A proper introduction to determinants

Most linear algebra classes introduce determinants as a way to test invertibility, compute inverses, and find eigenvalues of small matrices. Not only does this strategy fail to generalize to larger matrices, it also fails to identify determinants as a fundamentally important mathematical object: it is the unique multilinear and alternating function \(F\) from the free \(R\)-module \((R^n)^n\) to \(R\) such that \(F(I)=1_R\), where \(R\) is a commutative ring and \(I\in (R^n)^n\) is the identity matrix. Here \(R\) can be any commutative ring, for example, a field such as \(\mathbb{R}\) where all non-zero elements can be divided, or the integer ring \(\mathbb{Z}\), \(\mathbb{Z}/m\), the polynomial rings \(\mathbb{Z}[x_1,\dots,x_t]\), etc., where division is undefined. This generalization has applications in number theory, algebraic geometry, and combinatorics.

Working over a general ring exposes a subtlety the real case hides: the difference between alternating and antisymmetric and demonstrates why the right hypothesis is alternating. This article is the rigorous companion to the determinant remarks in the row-reduction article.

Throughout, \(R\) is a commutative ring with identity \(1\), \(M_n(R)\) the set of all \(n\times n\) matrices over \(R\), and \(a_{ij}\) is the entry of \(A\in M_n(R)\) in row \(i\), column \(j\).

Setup: matrices over a commutative ring

The module \(R^n\) of column vectors is free with standard basis \(e_1,\dots,e_n\), where \(e_k\) has a \(1\) in position \(k\) and \(0\) elsewhere. We view an \(n\times n\) matrix as the ordered list of its \(n\) columns,

\[ A=\big(a_{\cdot 1},\,a_{\cdot 2},\,\dots,\,a_{\cdot n}\big), \qquad a_{\cdot j}=\sum_{k=1}^{n} a_{kj}\,e_k\in R^n, \]

so a function of a matrix is the same thing as a function of \(n\) vectors, \(F:(R^n)^n\to R\).

Alternating multilinear maps

Before any formula, we pin down the two adjectives in the title.

Definition (multilinear). \(F:(R^n)^n\to R\) is \(R\)-multilinear (or \(n\)-linear) if it is \(R\)-linear in each column when the others are held fixed: for every position \(j\), all vectors, and all \(r,s\in R\),

\[ F(\dots,\,r\,u+s\,v,\,\dots)=r\,F(\dots,u,\dots)+s\,F(\dots,v,\dots). \]

Definition (alternating). \(F\) is alternating if it vanishes whenever two columns coincide: \(a_{\cdot p}=a_{\cdot q}\) for some \(p\neq q\) implies \(F(A)=0\).

Definition (normalized). \(F\) is normalized if \(F(e_1,\dots,e_n)=F(I)=1\).

The choice of alternating (rather than the weaker-looking "a swap negates the value") is the crux of working over a ring. The next lemma shows alternating is at least as strong; the remark after shows it is strictly stronger in general.

Lemma (alternating \(\Rightarrow\) antisymmetric). If \(F\) is multilinear and alternating, then swapping any two columns negates it:

\[ F(\dots,\underset{p}{v},\dots,\underset{q}{w},\dots)=-\,F(\dots,\underset{p}{w},\dots,\underset{q}{v},\dots). \]

Proof. Put the sum \(v+w\) in both slots \(p\) and \(q\) and use that \(F\) vanishes there (two columns equal):

\[ 0=F(\dots,\underset{p}{v+w},\dots,\underset{q}{v+w},\dots). \]

Expand by multilinearity in slot \(p\), then slot \(q\):

\[ 0=\underbrace{F(\dots,v,\dots,v,\dots)}_{=\,0} +F(\dots,v,\dots,w,\dots) +F(\dots,w,\dots,v,\dots) +\underbrace{F(\dots,w,\dots,w,\dots)}_{=\,0}. \]

The first and last vanish (a repeated column), leaving \(F(\dots,v,\dots,w,\dots)+F(\dots,w,\dots,v,\dots)=0\), which is the claim. \(\blacksquare\)

Why "alternating," not "antisymmetric"? The converse of the lemma is false over a general ring. Suppose we assumed only antisymmetry (a swap negates). Setting the two swapped columns equal gives \(F(A)=-F(A)\), i.e. \(2\,F(A)=0\). Over a field of characteristic \(\neq 2\) we divide by \(2\) to conclude \(F(A)=0\), recovering alternating. But over \(\mathbb{Z}/2\), where \(2=0\), the equation \(2F(A)=0\) says nothing about \(A\), and antisymmetric maps that are not alternating genuinely exist. For a concrete example, consider the bilinear form \(F(u,v):=u_1v_1\) defined on \((\mathbb{Z}/2)^2\oplus(\mathbb{Z}/2)^2\): it is multilinear, and \(F(u,v)=u_1v_1=v_1u_1=F(v,u)=-F(v,u)\), so \(F\) is antisymmetric. But \(F(e_1,e_1)=1\neq0\) on two equal columns, so \(F\) is not alternating.

Sign of a permutation (skip if already familiar with symmetric groups)

The Leibniz formula below attaches a sign \(\pm1\) to every permutation, so we owe a precise account of what that sign is and why it is well-defined. Throughout, \(S_n\) is the group of bijections \(\sigma:\{1,\dots,n\}\to\{1,\dots,n\}\), and the signs live in \(\{+1,-1\}\) under multiplication, a group isomorphic to \(\mathbb{Z}/2\mathbb{Z}\). The sign is computed once in \(\mathbb{Z}\) and read into any commutative ring \(R\) as \(\pm1_R\); it is a property of the permutation, not of \(R\).

Notation and vocabulary

There are two notations for a permutation, and the comma is not decoration. One-line notation lists the images in order: \(\sigma=(\sigma(1),\dots,\sigma(n))\), so \((1,2,3)\) is the identity and \((2,3,1)\) is the map \(1\mapsto2,\ 2\mapsto3,\ 3\mapsto1\). Cycle notation uses spaces: \((a\ b\ c)\) means \(a\mapsto b\mapsto c\mapsto a\) and fixes everything else, so the cycle \((1\ 2\ 3)\) is that same map \((2,3,1)\), while the comma version \((1,2,3)\) is the identity. A transposition \((k\ \ell)\) is a 2-cycle: it swaps \(k\) and \(\ell\) and fixes the rest.

An inversion of \(\sigma\) is a pair \(\{i,j\}\) with \(i\lt j\) but \(\sigma(i)\gt\sigma(j)\): two positions whose values come out of order. Write \(\operatorname{inv}(\sigma)\) for their number. The word \((2,3,1)\) has two inversions, \(\{1,3\}\) and \(\{2,3\}\).

Every permutation is a product of transpositions. This is exactly what lets Definition B below apply to every \(\sigma\). The argument is short and self-contained; it is spelled out in its own collapsible so the main thread can move on to the definitions.

Proof that every permutation is a product of transpositions

Start at any point \(i\) and follow \(i,\sigma(i),\sigma^2(i),\dots\); the list lives in a finite set, so a value must recur, and since \(\sigma\) is injective the first to recur is \(i\) itself. That traces a cycle \((i\ \ \sigma(i)\ \ \sigma^2(i)\ \cdots)\); deleting its points and repeating on what remains writes \(\sigma\) as a product of disjoint cycles. A single cycle splits into transpositions,

\[ (a_1\ \ a_2\ \cdots\ a_k)=(a_1\ \ a_k)(a_1\ \ a_{k-1})\cdots(a_1\ \ a_2), \]

as you can check by following each \(a_t\) through the right-hand side (composing right to left, \(a_t\mapsto a_{t+1}\) and \(a_k\mapsto a_1\)). Concatenating the cycles' factorizations writes \(\sigma\) itself as a product of transpositions. \(\blacksquare\)

Two definitions, one of which needs an argument

Definition A (inversions). \(\operatorname{sgn}(\sigma):=(-1)^{\operatorname{inv}(\sigma)}\), the parity of the number of out-of-order pairs. This is well-defined since counting inversions involves no choices.

Definition B (transpositions). Write \(\sigma=\tau_1\cdots\tau_m\) as a product of \(m\) transpositions and set \(\operatorname{sgn}(\sigma):=(-1)^{m}\). For this definition to make sense, we have to show the parity of the number of transpositions does not depend on a permutation's factorizations which are not unique in general. Furthermore,

Theorem. For every factorization \(\sigma=\tau_1\cdots\tau_m\) into transpositions, \((-1)^{m}=(-1)^{\operatorname{inv}(\sigma)}\).

Strategy. Rather than tracking inversions through the factorization one transposition at a time, we route everything through a single structural fact: among all group homomorphisms \(S_n\to\{+1,-1\}\) there is exactly one non-trivial map, and it sends every transposition to \(-1\). The Vandermonde construction below builds such a homomorphism and checks that it is \(-1\) on every transposition, so it is that unique map. Evaluating it on a factorization \(\sigma=\tau_1\cdots\tau_m\) gives \(\operatorname{sgn}(\sigma)=(-1)^m\), while reading it off the Vandermonde polynomial gives \(\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}\); this proves the theorem above.

The Vandermonde construction: a sign homomorphism

First define the Vandermonde polynomial

\[ \Delta:=\prod_{1\le i\lt j\le n}(x_j-x_i). \]

\(S_n\) acts on the set \(\{\pm\Delta\}\) via \(\sigma\cdot\Delta=\prod_{i\lt j}\big(x_{\sigma(j)}-x_{\sigma(i)}\big)\). This defines a group homomorphism: \(\sigma\cdot\Delta=\operatorname{sgn}(\sigma)\,\Delta\) with \(\operatorname{sgn}(\sigma)\in\{+1,-1\}\). To see that it is indeed a homomorphism:

\[ \operatorname{sgn}(\sigma\rho)\,\Delta=(\sigma\rho)\cdot\Delta =\sigma\cdot(\rho\cdot\Delta) =\sigma\cdot\big(\operatorname{sgn}(\rho)\Delta\big) =\operatorname{sgn}(\rho)\,(\sigma\cdot\Delta) =\operatorname{sgn}(\sigma)\operatorname{sgn}(\rho)\,\Delta, \]

so \(\operatorname{sgn}(\sigma\rho)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\rho)\). Fun fact: this is the general fact (Cayley's theorem) that an action on a set \(X\) is a homomorphism \(G\to\operatorname{Sym}(X)\).

Every transposition negates \(\Delta\), and \(\operatorname{sgn}(\sigma)\) tracks the parity of the number of inversions of \(\sigma\). This is because the Vandermonde polynomial contains a factor \((x_j-x_i)\) for every pair of indices \(\{i,j\}\).

Parity is the only non-trivial homomorphism

Proposition. For \(n\ge2\), \(\operatorname{sgn}\) is the only non-trivial group homomorphism \(S_n\to\{+1,-1\}\): any non-trivial homomorphism is forced to send every transposition to \(-1\), and the sole other one is the trivial map \(\sigma\mapsto+1\).

Proof that parity is the unique non-trivial homomorphism

Conjugation. For any \(\sigma\) and transposition \((a\ \ b)\), \(\sigma\,(a\ \ b)\,\sigma^{-1}=(\sigma(a)\ \ \sigma(b))\): the left side sends \(\sigma(a)\mapsto\sigma(b)\), \(\sigma(b)\mapsto\sigma(a)\), and fixes every other point.

All transpositions are conjugate. Given \((a\ \ b)\) and \((c\ \ d)\), choose \(\sigma\) with \(\sigma(a)=c\) and \(\sigma(b)=d\); then \(\sigma\,(a\ \ b)\,\sigma^{-1}=(c\ \ d)\).

A homomorphism is constant on transpositions. Let \(\phi:S_n\to\{\pm1\}\) be any homomorphism and \(\tau,\tau'\) transpositions with \(\rho\tau\rho^{-1}=\tau'\). Because \(\{\pm1\}\) is abelian, \(\phi(\tau')=\phi(\rho)\phi(\tau)\phi(\rho)^{-1}=\phi(\tau)\). Call the common value \(\varepsilon\).

That value decides everything. Transpositions generate \(S_n\) (shown above), so for any factorization \(\sigma=\tau_1\cdots\tau_m\), \(\phi(\sigma)=\varepsilon^{m}\). If \(\varepsilon=+1\) then \(\phi\) is the trivial map. If \(\varepsilon=-1\) then \(\phi(\sigma)=(-1)^m\); since \(\phi\) is a single-valued function of \(\sigma\), every non-trivial homomorphism must return this same value, so there is at most one. The Vandermonde homomorphism above is non-trivial — it sends each transposition to \(-1\) — so it exists and is precisely this map. Hence the trivial map and \(\operatorname{sgn}\) are the only two homomorphisms. \(\blacksquare\)

This is what makes the sign the right bookkeeping for the determinant. Swapping two columns of a matrix composes its permutation index with a transposition \(\tau\), and \(\operatorname{sgn}(\sigma\tau)=-\operatorname{sgn}(\sigma)\), so \(\operatorname{sgn}\) tracks the parity of how many column swaps return \(\sigma\) to the identity. Each swap costs a factor of \(-1\), and the proposition says no other valid \(\pm1\)-valued invariant of that swapping exists.

Transposition parity equals inversion parity

This completes the proof of the theorem.

Proof that transposition parity is well-defined and equals inversion parity

Through a factorization. Every \(\sigma\) is a product of transpositions, \(\sigma=\tau_1\cdots\tau_m\). Apply the homomorphism \(\operatorname{sgn}\) and use \(\operatorname{sgn}(\tau_i)=-1\) for each factor:

\[ \operatorname{sgn}(\sigma) =\operatorname{sgn}(\tau_1)\cdots\operatorname{sgn}(\tau_m)=(-1)^m. \]

The left-hand side depends only on \(\sigma\), so \((-1)^m\) does too: the parity of \(m\) is the same for every factorization.

Off the Vandermonde polynomial. Read \(\operatorname{sgn}(\sigma)\) directly from \(\sigma\cdot\Delta=\prod_{i\lt j}\big(x_{\sigma(j)}-x_{\sigma(i)}\big)\). As \(\{i,j\}\) runs over all pairs of positions, \(\{\sigma(i),\sigma(j)\}\) runs over all pairs of values, so these are the factors of \(\Delta\) up to sign. The factor for \(i\lt j\) keeps its sign when \(\sigma(i)\lt\sigma(j)\) and flips when \(\sigma(i)\gt\sigma(j)\) — that is, exactly at the inversions of \(\sigma\). One flip per inversion gives

\[ \operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}. \]

Definition B is well-defined and agrees with Definition A. \(\blacksquare\)

Definition of the determinant

Definition (Leibniz formula). For \(A\in M_n(R)\),

\[ \boxed{\ \det A:=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^{n} a_{\sigma(i),\,i}\ }\ \in R. \]

Each summand picks one entry from every column of \(A\): \(a_{\sigma(i),i}\) from column \(i\), multiplies them, and weights the product by the sign of \(\sigma\). The formula is built entirely from \(R\)'s addition, multiplication, and the constants \(\pm1_R\), so \(\det A\) is defined over any commutative ring. The sum has \(n!\) terms.

The Leibniz formula for a \(3\times 3\) matrix

To keep our feet on the ground, let's consider the familiar case with \(n=3\). Consider each permutation \(\sigma\in S_3\). Two notations meet in the table below: the comma-separated one-line form \((\sigma(1),\sigma(2),\sigma(3))\) names \(\sigma\) by listing its values, while the space-separated cycle form \((2\ 3)\) in the "type" column names which indices get swapped — so e.g. the one-line \((1,3,2)\) is the cycle \((2\ 3)\).

\(\sigma\)	type	\(\operatorname{sgn}\)	term
\((1,2,3)\)	identity	\(+\)	\(a_{11}a_{22}a_{33}\)
\((2,3,1)\)	3-cycle	\(+\)	\(a_{21}a_{32}a_{13}\)
\((3,1,2)\)	3-cycle	\(+\)	\(a_{31}a_{12}a_{23}\)
\((1,3,2)\)	transposition \((2\,3)\)	\(-\)	\(a_{11}a_{32}a_{23}\)
\((3,2,1)\)	transposition \((1\,3)\)	\(-\)	\(a_{31}a_{22}a_{13}\)
\((2,1,3)\)	transposition \((1\,2)\)	\(-\)	\(a_{21}a_{12}a_{33}\)

Collecting them (and reordering factors, legal since \(R\) is commutative) gives the familiar expansion:

\[ \det A = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32} -a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}. \]

This gives us the determinant formula for a \(3\times 3\) matrix which we are all familiar with.

The determinant is the unique alternating multilinear normalized map

Theorem. Over any commutative ring \(R\), the map \(\det:M_n(R)\to R\) above is multilinear, alternating and normalized; and it is the only function \((R^n)^n\to R\) with those three properties.

We prove existence (the Leibniz formula has the properties) and uniqueness (any map with the properties equals it) separately.

Existence

Multilinear. Fix a column index \(j\). In each term \(\operatorname{sgn}(\sigma)\prod_{i} a_{\sigma(i),i}\), the only factor drawn from column \(j\) is \(a_{\sigma(j),j}\), appearing to the first power. So each term is \(R\)-linear in column \(j\), and a sum of \(R\)-linear functions is \(R\)-linear. As \(j\) was arbitrary, \(\det\) is multilinear.

Normalized. For \(A=I\) we have \(a_{ki}=\delta_{ki}\), so \(a_{\sigma(i),i}=\delta_{\sigma(i),i}\), which is \(0\) unless \(\sigma(i)=i\) for every \(i\) — i.e. unless \(\sigma=\mathrm{id}\). The single surviving term is \(\operatorname{sgn}(\mathrm{id})\prod_i 1=1\). Thus \(\det I=1\).

Alternating (the ring-sensitive step). Suppose columns \(p\) and \(q\) are equal, \(p\neq q\): that is, \(a_{k,p}=a_{k,q}\) for every row \(k\) — call this condition \((\dagger)\). Let \(\tau=(p\ q)\), and pair each \(\sigma\in S_n\) with \(\sigma'=\sigma\circ\tau\). The map \(\sigma\mapsto\sigma\circ\tau\) is an involution with no fixed points, so \(S_n\) splits into disjoint pairs \(\{\sigma,\sigma'\}\). We show the two terms of each pair cancel. Since \(\sigma'(i)=\sigma(\tau(i))\) and \(\tau\) fixes every index except \(p,q\),

\[ \prod_{i} a_{\sigma'(i),i}=\Big(\prod_{i\neq p,q} a_{\sigma(i),i}\Big)\cdot a_{\sigma(q),\,p}\cdot a_{\sigma(p),\,q}. \]

Now apply \((\dagger)\) to the two special factors: reading the entry in row \(\sigma(q)\), column \(p\) off the equal columns gives \(a_{\sigma(q),p}=a_{\sigma(q),q}\); likewise \(a_{\sigma(p),q}=a_{\sigma(p),p}\). Therefore

\[ a_{\sigma(q),p}\,a_{\sigma(p),q}=a_{\sigma(q),q}\,a_{\sigma(p),p}=a_{\sigma(p),p}\,a_{\sigma(q),q}, \]

where the last equality uses commutativity of \(R\) to reorder the two factors. Substituting back, the \(\sigma'\) monomial equals the \(\sigma\) monomial, while the signs are opposite (\(\operatorname{sgn}(\sigma')=-\operatorname{sgn}(\sigma)\)). The two terms sum to zero; summing over all pairs, \(\det A=0\). So \(\det\) is alternating. \(\blacksquare\)

Uniqueness

Let \(F:(R^n)^n\to R\) be any multilinear, alternating, normalized map. Expand each column in the basis, \(a_{\cdot i}=\sum_{k_i} a_{k_i,i}\,e_{k_i}\), and pull all the scalars out using multilinearity, one column at a time:

\[ F(A)=\sum_{k_1=1}^n\cdots\sum_{k_n=1}^n \Big(\prod_{i=1}^n a_{k_i,i}\Big)\,F(e_{k_1},\dots,e_{k_n}). \]

This is a sum of \(n^n\) terms. Because \(F\) is alternating, \(F(e_{k_1},\dots,e_{k_n})=0\) whenever two indices coincide. So only the tuples that are a permutation of \((1,\dots,n)\) survive — write \(k_i=\sigma(i)\). For such a tuple, repeatedly apply antisymmetry (the lemma, valid over any \(R\)) to sort the arguments back into the order \(e_1,\dots,e_n\); the total factor is \(\operatorname{sgn}(\sigma)\):

\[ F(e_{\sigma(1)},\dots,e_{\sigma(n)})=\operatorname{sgn}(\sigma)\,F(e_1,\dots,e_n)=\operatorname{sgn}(\sigma), \]

the last step using normalization. Substituting,

\[ F(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{\sigma(i),i}=\det A. \]

Thus every alternating multilinear normalized map equals the Leibniz determinant — it is unique. \(\blacksquare\)

Since we never assume \(R\) to be a field, the theorem holds verbatim over any commutative ring: \(\mathbb{Z}\), over \(\mathbb{Z}/m\) for any \(m\) (including \(\mathbb{Z}/2\)), over polynomial rings \(R[x_1,\dots,x_t]\), etc. The single load-bearing hypothesis beyond the ring axioms is commutativity; drop it and the alternating step collapses.

Fundamental properties of the determinant

We can now prove all of the familiar determinant formulae with ease, generalized to an arbitrary commutative ring \(R\).

A rigidity lemma

The uniqueness proof establishes slightly more than uniqueness. Drop normalization, and the same computation gives:

Lemma (rigidity). Every multilinear alternating \(F:(R^n)^n\to R\) is a constant multiple of the determinant: \(F(A)=F(I)\cdot\det A\) for all \(A\).

Proof. Expand each column and pull the scalars out by multilinearity exactly as in the uniqueness proof; the alternating property kills every tuple with a repeated index, and antisymmetry sorts each surviving permutation \(\sigma\) back to identity order at the cost of \(\operatorname{sgn}(\sigma)\):

\[ F(A)=\Big(\sum_\sigma\operatorname{sgn}(\sigma)\prod_i a_{\sigma(i),i}\Big)F(I)=\det(A)\,F(I). \qquad\blacksquare \]

This single lemma powers the product theorem below.

Transpose: the row–column symmetry

Proposition. \(\det(A^{\mathsf T})=\det(A)\).

Proof. By the Leibniz formula, \(\det(A^{\mathsf T})=\sum_\sigma\operatorname{sgn}(\sigma)\prod_i a_{i,\sigma(i)}\). In each term, reindex the product by \(j=\sigma(i)\) (so \(i=\sigma^{-1}(j)\)), using commutativity to reorder: \(\prod_i a_{i,\sigma(i)}=\prod_j a_{\sigma^{-1}(j),j}\). Since \(\operatorname{sgn}(\sigma)=\operatorname{sgn}(\sigma^{-1})\) and \(\sigma\mapsto\sigma^{-1}\) is a bijection of \(S_n\),

\[ \det(A^{\mathsf T})=\sum_{\rho}\operatorname{sgn}(\rho)\prod_j a_{\rho(j),j}=\det A. \qquad\blacksquare \]

Corollary (rows = columns). \(\det\) is multilinear and alternating in the rows of \(A\) as well: every column property holds verbatim for rows, since the rows of \(A\) are the columns of \(A^{\mathsf T}\) and \(\det A=\det A^{\mathsf T}\).

Elementary row (and column) operations

Write \(R_1,\dots,R_n\) for the rows of \(A\).

Proposition. (1) Swap — exchanging two rows multiplies the determinant by \(-1\); (2) Scale — multiplying one row by \(c\in R\) multiplies the determinant by \(c\); (3) Replacement — adding a multiple of one row to a different row leaves the determinant unchanged.

Proof. (1) is antisymmetry in the rows. (2) is multilinearity in the scaled row. (3) For \(i\neq j\), multilinearity in row \(i\) gives \(\det(\dots,R_i+cR_j,\dots)=\det(\dots,R_i,\dots)+c\,\det(\dots,R_j,\dots)\), and the second determinant has row \(R_j\) in two places, so it vanishes (alternating). \(\blacksquare\)

Equivalently, each operation is left-multiplication by an elementary matrix \(E\) (the result of performing it on \(I\)), and the proposition computes \(\det E\):

operation on \(A\)	elementary matrix \(E\)	\(\det E\)
swap rows \(i,j\)	\(E_{\text{swap}}\)	\(-1\)
scale row \(i\) by \(c\)	\(E_{\text{scale}}(c)\)	\(c\)
add \(c\times\) row \(j\) to row \(i\)	\(E_{\text{add}}(c)\)	\(1\)

Performing the operation is \(A\mapsto EA\), so the proposition says \(\det(EA)=\det(E)\det(A)\) — the special case of the product theorem from which row reduction bootstraps the general computation.

Triangular and diagonal matrices

Proposition. If \(A\) is upper- or lower-triangular, then \(\det A=\prod_{i=1}^n a_{ii}\). In particular the determinant of a diagonal matrix is the product of its diagonal entries.

Proof. Take \(A\) upper-triangular, so \(a_{k\ell}=0\) whenever \(k\gt\ell\). In a Leibniz term the factor \(a_{\sigma(i),i}\) vanishes unless \(\sigma(i)\le i\). If \(\sigma(i)\le i\) for every \(i\) then \(\sum_i\sigma(i)\le\sum_i i\); but \(\sigma\) is a permutation, so \(\sum_i\sigma(i)=\sum_i i\), forcing \(\sigma(i)=i\) for all \(i\). Only \(\sigma=\mathrm{id}\) survives, contributing \(\prod_i a_{ii}\). The lower-triangular case is its transpose. \(\blacksquare\)

This is the engine of practical computation: Gaussian elimination reduces \(A\) to triangular form by the operations above — multiplying \(\det\) by the known factors \(-1\) and \(c\), and leaving it unchanged on replacements — after which \(\det\) is read straight off the diagonal. That is the \(O(n^3)\) method of the row-reduction article, in place of the \(n!\)-term sum.

The multiplicative law

Theorem. For all \(A,B\in M_n(R)\), \(\det(AB)=\det(A)\,\det(B)\).

Proof. Fix \(A\) and regard \(G(B):=\det(AB)\) as a function of the \(n\) columns of \(B\). The \(j\)-th column of \(AB\) is \(A\,b_{\cdot j}\), which is \(R\)-linear in \(b_{\cdot j}\); composing with the multilinearity of \(\det\), the map \(G\) is multilinear in the columns of \(B\). If two columns of \(B\) coincide, then \(AB\) has two equal columns and \(\det(AB)=0\); thus \(G\) is alternating. By the rigidity lemma,

\[ \det(AB)=G(B)=G(I)\cdot\det B=\det(AI)\cdot\det B=\det(A)\,\det(B). \qquad\blacksquare \]

Corollaries. Immediately:

• \(\det(A_1\cdots A_k)=\det A_1\cdots\det A_k\), and \(\det(A^m)=(\det A)^m\).
• If \(A\) is invertible then \(\det A\in R^\times\) and \(\det(A^{-1})=(\det A)^{-1}\), from \(\det A\,\det(A^{-1})=\det I=1\).
• Similarity invariance: \(\det(P^{-1}AP)=\det A\) for invertible \(P\), so \(\det\) is an invariant of the linear operator, independent of the basis chosen to represent it.

The invertibility criterion

The converse of the second corollary holds, characterizing exactly which matrices are invertible over \(R\).

Theorem. \(A\in M_n(R)\) is invertible \(\iff\) \(\det A\) is a unit of \(R\). Over a field this reads: \(A\) is invertible \(\iff \det A\neq0\).

The forward direction is the corollary above. The converse rests on the adjugate identity.

Adjugate, Cramer's rule, and the proof of the invertibility criterion

For each \(i,j\) let \(C_{ij}=(-1)^{i+j}\det\big(A^{(i,j)}\big)\) be the \((i,j)\) cofactor, where \(A^{(i,j)}\) is \(A\) with row \(i\) and column \(j\) deleted, and define the adjugate by \(\operatorname{adj}(A)_{ij}=C_{ji}\). The claim is

\[ A\,\operatorname{adj}(A)=\operatorname{adj}(A)\,A=(\det A)\,I. \tag{\(\star\)} \]

Cofactor (Laplace) expansion. Expand \(\det\) by multilinearity in column \(j\), writing \(a_{\cdot j}=\sum_i a_{ij}e_i\): \(\det A=\sum_i a_{ij}\,\det(\dots,\underset{j}{e_i},\dots)=\sum_i a_{ij}\,C_{ij}\). The last equality holds because, in the matrix whose \(j\)-th column is \(e_i\), moving that column left past \(j-1\) columns and its single nonzero entry up past \(i-1\) rows costs \((-1)^{(j-1)+(i-1)}=(-1)^{i+j}\) and exposes a leading \(1\) over the block \(A^{(i,j)}\), whose determinant is the minor. This is the \((j,j)\) entry of \(\operatorname{adj}(A)A\), equal to \(\det A\).

Off-diagonal entries vanish. For \(k\neq j\), the entry \((\operatorname{adj}(A)A)_{kj}=\sum_i C_{ik}\,a_{ij}\) is the cofactor expansion along column \(k\) of the matrix \(A'\) got from \(A\) by replacing column \(k\) with column \(j\) (changing column \(k\) does not affect the cofactors \(C_{ik}\)). But \(A'\) then has columns \(j\) and \(k\) both equal to \(a_{\cdot j}\), so \(\det A'=0\). Hence \((\operatorname{adj}(A)A)_{kj}=\delta_{kj}\det A\), which is \((\star)\) for \(\operatorname{adj}(A)A\). For the other product, run the same argument on \(A^{\mathsf T}\): since \(C_{ij}(A^{\mathsf T})=C_{ji}(A)\) (deleting row \(i\), column \(j\) of \(A^{\mathsf T}\) is the transpose of deleting row \(j\), column \(i\) of \(A\), and \(\det\) is transpose-invariant), one has \(\operatorname{adj}(A^{\mathsf T})=\operatorname{adj}(A)^{\mathsf T}\); applying \(\operatorname{adj}(B)B=(\det B)I\) at \(B=A^{\mathsf T}\) and transposing gives \(A\,\operatorname{adj}(A)=(\det A)I\).

Conclusion. If \(\det A\) is a unit, \((\star)\) gives \(A^{-1}=(\det A)^{-1}\operatorname{adj}(A)\), so \(A\) is invertible; conversely an invertible \(A\) has \(\det A\in R^\times\) by the corollary. Over a field the units are the nonzero elements, recovering the classical \(\det A\neq0\) test. The same identity yields Cramer's rule: when \(\det A\) is a unit, the solution of \(Ax=b\) has \(i\)-th coordinate \(x_i=(\det A)^{-1}\det(A_i\!\to\! b)\), where \(A_i\!\to\! b\) is \(A\) with column \(i\) replaced by \(b\). \(\blacksquare\)

Why commutativity is the load-bearing hypothesis

We named commutativity as the one assumption, beyond the bare ring axioms, that the determinant cannot do without. This section makes that precise: it isolates the single step where commutativity is spent, shows the whole edifice collapses without it, and points to the substitutes algebraists use when the ring genuinely fails to commute. Throughout, let \(R\) be a (possibly noncommutative) ring with \(1\).

The one place it is used: the alternating step

Over a noncommutative ring the product \(\prod_{i} a_{\sigma(i),i}\) is ambiguous until we fix an order; take the natural one, increasing column index: \(\operatorname{term}(\sigma)=A_1A_2\cdots A_n\) with \(A_i:=a_{\sigma(i),i}\). Re-examine the alternating proof: with columns \(p\lt q\) equal (condition \((\dagger)\)), \(\sigma'=\sigma\tau\) with \(\tau=(p\ q)\), and the proof needs the two ordered monomials to be equal so their opposite signs cancel. Splitting around the two moving slots,

\[ \operatorname{term}(\sigma)=\underbrace{(A_1\cdots A_{p-1})}_{P}\;A_p\;\underbrace{(A_{p+1}\cdots A_{q-1})}_{M}\;A_q\;\underbrace{(A_{q+1}\cdots A_n)}_{S}, \]

and \((\dagger)\) turns \(\operatorname{term}(\sigma')\) into \(P\,A_q\,M\,A_p\,S\). The two monomials coincide as soon as

\[ \boxed{\,A_p\,M\,A_q=A_q\,M\,A_p\,.} \]

Over a commutative ring both sides collapse to \(M\,A_p\,A_q\), every pair cancels, and \(\det\) is alternating. Over a noncommutative ring the identity fails in general, the paired terms no longer cancel, and \(\det\) is not alternating. The simplest counterexample is already \(2\times2\): over the quaternions, the matrix with two identical columns

\[ \begin{bmatrix} i & i\\[2pt] j & j\end{bmatrix}\qquad\text{has}\qquad \det=a_{11}a_{22}-a_{21}a_{12}=i\,j-j\,i=k-(-k)=2k\neq0. \]

What else breaks

• Well-definedness of the formula itself. The same \(2\times2\) example shows the Leibniz sum depends on the order chosen for each product: \(a_{21}a_{12}\) and \(a_{12}a_{21}\) differ. Without commutativity there is no canonical scalar "\(\det\)".
• Multiplicativity \(\det(AB)=\det A\,\det B\). The proof runs \(\det(AB)\) through the rigidity lemma, which needs \(\det\) to be alternating; the classical entry-by-entry proof likewise reorders products of entries. Either way it uses commutativity throughout.
• The geometric meaning. Signed volume is multilinear only in the commutative world; the volume interpretation below has no noncommutative analogue.

The repair: relocate commutativity, don't remove it

Because the obstruction is intrinsic, noncommutative settings replace the Leibniz determinant with invariants engineered to control noncommutativity rather than ignore it:

• Dieudonné determinant over a division ring \(D\): it lands in the abelianization \(D^\times/[D^\times,D^\times]\) — one quotients out commutators precisely to restore multiplicativity.
• Quasideterminants (Gelfand–Retakh): an \(n\times n\) array of "partial" determinants, since no single scalar can do the job.
• Quantum determinant for quantum matrix groups: a genuine central \(\det\)-element does exist, but only because the entries obey prescribed \(q\)-commutation relations: controlled noncommutativity tuned so the cancellation goes through up to a \(q\)-factor.

The moral: the determinant is a fundamentally commutative construction. Multilinearity, the alternating law, and multiplicativity all rest on the freedom to reorder a product of entries, which is granted by commutativity.

The Leibniz formula is signed volume

Specialize to \(R=\mathbb{R}\). The determinant then has a geometric identity: writing \(P_A=\big\{\sum_i t_i v_i:t_i\in[0,1]\big\}\) for the parallelepiped spanned by the columns \(v_1,\dots,v_n\) of \(A\), and \(\lambda\) for \(n\)-dimensional Lebesgue measure,

\[ \boxed{\ \det A=\varepsilon(A)\cdot\lambda(P_A)\ },\qquad\text{equivalently}\qquad \lambda\big(A([0,1]^n)\big)=|\det A|, \]

where \(\lambda(P_A)\ge0\) is the volume of the parallelepiped and \(\varepsilon(A)\in\{\pm1\}\) is the orientation of the frame \((v_1,\dots,v_n)\): which of the two connected components of \(GL_n(\mathbb{R})\) the matrix \(A\) lies in without any reference to the determinant (see the deferred construction below). The boxed line is an assertion, not a definition: it asserts that this independently-defined signed volume coincides with the Leibniz polynomial. The sign tells you whether \(A\) preserves or reverses orientation, the magnitude tells you how much \(A\) rescales the unit cube.

For a concrete \(2\times2\) example, take \(A=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), with \(\det A=2\cdot2-1\cdot1=3\). Its columns \((2,1)\) and \((1,2)\) span a parallelogram whose area is exactly \(3\) times that of the unit square, with orientation preserved since \(\det A\gt0\). The animation below interpolates linearly from the identity to \(A\), i.e. \(A(t)=I+t(A-I)\); the signed area is \(\det A(t)=(1+t)^2-t^2=1+2t\). Swapping the two columns gives \(A'=\begin{bmatrix}1&2\\2&1\end{bmatrix}\), with \(\det A'=1\cdot1-2\cdot2=-3\): the same parallelogram, but the frame \((1,2),(2,1)\) is the mirror image of the standard one, so orientation is reversed and the signed area turns negative — exactly the \(-1\) the swap rule predicts. The second animation below interpolates \(A'(t)=I+t(A'-I)\), whose signed area \(\det A'(t)=1-4t^2\) starts at \(1\), collapses to \(0\) at \(t=\tfrac12\) (the frame momentarily flattens onto a line), then reopens with the orientation flipped, finishing at \(-3\).

Important observations: the unit cube has volume \(1\) (normalized); a flattened box has volume \(0\) (alternating); volume scales and adds in each edge separately (multilinear). Signed volume, as a function of the \(n\) edge-vectors, therefore is a multilinear alternating normalized map, and by the uniqueness theorem there is exactly one such map, namely \(\det\). The sign \(\operatorname{sgn}(\sigma)\) is the orientation bookkeeping: swapping two edge-vectors reflects the frame and flips the sign, the geometric shadow of \(\operatorname{sgn}(\sigma\tau)=-\operatorname{sgn}(\sigma)\) where \(\tau\) is a transposition.

Full construction deferred. Turning this picture into a proof takes care on two points: orientation must be defined without circular appeal to the determinant (it is the choice of one of the two path-components of \(GL_n(\mathbb{R})\)), and the magnitude \(\lambda(P_A)=|\det A|\) must be obtained without assuming the volume-scaling law it is meant to prove (via a base × height / Fubini argument using only the normalization \(\lambda([0,1]^n)=1\), translation invariance, the nullity of proper subspaces, and the product structure of Lebesgue measure). The geometric story is logically independent of the noncommutative discussion above, which returns to the algebra over a general ring.