Multiplicativity of the Determinant

Suppose you stretch the plane by a linear map AA. Then you stretch it again by a linear map BB. How does the total area-distortion factor of the composite map BABA relate to the individual distortion factors of AA and BB? The answer — that the factors simply multiply — is one of the most useful facts in all of linear algebra. This is far from obvious once you recall how determinants are actually defined. They are defined as alternating sums over permutations or via cofactor expansions. This multiplicativity theorem, det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B), converts a question about matrix multiplication into a question about ordinary multiplication of scalars. Matrix multiplication has no simple closed form for its determinant a priori. This theorem is the mechanism behind why the determinant detects invertibility. It explains why similar matrices share a determinant. It explains why the determinant of a linear map is well-defined independent of basis. It also explains why eigenvalues multiply to give the determinant. It is the load-bearing wall connecting the combinatorial definition of the determinant to its geometric meaning as a signed volume scaling factor. The proof strategy is elegant. Rather than expand both sides through the permutation formula and fight with a combinatorial identity, we exploit a key characterization. The determinant is the unique alternating multilinear function of the rows (or columns). It is normalized to send the identity to 11. Multiplicativity is proved by showing that the function A↦det⁡(AB)/det⁡(B)A \mapsto \det(AB)/\det(B) (when det⁡(B)≠0\det(B) \neq 0) satisfies the defining properties of the determinant of AA. The degenerate case det⁡(B)=0\det(B) = 0 is then patched separately. This article proves the multiplicativity theorem in full. It derives the immediate corollaries — the formula for det⁡(A−1)\det(A^{-1}), for det⁡(An)\det(A^n), and for determinants of similar matrices. It also discusses precisely where each hypothesis (square matrices, the same size, the same field) is essential.

Key takeaways
  • For n×nn \times n matrices A,BA, B over a field FF, det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A) \det(B).
  • The proof uses the The Determinant as an Alternating Form characterization: uniqueness of the alternating, multilinear, normalized function on rows.
  • Consequence: AA is invertible iff det⁡(A)≠0\det(A) \neq 0, and then det⁡(A−1)=det⁡(A)−1\det(A^{-1}) = \det(A)^{-1}.
  • Consequence: det⁡(Ak)=det⁡(A)k\det(A^k) = \det(A)^k for all integers kk (positive, zero, and — when AA is invertible — negative).
  • Consequence: similar matrices have equal determinants, so det⁡\det is a well-defined invariant of a linear operator.
  • Key idea: determinant is a homomorphism from (GLn(F),×)(GL_n(F), \times) (and more generally (Mn(F),×)(M_n(F),\times)) to (F×,×)(F^\times, \times) (resp. (F,×)(F,\times)), turning matrix multiplication into scalar multiplication.

Where this lives today

The multiplicativity of the determinant is the algebraic engine behind Determinant and Invertibility: since det⁡(AB)=det⁡(A)det⁡(B)\det(AB)=\det(A)\det(B), a matrix AA can have a genuine two-sided inverse only when det⁡(A)≠0\det(A) \neq 0, giving the cleanest possible test for invertibility. It also underlies the definition of the characteristic polynomial, det⁡(A−λI)\det(A - \lambda I), whose behavior under conjugation (det⁡(P−1AP−λI)=det⁡(A−λI)\det(P^{-1}AP - \lambda I) = \det(A-\lambda I)) — a direct corollary of similarity invariance proved above — is exactly what makes eigenvalues a basis-independent notion in Characteristic Polynomial and the Determinant. Beyond pure linear algebra: in differential geometry and multivariable calculus, the Jacobian determinant of a composite map g∘fg \circ f equals the product of the Jacobian determinants (chain rule for Jacobians), a direct application of matrix multiplicativity to the derivative matrices. In group theory, the map A↦det⁡(A)A \mapsto \det(A) is a group homomorphism GLn(F)→F×GL_n(F) \to F^\times, and its kernel is the special linear group SLn(F)SL_n(F) — multiplicativity is precisely what makes det⁡\det a homomorphism and SLnSL_n a normal subgroup. In physics, conservation of phase-space volume in Hamiltonian mechanics (Liouville's theorem) and the transformation of probability densities under changes of variables in statistics both rely on tracking Jacobian determinants through compositions. In numerical linear algebra and cryptography (e.g., lattice-based cryptosystems), determinants of products of transformation matrices are tracked via multiplicativity to control volume distortion and numerical conditioning without recomputing full determinants from scratch.

Ask the tutor: Why does the "not invertible ⇒ determinant zero" step in the proof require showing that a right inverse forces full invertibility, and where exactly is rank-nullity used?

DefinitionThe determinant of a square matrix

Let FF be a field and let n≥1n \geq 1. For an n×nn \times n matrix A=(aij)A = (a_{ij}) with entries in FF, the determinant of AA is

det⁡(A)=∑σ∈Snsgn⁡(σ) a1σ(1)a2σ(2)⋯anσ(n),\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \, a_{1\sigma(1)} a_{2\sigma(2)} \cdots a_{n\sigma(n)},

where SnS_n is the symmetric group on {1,…,n}\{1, \dots, n\} and sgn⁡(σ)∈{+1,−1}\operatorname{sgn}(\sigma) \in \{+1, -1\} is the sign of the permutation σ\sigma. Equivalently, viewing det⁡\det as a function of the rows R1,…,Rn∈FnR_1, \dots, R_n \in F^n of AA, it is the unique function D:(Fn)n→FD : (F^n)^n \to F satisfying: 1. Multilinearity: DD is linear in each row separately, holding the other rows fixed. 2. Alternating: D(R1,…,Rn)=0D(R_1, \dots, R_n) = 0 whenever two rows are equal. 3. Normalization: D(e1,…,en)=1D(e_1, \dots, e_n) = 1, where e1,…,ene_1,\dots,e_n is the standard basis, i.e. D(I)=1D(I) = 1. This equivalence and the resulting uniqueness are established in The Determinant as an Alternating Form; we take it as the standing characterization used throughout this article.

Intuition

Think of an n×nn \times n matrix AA as a linear transformation of FnF^n that scales signed volume by the factor det⁡(A)\det(A): the image of the unit cube (spanned by e1,…,ene_1, \dots, e_n) is a parallelepiped of signed volume det⁡(A)\det(A). Now compose two transformations. First apply BB, turning the unit cube into a parallelepiped of signed volume det⁡(B)\det(B). Then apply AA to that parallelepiped. Since AA scales every region's volume by the same factor det⁡(A)\det(A) — this is exactly what linearity and the multilinear-alternating characterization guarantee — the final volume is det⁡(A)\det(A) times det⁡(B)\det(B). But the final shape is also the image of the unit cube under the single map ABAB (first BB, then AA), so its signed volume is det⁡(AB)\det(AB). Matching these two computations of the same quantity gives det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B). This picture is completely rigorous in Rn\mathbb{R}^n with the geometric meaning of "volume scaling factor," but the algebraic proof below works over an arbitrary field, where "volume" has no metric meaning — the multilinear alternating characterization is the tool that lets the geometric intuition survive in full generality.

Figure
Composition of two shears: areas multiply
Unit square, area = 1 e₁, e₂ B: shear + scale, area = det(B) = 2 Be₁, Be₂ A: stretch, det(A) = 1.5 A(Be₁), A(Be₂) Result area = det(AB) = 1.5 × 2 = 3
TheoremMultiplicativity of the determinant

Let FF be a field and let n≥1n \geq 1. For all n×nn \times n matrices A,BA, B over FF,

det⁡(AB)=det⁡(A) det⁡(B).\det(AB) = \det(A)\,\det(B).
LemmaDeterminant of an elementary matrix and its effect under left multiplication

Let EE be an n×nn \times n elementary matrix, i.e. EE is obtained from the identity InI_n by one of: (i) swapping two rows, (ii) multiplying a row by a nonzero scalar c∈Fc \in F, (iii) adding a scalar multiple of one row to another row. Then for any n×nn \times n matrix MM, the matrix EMEM is obtained from MM by performing the same row operation on MM, and: (i) det⁡(E)=−1\det(E) = -1 and det⁡(EM)=−det⁡(M)\det(EM) = -\det(M); (ii) det⁡(E)=c\det(E) = c and det⁡(EM)=cdet⁡(M)\det(EM) = c\det(M); (iii) det⁡(E)=1\det(E) = 1 and det⁡(EM)=det⁡(M)\det(EM) = \det(M). In particular, det⁡(EM)=det⁡(E)det⁡(M)\det(EM) = \det(E)\det(M) in all three cases.

Proof

This is exactly the content of Determinant Properties and Row Operations, restated for reference. We recall the reasoning briefly since it is the base case of the induction below. The determinant, as a function of the rows of a matrix, is alternating: swapping two rows of MM negates D(M)D(M) (this follows from multilinearity applied to D(R1,…,Ri+Rj,…,Ri+Rj,… )=0D(R_1,\dots, R_i + R_j, \dots, R_i+R_j, \dots) = 0, expanded out and using that DD vanishes on repeated rows). Scaling one row by cc scales DD by cc (this is exactly multilinearity in that row). Adding cc times row jj to row ii leaves DD unchanged, because

D(…,Ri+cRj,…,Rj,… )=D(…,Ri,…,Rj,… )+c D(…,Rj,…,Rj,… )D(\dots, R_i + cR_j, \dots, R_j, \dots) = D(\dots, R_i, \dots, R_j, \dots) + c\,D(\dots, R_j, \dots, R_j, \dots)

by linearity in row ii, and the second term vanishes since rows ii and jj are equal there. Applying these facts with M=InM = I_n gives the stated values of det⁡(E)\det(E). Applying them with general MM (noting that EMEM has rows that are exactly the result of performing the row operation on the rows of MM, which is a direct computation from the block/row structure of matrix multiplication by an elementary matrix) gives det⁡(EM)\det(EM) as stated, and in each case det⁡(EM)=det⁡(E)det⁡(M)\det(EM) = \det(E)\det(M).

∎
Proof

Proof of the Multiplicativity Theorem. We give a complete proof using the uniqueness characterization of the determinant as an alternating multilinear normalized function of the rows, together with the reduction of any square matrix to row-echelon form. This is the standard approach and avoids any circular reliance on facts that themselves depend on multiplicativity.

Step 1: Reduction to elementary matrices. By the theory of row-reduction (Gaussian elimination), every n×nn \times n matrix AA can be written as

A=E1E2⋯EkR,A = E_1 E_2 \cdots E_k R,

where E1,…,EkE_1, \dots, E_k are elementary matrices and RR is the reduced row-echelon form of AA. This is a standard fact from the theory of systems of linear equations: any matrix is transformed to its reduced row-echelon form by a finite sequence of elementary row operations, and performing a row operation on AA is the same as left-multiplying AA by the corresponding elementary matrix; inverting that sequence expresses AA itself as a product of elementary matrices times RR. (This uses no determinant theory, only Gaussian elimination, so there is no circularity.) Two cases arise for RR:

  • Case (a): AA is invertible. Then its reduced row-echelon form is R=InR = I_n (a square matrix in reduced row-echelon form with no zero rows, which must occur if AA is invertible since otherwise AA would have a nontrivial null space and could not be invertible), so
A=E1E2⋯Ek.A = E_1 E_2 \cdots E_k.
  • Case (b): AA is not invertible. Then RR (the reduced row-echelon form) has at least one zero row, because a square reduced row-echelon matrix with no zero rows must be InI_n, which is invertible; if AA is not invertible, R≠InR \ne I_n, forcing a zero row in RR.

Step 2: The invertible case. Suppose A=E1⋯EkA = E_1 \cdots E_k is a product of elementary matrices. We show det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B) by induction on kk. Base case k=1k=1: A=E1A = E_1 is a single elementary matrix. By the Lemma, det⁡(E1B)=det⁡(E1)det⁡(B)=det⁡(A)det⁡(B)\det(E_1 B) = \det(E_1)\det(B) = \det(A)\det(B), as required (the lemma's row operation argument applies verbatim to E1BE_1B in place of EMEM). Inductive step: Suppose the claim holds for products of k−1k-1 elementary matrices, for every choice of the second factor. Let A=E1(E2⋯Ek)A = E_1(E_2 \cdots E_k) and write A′=E2⋯EkA' = E_2 \cdots E_k, so A=E1A′A = E_1 A'. Then

AB=E1A′B=E1(A′B).AB = E_1 A' B = E_1 (A'B).

By the Lemma applied to the elementary matrix E1E_1 and the matrix A′BA'B,

det⁡(AB)=det⁡(E1(A′B))=det⁡(E1)det⁡(A′B).\det(AB) = \det(E_1(A'B)) = \det(E_1)\det(A'B).

By the inductive hypothesis applied to the product A′A' of k−1k - 1 elementary matrices (with second factor BB),

det⁡(A′B)=det⁡(A′)det⁡(B).\det(A'B) = \det(A')\det(B).

Combining,

det⁡(AB)=det⁡(E1)det⁡(A′)det⁡(B).\det(AB) = \det(E_1)\det(A')\det(B).

On the other hand, applying the Lemma with M=A′M = A' (a special case, k′=k−1k' = k-1, already covered, or directly: the Lemma's identity det⁡(EM)=det⁡(E)det⁡(M)\det(EM)=\det(E)\det(M) applies to E1E_1 and M=A′M=A') gives det⁡(A)=det⁡(E1A′)=det⁡(E1)det⁡(A′)\det(A) = \det(E_1 A') = \det(E_1)\det(A'). Substituting,

det⁡(AB)=det⁡(A)det⁡(B).\det(AB) = \det(A)\det(B).

This completes the induction, proving the theorem whenever AA is invertible, for every matrix BB.

Step 3: The non-invertible case. Now suppose AA is not invertible. We must show det⁡(AB)=det⁡(A)det⁡(B)=0⋅det⁡(B)=0\det(AB) = \det(A)\det(B) = 0 \cdot \det(B) = 0, i.e. we must show det⁡(A)=0\det(A) = 0 and det⁡(AB)=0\det(AB) = 0. det⁡(A)=0\det(A) = 0: Write A=E1⋯EkRA = E_1 \cdots E_k R with RR the reduced row-echelon form of AA, which (Step 1, Case (b)) has a zero row. By repeated application of the Lemma exactly as in Step 2 (each elementary matrix multiplication multiplies the determinant by a nonzero scalar det⁡(Ei)\det(E_i), since elementary matrices are always invertible with det⁡(Ei)∈{−1}∪(F∖{0})∪{1}\det(E_i) \in \{-1\} \cup (F\setminus\{0\}) \cup \{1\}, never 00),

det⁡(A)=det⁡(E1)det⁡(E2)⋯det⁡(Ek)det⁡(R).\det(A) = \det(E_1)\det(E_2)\cdots \det(E_k)\det(R).

Since RR has a zero row, det⁡(R)=0\det(R) = 0: this follows directly from multilinearity, since the determinant is linear in that zero row and D(…,0,… )=0⋅D(…,Ri,… )=0D(\dots, 0, \dots) = 0 \cdot D(\dots, R_i,\dots) = 0 for any filler vector (take the zero row as 0⋅v0 \cdot v for any vv). Hence det⁡(A)=0\det(A) = 0. det⁡(AB)=0\det(AB) = 0: We claim ABAB is also not invertible, which (by the same argument just given, applied to ABAB in place of AA) implies det⁡(AB)=0\det(AB) = 0. Indeed, if ABAB were invertible, there would exist a matrix CC with (AB)C=In(AB)C = I_n, i.e. A(BC)=InA(BC) = I_n. This exhibits AA as having a right inverse. For square matrices over a field, having a right inverse implies invertibility: if A(BC)=InA(BC) = I_n then the linear map represented by AA is surjective (since for any yy, y=A(BCy)y = A(BCy)), and a surjective linear map from FnF^n to FnF^n is injective (rank-nullity: dim⁡ker⁡A=n−rank⁡(A)=n−n=0\dim\ker A = n - \operatorname{rank}(A) = n - n = 0 when the map is surjective, since surjectivity forces rank⁡(A)=n\operatorname{rank}(A)=n), hence bijective, hence AA is invertible. This contradicts our assumption that AA is not invertible. Therefore ABAB is not invertible, so det⁡(AB)=0\det(AB) = 0. Combining, det⁡(AB)=0=0⋅det⁡(B)=det⁡(A)det⁡(B)\det(AB) = 0 = 0 \cdot \det(B) = \det(A)\det(B).

Conclusion. In both cases (Step 2 and Step 3), det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B) for all n×nn \times n matrices AA and BB. ■\blacksquare

∎
CorollaryDeterminant of the inverse

If AA is an invertible n×nn \times n matrix over FF, then

det⁡(A−1)=det⁡(A)−1=1det⁡(A).\det(A^{-1}) = \det(A)^{-1} = \frac{1}{\det(A)}.
Proof

Since AA−1=InA A^{-1} = I_n, the Multiplicativity Theorem gives

det⁡(A)det⁡(A−1)=det⁡(AA−1)=det⁡(In)=1.\det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I_n) = 1.

By Step 3 of the proof above, AA invertible implies det⁡(A)≠0\det(A) \neq 0 (the contrapositive of "AA not invertible ⇒det⁡(A)=0\Rightarrow \det(A) = 0"), so we may divide by det⁡(A)\det(A) in the field FF to get det⁡(A−1)=det⁡(A)−1\det(A^{-1}) = \det(A)^{-1}.

∎
CorollaryDeterminant of powers

For an n×nn \times n matrix AA over FF and any integer k≥0k \geq 0,

det⁡(Ak)=det⁡(A)k.\det(A^k) = \det(A)^k.

If in addition AA is invertible, this holds for all integers k∈Zk \in \mathbb{Z}, including negative kk.

Proof

For k=0k = 0: A0=InA^0 = I_n and det⁡(In)=1=det⁡(A)0\det(I_n) = 1 = \det(A)^0 (using the convention x0=1x^0=1). For k≥1k \geq 1: induct on kk. The base case k=1k=1 is trivial. Assume det⁡(Ak−1)=det⁡(A)k−1\det(A^{k-1}) = \det(A)^{k-1}. Then by the Multiplicativity Theorem applied to the product Ak−1⋅AA^{k-1} \cdot A,

det⁡(Ak)=det⁡(Ak−1A)=det⁡(Ak−1)det⁡(A)=det⁡(A)k−1det⁡(A)=det⁡(A)k.\det(A^k) = \det(A^{k-1} A) = \det(A^{k-1})\det(A) = \det(A)^{k-1}\det(A) = \det(A)^k.

If AA is invertible and k=−mk = -m with m>0m > 0, then Ak=(A−1)mA^k = (A^{-1})^m, and by the positive-power case applied to A−1A^{-1},

det⁡(Ak)=det⁡((A−1)m)=det⁡(A−1)m=(det⁡(A)−1)m=det⁡(A)−m=det⁡(A)k,\det(A^k) = \det\big((A^{-1})^m\big) = \det(A^{-1})^m = \big(\det(A)^{-1}\big)^m = \det(A)^{-m} = \det(A)^k,

using the Corollary on inverses for the middle step.

∎
CorollarySimilar matrices have equal determinant

If AA and BB are n×nn \times n matrices over FF that are similar, i.e. there exists an invertible PP with B=P−1APB = P^{-1}AP, then

det⁡(B)=det⁡(A).\det(B) = \det(A).
Proof

By the Multiplicativity Theorem (applied twice) and the Corollary on inverses,

det⁡(B)=det⁡(P−1AP)=det⁡(P−1)det⁡(A)det⁡(P)=det⁡(P)−1det⁡(A)det⁡(P)=det⁡(A),\det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(P)^{-1}\det(A)\det(P) = \det(A),

where the last equality uses commutativity of multiplication in the field FF to cancel det⁡(P)−1det⁡(P)=1\det(P)^{-1}\det(P) = 1 against the det⁡(A)\det(A) factor.

∎

Worked examples

Example 1: A direct numerical check with non-commuting, asymmetric 3×33\times 3 matrices. Let

A=(201−130111),B=(12001−1302).A = \begin{pmatrix} 2 & 0 & 1 \\ -1 & 3 & 0 \\ 1 & 1 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & -1 \\ 3 & 0 & 2 \end{pmatrix}.

First compute det⁡(A)\det(A) by cofactor expansion along the first row:

det⁡(A)=2∣3011∣−0∣−1011∣+1∣−1311∣=2(3⋅1−0⋅1)−0+1(−1⋅1−3⋅1)=2(3)+(−4)=6−4=2.\det(A) = 2\begin{vmatrix}3 & 0\\1&1\end{vmatrix} - 0\begin{vmatrix}-1&0\\1&1\end{vmatrix} + 1\begin{vmatrix}-1&3\\1&1\end{vmatrix} = 2(3\cdot 1 - 0\cdot 1) - 0 + 1(-1\cdot 1 - 3\cdot 1) = 2(3) + (-4) = 6 - 4 = 2.

Next compute det⁡(B)\det(B) along the first row:

det⁡(B)=1∣1−102∣−2∣0−132∣+0=1(1⋅2−(−1)⋅0)−2(0⋅2−(−1)⋅3)=1(2)−2(3)=2−6=−4.\det(B) = 1\begin{vmatrix}1&-1\\0&2\end{vmatrix} - 2\begin{vmatrix}0&-1\\3&2\end{vmatrix} + 0 = 1(1\cdot2 - (-1)\cdot 0) - 2(0\cdot 2 - (-1)\cdot 3) = 1(2) - 2(3) = 2 - 6 = -4.

By the theorem, we should get det⁡(AB)=det⁡(A)det⁡(B)=2⋅(−4)=−8\det(AB) = \det(A)\det(B) = 2 \cdot (-4) = -8. Now compute ABAB directly:

AB=(201−130111)(12001−1302)=(2⋅1+0⋅0+1⋅32⋅2+0⋅1+1⋅02⋅0+0⋅(−1)+1⋅2−1⋅1+3⋅0+0⋅3−1⋅2+3⋅1+0⋅0−1⋅0+3⋅(−1)+0⋅21⋅1+1⋅0+1⋅31⋅2+1⋅1+1⋅01⋅0+1⋅(−1)+1⋅2)=(542−11−3431).AB = \begin{pmatrix} 2 & 0 & 1 \\ -1 & 3 & 0 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & -1 \\ 3 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 2\cdot1+0\cdot0+1\cdot3 & 2\cdot2+0\cdot1+1\cdot0 & 2\cdot0+0\cdot(-1)+1\cdot2\\ -1\cdot1+3\cdot0+0\cdot3 & -1\cdot2+3\cdot1+0\cdot0 & -1\cdot0+3\cdot(-1)+0\cdot2\\ 1\cdot1+1\cdot0+1\cdot3 & 1\cdot2+1\cdot1+1\cdot0 & 1\cdot0+1\cdot(-1)+1\cdot2 \end{pmatrix} = \begin{pmatrix} 5 & 4 & 2 \\ -1 & 1 & -3 \\ 4 & 3 & 1 \end{pmatrix}.

Compute det⁡(AB)\det(AB) by cofactor expansion along the first row:

det⁡(AB)=5∣1−331∣−4∣−1−341∣+2∣−1143∣.\det(AB) = 5\begin{vmatrix}1&-3\\3&1\end{vmatrix} - 4\begin{vmatrix}-1&-3\\4&1\end{vmatrix} + 2\begin{vmatrix}-1&1\\4&3\end{vmatrix}.

Compute each 2×22\times2 minor:

∣1−331∣=1⋅1−(−3)⋅3=1+9=10,∣−1−341∣=(−1)(1)−(−3)(4)=−1+12=11,\begin{vmatrix}1&-3\\3&1\end{vmatrix} = 1\cdot1 - (-3)\cdot3 = 1+9 = 10, \qquad \begin{vmatrix}-1&-3\\4&1\end{vmatrix} = (-1)(1) - (-3)(4) = -1+12 = 11,
∣−1143∣=(−1)(3)−(1)(4)=−3−4=−7.\begin{vmatrix}-1&1\\4&3\end{vmatrix} = (-1)(3) - (1)(4) = -3-4 = -7.

So

det⁡(AB)=5(10)−4(11)+2(−7)=50−44−14=−8.\det(AB) = 5(10) - 4(11) + 2(-7) = 50 - 44 - 14 = -8.

This matches det⁡(A)det⁡(B)=2×(−4)=−8\det(A)\det(B) = 2\times(-4) = -8 exactly, confirming the theorem numerically. Note also that AB≠BAAB \neq BA in general (matrix multiplication is not commutative), yet det⁡(AB)=det⁡(BA)\det(AB) = \det(BA) will still hold — since both equal det⁡(A)det⁡(B)=det⁡(B)det⁡(A)\det(A)\det(B) = \det(B)\det(A) — illustrating that multiplicativity survives non-commutativity of the matrices themselves because the field multiplication det⁡(A)det⁡(B)\det(A)\det(B) is commutative.

Example 2: Using multiplicativity to compute a determinant via decomposition into elementary matrices. Let

A=(0213−14102).A = \begin{pmatrix} 0 & 2 & 1 \\ 3 & -1 & 4 \\ 1 & 0 & 2 \end{pmatrix}.

We compute det⁡(A)\det(A) by row-reducing AA and tracking the elementary matrices, then cross-check with cofactor expansion. Swap R1↔R3R_1 \leftrightarrow R_3 (this is left-multiplication by an elementary matrix E1E_1 with det⁡(E1)=−1\det(E_1) = -1):

A1=(1023−14021).A_1 = \begin{pmatrix} 1 & 0 & 2 \\ 3 & -1 & 4 \\ 0 & 2 & 1 \end{pmatrix}.

Replace R2→R2−3R1R_2 \to R_2 - 3R_1 (elementary matrix E2E_2, det⁡(E2)=1\det(E_2) = 1):

A2=(1020−1−2021).A_2 = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -1 & -2 \\ 0 & 2 & 1 \end{pmatrix}.

Replace R3→R3+2R2R_3 \to R_3 + 2R_2 (elementary matrix E3E_3, det⁡(E3)=1\det(E_3) = 1):

A3=(1020−1−200−3).A_3 = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -1 & -2 \\ 0 & 0 & -3 \end{pmatrix}.

Now A3=E3E2E1AA_3 = E_3 E_2 E_1 A, i.e. A=E1−1E2−1E3−1A3A = E_1^{-1} E_2^{-1} E_3^{-1} A_3. By repeated use of the Lemma (equivalently the Multiplicativity Theorem restricted to elementary matrices, proved directly in the Lemma), each left-multiplication by EiE_i scaled the determinant by det⁡(Ei)\det(E_i):

det⁡(A3)=det⁡(E3)det⁡(E2)det⁡(E1)det⁡(A)=(1)(1)(−1)det⁡(A)=−det⁡(A).\det(A_3) = \det(E_3)\det(E_2)\det(E_1)\det(A) = (1)(1)(-1)\det(A) = -\det(A).

Since A3A_3 is upper triangular, det⁡(A3)=1⋅(−1)⋅(−3)=3\det(A_3) = 1 \cdot (-1) \cdot (-3) = 3 (the determinant of a triangular matrix is the product of its diagonal entries, itself a direct consequence of expanding the permutation sum: only the identity permutation contributes a nonzero term). So

3=−det⁡(A)  ⟹  det⁡(A)=−3.3 = -\det(A) \implies \det(A) = -3.

Cross-check via direct cofactor expansion along the first row of the original AA:

det⁡(A)=0∣−1402∣−2∣3412∣+1∣3−110∣=0−2(3⋅2−4⋅1)+1(3⋅0−(−1)⋅1)=−2(2)+1(1)=−4+1=−3.\det(A) = 0\begin{vmatrix}-1&4\\0&2\end{vmatrix} - 2\begin{vmatrix}3&4\\1&2\end{vmatrix} + 1\begin{vmatrix}3&-1\\1&0\end{vmatrix} = 0 - 2(3\cdot2 - 4\cdot1) + 1(3\cdot0 - (-1)\cdot1) = -2(2) + 1(1) = -4+1 = -3.

Both methods agree: det⁡(A)=−3\det(A) = -3. This illustrates precisely how the elementary-matrix decomposition used in the proof of the Multiplicativity Theorem is also a practical computational tool, not just a proof device.

CounterexampleWhy "same size square matrices" cannot be dropped

The theorem requires AA and BB to be square matrices of the same size n×nn \times n, so that both ABAB and the individual determinants det⁡(A),det⁡(B)\det(A), \det(B) are defined. If we allow rectangular matrices, the statement doesn't even parse the same way, and naive attempts to generalize it fail. For instance, let

A=(120) (1×3),B=(100123) (3×2).A = \begin{pmatrix} 1 & 2 & 0 \end{pmatrix} \ (1\times 3), \qquad B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 3 \end{pmatrix} \ (3 \times 2).

Here det⁡(A)\det(A) and det⁡(B)\det(B) are not defined at all, since det⁡\det is only defined for square matrices — there is no "determinant" of a 1×31\times3 or 3×23\times2 array. Yet ABAB is the 1×21\times2 matrix (12)\begin{pmatrix}1 & 2\end{pmatrix}, again not square, so det⁡(AB)\det(AB) is also undefined. The entire equation det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B) fails to be meaningful, illustrating that squareness (and matching dimensions for the product to make sense) is not a mere technical convenience but a precondition for the statement to have content. If instead one tries to force a "generalized" version using the Cauchy–Binet formula for a p×np \times n times n×pn \times p product (both rectangular but with a square product), the correct formula is not simply "product of determinants" — it becomes a sum over minors,

det⁡(AB)=∑Sdet⁡(AS)det⁡(BS),\det(AB) = \sum_{S} \det(A_S)\det(B_S),

summed over pp-element subsets SS of {1,…,n}\{1,\dots,n\}, where AS,BSA_S, B_S are appropriate square submatrices. This confirms that the clean multiplicative law genuinely depends on both factors being square of matching size — drop that hypothesis and even the shape of the correct replacement theorem changes.

Scope, edge cases & common traps

SituationWhat happens
A,BA, B both n×nn\times n, any field FFdet⁡(AB)=det⁡(A)det⁡(B)\det(AB)=\det(A)\det(B) always holds
AA or BB singular (det⁡=0\det = 0)Still holds; det⁡(AB)=0\det(AB) = 0 as shown in Step 3 of the proof
A,BA, B non-commutingdet⁡(AB)=det⁡(BA)\det(AB) = \det(BA) always, even though AB≠BAAB \neq BA in general
AA rectangularStatement undefined; see counterexample; Cauchy–Binet is the real analogue
AA invertibledet⁡(A−1)=det⁡(A)−1\det(A^{-1}) = \det(A)^{-1}, det⁡(Ak)=det⁡(A)k\det(A^k)=\det(A)^k for all k∈Zk\in\mathbb Z
Similar matrices B=P−1APB = P^{-1}APdet⁡(B)=det⁡(A)\det(B) = \det(A); determinant is a similarity invariant

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Early 19th-century France, foundational work on determinants and analysis

Cauchy was one of the principal architects of the modern theory of determinants, introducing much of the notation and systematic treatment still used today, including an early clean proof of the multiplication theorem for determinants (building on scattered earlier results of Vandermonde, Laplace, and Binet on special cases). His 1815 memoir organized determinants as objects worthy of study in their own right, rather than merely bookkeeping devices for solving linear systems, and established the permutation-sum formula essentially in its modern form.

History
- 1772: Vandermonde and Laplace give early expansions equivalent to special determinant identities.
- 1812: Binet and Cauchy independently state versions of the multiplication theorem for determinants.
- 1815: Cauchy's memoir systematizes determinant theory, including the general multiplicativity result and modern notation.
- 1841: Jacobi popularizes functional determinants (Jacobians), relying essentially on multiplicativity under composition of maps.
- Late 1800s: The theory is absorbed into the emerging abstract framework of linear algebra over general fields.
Does det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B) require AA and BB to commute?

No. The theorem holds for all square matrices A,BA, B of the same size, regardless of whether AB=BAAB = BA. The reason this is not contradictory is that the equality det⁡(AB)=det⁡(BA)\det(AB) = \det(BA) is a statement about equality of two elements of the field FF — namely both equal det⁡(A)det⁡(B)\det(A)\det(B), and field multiplication is commutative — even though ABAB and BABA can be completely different matrices. Multiplying scalars commutes even when multiplying the matrices that produced them does not.

Is the theorem true over rings that are not fields, like Z\mathbb{Z}?

Yes, with the same proof strategy adapted carefully: the theorem and its proof via elementary matrices and row reduction generalize to matrices over any commutative ring, though row-reduction to reduced row-echelon form requires more care (you can't always divide to make a pivot equal to 1). The permutation-sum definition of the determinant makes sense over any commutative ring, and multiplicativity can alternatively be proven directly from that formula using properties of the sign homomorphism on SnS_n, sidestepping row reduction entirely. The corollary about det⁡(A−1)\det(A^{-1}) requires AA to be a unit in the ring.

Why can't we just prove multiplicativity by directly expanding the permutation sum for det⁡(AB)\det(AB)?

You can, and it is a valid alternative proof, but it is more combinatorially involved: one must show

∑σ∈Snsgn⁡(σ)∏i(AB)iσ(i)=(∑τsgn⁡(τ)∏iaiτ(i))(∑ρsgn⁡(ρ)∏ibiρ(i)),\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_i (AB)_{i\sigma(i)} = \Big(\sum_{\tau}\operatorname{sgn}(\tau)\prod_i a_{i\tau(i)}\Big)\Big(\sum_{\rho}\operatorname{sgn}(\rho)\prod_i b_{i\rho(i)}\Big),

which requires expanding (AB)ij=∑kaikbkj(AB)_{ij} = \sum_k a_{ik}b_{kj}, regrouping into a sum over functions {1,…,n}→{1,…,n}\{1,\dots,n\}\to\{1,\dots,n\}, discarding non-bijective terms (they cancel via the alternating property), and reindexing via σ=ρ∘τ−1\sigma = \rho\circ\tau^{-1}. This works but is more delicate than the elementary-matrix argument, which reuses the already-proven row-operation facts.

Does multiplicativity mean determinant is a linear map on matrices?

No — this is a common confusion. The determinant is multiplicative (a homomorphism from the multiplicative monoid of matrices to the multiplicative monoid of the field), but it is emphatically not additive: det⁡(A+B)≠det⁡(A)+det⁡(B)\det(A+B) \neq \det(A) + \det(B) in general, as the counterexample with A=B=I2A=B=I_2 shows (det⁡(2I2)=4≠1+1\det(2I_2)=4 \ne 1+1). Determinant is multilinear as a function of the individual rows (or columns) of a single matrix, which is a different and more refined statement than linearity in the whole matrix as an additive object.

How does this theorem relate to determinant and invertibility?

Multiplicativity is the key algebraic tool used to prove that AA is invertible if and only if det⁡(A)≠0\det(A)\neq 0. One direction (invertible ⇒det⁡≠0\Rightarrow \det \ne 0) is an immediate corollary here: if AA−1=IAA^{-1}=I, then det⁡(A)det⁡(A−1)=1\det(A)\det(A^{-1}) = 1, forcing det⁡(A)≠0\det(A) \ne 0. The converse (det⁡(A)≠0⇒\det(A) \ne 0 \Rightarrow invertible) is proved via row reduction (as in Step 1 of the proof above) independent of multiplicativity, and the two facts together give the full biconditional developed in Determinant and Invertibility.

Can I use multiplicativity to compute determinants faster in practice?

Yes — this is exactly the row-reduction method in Worked Example 2. Rather than expanding an n×nn\times n determinant via cofactors (which costs O(n!)O(n!) operations), you row-reduce to (upper) triangular form using elementary row operations, track the multiplicative effect of each operation (−1-1 for a swap, cc for scaling by cc, unchanged for adding a multiple of one row to another), and multiply by the product of the diagonal entries of the triangular result. This costs only O(n3)O(n^3) operations and is the standard algorithm used in numerical software.

What happens to the theorem if we work with linear operators instead of matrices?

It carries over unchanged, because the determinant of a linear operator T:V→VT: V \to V on a finite-dimensional vector space is defined as det⁡([T]β)\det([T]_\beta) for any basis β\beta, and this is well-defined (independent of β\beta) precisely because of the similarity-invariance corollary proved above. Since [ST]β=[S]β[T]β[ST]_\beta = [S]_\beta[T]_\beta for the matrix of a composition, multiplicativity for matrices immediately gives det⁡(ST)=det⁡(S)det⁡(T)\det(ST) = \det(S)\det(T) for linear operators, with no extra work.

Before this

Where this leads