n}$ is a finite nonempty set by Finite Sets: Removing an Element, Injective Self-Maps, Infinite Sets, Sets of Permutations, Pairs and Extended Tuples §permutations, whose elements are exactly the bijections $[n]\to[n]$, by Permutations of a Set and the Set of Permutations of [n] §membership. For $\sigma,\tau\in S_{n}$: $\mathrm{id}_{[n]}$, $\sigma\circ\tau$ and $\sigma^{-1}$ lie in $S_{n}$, composition is associative, $\mathrm{id}_{[n]}$ is neutral, and $\sigma^{-1}\circ\sigma=\sigma\circ\sigma^{-1}=\mathrm{id}_{[n]}$, by Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §identity, Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §composition and Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §inverse.

(D) Let $B\in\mathcal{M}_{n}(\mathbb{R})$. For $\sigma\in S_{n}$, let $q^{B}_{\sigma}:[n]\to\mathbb{R}$ be the map $k\mapsto B_{k\sigma(k)}$ and $p_{B}(\sigma)=\prod_{k\in[n]}B_{k\sigma(k)}$ its product, and let $t_{B}:S_{n}\to\mathbb{R}$, $t_{B}(\sigma)=\operatorname{sgn}(\sigma)\,p_{B}(\sigma)$, all as in The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §terms with $B$ in place of $A$. By Determinant of a Real Square Matrix §determinant and The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §determinant, $\det B=D_{n}(B)=\sum_{\sigma\in S_{n}}t_{B}(\sigma)$.

(R) Reindexing a product. Let $B,C\in\mathcal{M}_{n}(\mathbb{R})$, $\sigma,\tau\in S_{n}$ and let $g:[n]\to[n]$ be a bijection with $C_{i\sigma(i)}=B_{g(i)\tau(g(i))}$ for every $i\in[n]$. Then $p_{C}(\sigma)=p_{B}(\tau)$. Indeed, $q^{C}_{\sigma}=q^{B}_{\tau}\circ g$, as their values agree at every $i\in[n]$; and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing, with the finite set $[n]$ in both roles, the bijection $g$ and the map $q^{B}_{\tau}$, gives $\prod_{i\in[n]}q^{B}_{\tau}(g(i))=\prod_{k\in[n]}q^{B}_{\tau}(k)$, that is, $p_{C}(\sigma)=p_{B}(\tau)$.

Clause identity. By Identity Matrix and Standard Basis Vectors §identity, $I_{n}\in\mathcal{M}_{n}(\mathbb{R})$ has the entries $\delta_{ij}$, with $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i\neq j$. Let $\sigma\in S_{n}$ with $\sigma\neq\mathrm{id}_{[n]}$. Then there is $i\in[n]$ with $\sigma(i)\neq i$, and $q^{I_{n}}_{\sigma}(i)=\delta_{i\sigma(i)}=0$; by Products over Finite Sets: a Single Factor, Vanishing in a Field, and Signs and Monotonicity in an Ordered Ring §zero, $\mathbb{R}$ being a field, $p_{I_{n}}(\sigma)=0$, so $t_{I_{n}}(\sigma)=0$. For $\sigma=\mathrm{id}_{[n]}$, $q^{I_{n}}_{\sigma}(k)=\delta_{kk}=1$ for every $k\in[n]$, so Products over Finite Sets: a Single Factor, Vanishing in a Field, and Signs and Monotonicity in an Ordered Ring §single with $x_{0}=1\in[n]$ gives $p_{I_{n}}(\mathrm{id}_{[n]})=1$; and $\operatorname{sgn}(\mathrm{id}_{[n]})=1$ by Rules for the Sign of a Permutation §identity, so $t_{I_{n}}(\mathrm{id}_{[n]})=1$. By (D) and Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §vanishing with $y=\mathrm{id}_{[n]}$, $\det I_{n}=t_{I_{n}}(\mathrm{id}_{[n]})=1$.

Clause row-linearity. Let $i$, $m$, $c$ and $u$ be as in the clause, and for $l\in[m]$ write $u^{(l)}=u(l)\in\mathbb{R}^{n}$, a map $[n]\to\mathbb{R}$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §membership, and $B^{(l)}=A[i\leftarrow u^{(l)}]$, so that $B^{(l)}_{ij}=u^{(l)}_{j}$ and $B^{(l)}_{kj}=A_{kj}$ for all $j,k\in[n]$ with $k\neq i$, by The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §row-replacement. Let $F=[n]\setminus\{i\}$; $F$ and $\{i\}$ are finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §subset and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small, they are disjoint, and their union is $[n]$.

(a) For $C\in\mathcal{M}_{n}(\mathbb{R})$ and $\sigma\in S_{n}$, $p_{C}(\sigma)=\bigl(\prod_{k\in F}C_{k\sigma(k)}\bigr)\cdot C_{i\sigma(i)}$, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §disjoint-union and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §singleton for the map $q^{C}_{\sigma}$, the products over $F$ and $\{i\}$ being those of its restrictions, by Sums and Products over a Finite Set and over an Interval §subsets.

(b) Let $Q:S_{n}\to\mathbb{R}$ be the map $Q(\sigma)=\prod_{k\in F}A_{k\sigma(k)}$. For $l\in[m]$, $\sigma\in S_{n}$ and $k\in F$, $B^{(l)}_{k\sigma(k)}=A_{k\sigma(k)}$, as $k\neq i$; so the restrictions of $q^{B^{(l)}}_{\sigma}$ and of $q^{A}_{\sigma}$ to $F$ coincide, and both products over $F$ are $Q(\sigma)$. With (a) and $B^{(l)}_{i\sigma(i)}=u^{(l)}_{\sigma(i)}$ this gives, for every $\sigma\in S_{n}$ and $l\in[m]$,

$$t_{A}(\sigma)=\operatorname{sgn}(\sigma)\bigl(Q(\sigma)\,A_{i\sigma(i)}\bigr),\qquad t_{B^{(l)}}(\sigma)=\operatorname{sgn}(\sigma)\bigl(Q(\sigma)\,u^{(l)}_{\sigma(i)}\bigr).$$

(c) Let $G:[m]\times S_{n}\to\mathbb{R}$ be the map $G(l,\sigma)=c_{l}\,t_{B^{(l)}}(\sigma)$. Fix $\sigma\in S_{n}$ and let $w=\operatorname{sgn}(\sigma)Q(\sigma)$. The hypothesis with $j=\sigma(i)$ gives $A_{i\sigma(i)}=\sum_{l\in[m]}c_{l}u^{(l)}_{\sigma(i)}$, so by (b) and Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive,

$$t_{A}(\sigma)=w\sum_{l\in[m]}c_{l}u^{(l)}_{\sigma(i)}=\sum_{l\in[m]}w\,\bigl(c_{l}u^{(l)}_{\sigma(i)}\bigr)=\sum_{l\in[m]}G(l,\sigma).$$

Fix $l\in[m]$. By Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive and (D),

$$\sum_{\sigma\in S_{n}}G(l,\sigma)=c_{l}\sum_{\sigma\in S_{n}}t_{B^{(l)}}(\sigma)=c_{l}\det B^{(l)}.$$

Hence, by (D) and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §product for $+$ on $\mathbb{R}$ with the finite sets $[m]$ and $S_{n}$ and the map $G$,

$$\det A=\sum_{\sigma\in S_{n}}\sum_{l\in[m]}G(l,\sigma)=\sum_{l\in[m]}\sum_{\sigma\in S_{n}}G(l,\sigma)=\sum_{l=1}^{m}c_{l}\det A[i\leftarrow u^{(l)}].$$

Clause row-reindexing. Let $\pi\in S_{n}$; $A^{\pi}$ has the entries $(A^{\pi})_{ij}=A_{\pi(i)j}$ by The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §row-reindexing. Let $\sigma\in S_{n}$ and $\tau=\sigma\circ\pi^{-1}\in S_{n}$. For $i\in[n]$, $\tau(\pi(i))=\sigma(\pi^{-1}(\pi(i)))=\sigma(i)$, so $(A^{\pi})_{i\sigma(i)}=A_{\pi(i)\tau(\pi(i))}$, and (R) with $g=\pi$ gives $p_{A^{\pi}}(\sigma)=p_{A}(\tau)$. Moreover $\tau\circ\pi=\sigma\circ(\pi^{-1}\circ\pi)=\sigma$, so $\operatorname{sgn}(\sigma)=\operatorname{sgn}(\tau)\operatorname{sgn}(\pi)$ by Rules for the Sign of a Permutation §multiplicative. Hence $t_{A^{\pi}}(\sigma)=\operatorname{sgn}(\pi)\,t_{A}(R(\sigma))$, where $R:S_{n}\to S_{n}$, $R(\sigma)=\sigma\circ\pi^{-1}$, is the bijection $R_{\pi^{-1}}$ of Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §translation. By (D), Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing (with the bijection $R$ and the map $t_{A}$),

$$\det A^{\pi}=\sum_{\sigma\in S_{n}}\operatorname{sgn}(\pi)\,t_{A}(R(\sigma))=\operatorname{sgn}(\pi)\sum_{\sigma\in S_{n}}t_{A}(R(\sigma))=\operatorname{sgn}(\pi)\sum_{\sigma\in S_{n}}t_{A}(\sigma)=\operatorname{sgn}(\pi)\det A.$$

Clause equal-rows. Let $p$ and $q$ be as in the clause and $\theta=\theta_{pq}\in S_{n}$ the transposition of $p$ and $q$: $\theta(p)=q$, $\theta(q)=p$, $\theta(k)=k$ for $k\notin\{p,q\}$, and $\theta\circ\theta=\mathrm{id}_{[n]}$ by Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §transposition; $\operatorname{sgn}(\theta)=-1$ by Rules for the Sign of a Permutation §transposition.

(a) For $k,j\in[n]$, $A_{\theta(k)j}=A_{kj}$: for $k=p$ this is $A_{qj}=A_{pj}$, for $k=q$ it is $A_{pj}=A_{qj}$, and otherwise $\theta(k)=k$.

(b) Let $\sigma\in S_{n}$ and $\tau=\sigma\circ\theta\in S_{n}$. For $i\in[n]$, $\tau(\theta(i))=\sigma(\theta(\theta(i)))=\sigma(i)$, so by (a) $A_{i\sigma(i)}=A_{\theta(i)\sigma(i)}=A_{\theta(i)\tau(\theta(i))}$, and (R) with $B=C=A$ and $g=\theta$ gives $p_{A}(\sigma)=p_{A}(\sigma\circ\theta)$. By Rules for the Sign of a Permutation §multiplicative, $\operatorname{sgn}(\sigma\circ\theta)=-\operatorname{sgn}(\sigma)$, and so $t_{A}(\sigma\circ\theta)=-t_{A}(\sigma)$.

(c) Let $E=\{\sigma\in S_{n}:\operatorname{sgn}(\sigma)=1\}$ and $E'=S_{n}\setminus E$; both are finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §subset, they are disjoint, and their union is $S_{n}$. By Rules for the Sign of a Permutation §values, and as $1\neq-1$ in $\mathbb{R}$ (if $1=-1$ then $1+1=1+(-1)=0$ by Negatives, Differences, Reciprocals and Quotients §negative, contradicting Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima §two, which gives $1+1=2\neq0$), $\sigma\in E'$ if and only if $\operatorname{sgn}(\sigma)=-1$. By (b), for $\sigma\in E$, $\operatorname{sgn}(\sigma\circ\theta)=-1$, so $\sigma\circ\theta\in E'$; and for $\sigma\in E'$, $\operatorname{sgn}(\sigma\circ\theta)=1$, so $\sigma\circ\theta\in E$. Let $R_{\theta}:S_{n}\to S_{n}$, $R_{\theta}(\sigma)=\sigma\circ\theta$, be the map of Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §translation; its restrictions are maps $g:E\to E'$ and $h:E'\to E$. For $\sigma\in E$, $h(g(\sigma))=(\sigma\circ\theta)\circ\theta=\sigma\circ(\theta\circ\theta)=\sigma$, and likewise $g(h(\sigma))=\sigma$ for $\sigma\in E'$; so $h\circ g=\mathrm{id}_{E}$ and $g\circ h=\mathrm{id}_{E'}$, and $g$ is a bijection by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §inverse-criterion.

(d) By (D) and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §disjoint-union, $\det A=s+s'$ with $s=\sum_{\sigma\in E}t_{A}(\sigma)$ and $s'=\sum_{\sigma\in E'}t_{A}(\sigma)$, the sums over $E$ and $E'$ being those of the restrictions of $t_{A}$, by Sums and Products over a Finite Set and over an Interval §subsets. By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing with the bijection $g$ and the map $t_{A}|_{E'}$, then (b), and Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive with $c=-1$,

$$s'=\sum_{\sigma\in E}t_{A}(g(\sigma))=\sum_{\sigma\in E}(-1)\cdot t_{A}(\sigma)=-s.$$

Hence $\det A=s+(-s)=0$.

Clause zero-row. Let $i$ be as in the clause. For every $\sigma\in S_{n}$, $q^{A}_{\sigma}(i)=A_{i\sigma(i)}=0$, as $\sigma(i)\in[n]$; so $p_{A}(\sigma)=0$ by Products over Finite Sets: a Single Factor, Vanishing in a Field, and Signs and Monotonicity in an Ordered Ring §zero and $t_{A}(\sigma)=0$. By (D) and Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §vanishing with $y=\mathrm{id}_{[n]}$, $\det A=t_{A}(\mathrm{id}_{[n]})=0$.

Clause dependent-rows. Let $x$ be as in the clause. The origin $0$ of $\mathbb{R}^{n}$ has every component $0$, by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §origin and Pointwise Operations on Real-Valued Maps §constants; as $x\neq0$, Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §equality gives $k\in[n]$ with $x_{k}\neq0$, which we fix. Since $x_{k}\neq0$, the reciprocal $x_{k}^{-1}\in\mathbb{R}$ of Negatives, Differences, Reciprocals and Quotients §reciprocal satisfies $x_{k}x_{k}^{-1}=1$. Let $d=-x_{k}^{-1}$, with the negative of Negatives, Differences, Reciprocals and Quotients §negative, so that $d\,x_{k}=-1$. Let $u:[n]\to\mathbb{R}^{n}$, $l\mapsto u^{(l)}$, be the map with $u^{(l)}_{j}=A_{lj}$ for all $l,j\in[n]$.

The coefficients. Let $c=d\,x+e_{k}\in\mathbb{R}^{n}$, formed with the operations of Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations and the standard basis vector $e_{k}$ of Identity Matrix and Standard Basis Vectors §basis; so $c_{l}=d\,x_{l}+\delta_{lk}$ for $l\in[n]$, and $c_{k}=d\,x_{k}+1=0$. For $j\in[n]$, by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §termwise for $+$, Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive, Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §vanishing with $y=k$ (as $\delta_{lk}A_{lj}=0$ for $l\neq k$ and $\delta_{kk}A_{kj}=A_{kj}$), and the hypothesis,

$$\sum_{l=1}^{n}c_{l}u^{(l)}_{j}=\sum_{l\in[n]}\bigl(d\,(x_{l}A_{lj})+\delta_{lk}A_{lj}\bigr)=d\sum_{l\in[n]}x_{l}A_{lj}+\sum_{l\in[n]}\delta_{lk}A_{lj}=d\cdot0+A_{kj}=A_{kj}.$$

So the clause row-linearity applies with $k$ in place of $i$, $n$ in place of $m$, and $c$ and $u$, and gives $\det A=\sum_{l=1}^{n}c_{l}\det A[k\leftarrow u^{(l)}]$. Let $l\in[n]$ and $C=A[k\leftarrow u^{(l)}]$. If $l\neq k$, then, by The Determinant Map and the Matrices Formed from a Square Matrix by Reindexing Rows, Replacing a Row and Deleting a Row and a Column Exist and Are Unique §row-replacement, $C_{kj}=u^{(l)}_{j}=A_{lj}=C_{lj}$ for every $j\in[n]$, so $\det C=0$ by the clause equal-rows, applied to $C$ with $k$ and $l$ in place of $p$ and $q$, and $c_{l}\det C=0$. By Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §vanishing with $y=k$, $\det A=c_{k}\det A[k\leftarrow u^{(k)}]=0$, as $c_{k}=0$.

Clause scalar. Let $\mu\in\mathbb{R}$. By Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations, $\mu A\in\mathcal{M}_{n}(\mathbb{R})$ and $(\mu A)_{ij}=\mu A_{ij}$. Let $\kappa:[n]\to\mathbb{R}$ be the constant map with value $\mu$, of Pointwise Operations on Real-Valued Maps §constants with $I=[n]$. For $\sigma\in S_{n}$ and $k\in[n]$, $q^{\mu A}_{\sigma}(k)=\mu A_{k\sigma(k)}=\kappa(k)\,q^{A}_{\sigma}(k)$, so by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §termwise for $\cdot$, $p_{\mu A}(\sigma)=\bigl(\prod_{k\in[n]}\kappa(k)\bigr)p_{A}(\sigma)$. Here $\prod_{k\in[n]}\kappa(k)=\prod_{k=1}^{n}\mu=\mu^{n}$, by Iterated Operations: Finite Sums and Finite Products §iterated and Powers with Exponents in the Natural Numbers with Zero §power. Hence $t_{\mu A}(\sigma)=\mu^{n}\,t_{A}(\sigma)$, and by (D) and Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §distributive, $\det(\mu A)=\sum_{\sigma\in S_{n}}\mu^{n}\,t_{A}(\sigma)=\mu^{n}\det A$.

Clause transpose. By Transpose of a Real Matrix §transpose with $m=n$, $A^{\top}\in\mathcal{M}_{n}(\mathbb{R})$ and $(A^{\top})_{ji}=A_{ij}$ for $i,j\in[n]$. Let $\sigma\in S_{n}$. For $i\in[n]$, $\sigma^{-1}(\sigma(i))=i$; so $(A^{\top})_{i\sigma(i)}=A_{\sigma(i)i}=A_{\sigma(i)\sigma^{-1}(\sigma(i))}$, and (R) with $C=A^{\top}$, $B=A$, $\tau=\sigma^{-1}$ and $g=\sigma$ gives $p_{A^{\top}}(\sigma)=p_{A}(\sigma^{-1})$. By Rules for the Sign of a Permutation §inverse, $\operatorname{sgn}(\sigma)=\operatorname{sgn}(\sigma^{-1})$; so $t_{A^{\top}}(\sigma)=t_{A}(J(\sigma))$, where $J:S_{n}\to S_{n}$, $J(\sigma)=\sigma^{-1}$, is the bijection of Permutations of [n]: Identity, Composition, Inverses, Translations and Transpositions §inversion. By (D) and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing with the bijection $J$ and the map $t_{A}$, $\det(A^{\top})=\sum_{\sigma\in S_{n}}t_{A}(J(\sigma))=\sum_{\sigma\in S_{n}}t_{A}(\sigma)=\det A$.