Determinants and Their Properties

The determinant collapses an entire square matrix into a single number, and that number carries an astonishing amount of the matrix's structure. It tells you at a glance whether the matrix is invertible, by what factor the linear map it represents stretches or shrinks volume, whether it preserves or flips orientation, what the product of its eigenvalues is, and what the constant term of its characteristic polynomial equals. A quantity that compact, doing that much work, is rare — and it is why the determinant threads through nearly every corner of linear algebra and mathematical physics, from Cramer's rule for small systems to the Jacobian that rescales every multidimensional integral you will ever compute. The reason one scalar can do so much is that it is forced to exist by three simple demands. Ask for a function of a matrix's columns that is linear in each column separately, that changes sign whenever two columns are swapped, and that equals one on the identity — and there is exactly one such function. That uniqueness is the engine of the whole subject. The explicit Leibniz sum over permutations, the cofactor expansion, the product rule, transpose invariance, Cramer's rule, the eigenvalue product, and the volume interpretation are not independent facts to memorize; each is a short consequence of those three axioms. This article takes the axioms as the ground truth and shows how everything else falls out, keeping the geometry — signed volume — always in view as the picture behind the algebra.

Key takeaways

The determinant det A is the unique scalar function of an n×n matrix that is multilinear and antisymmetric in its columns and normalized so that det I = 1.
Geometrically |det A| is the factor by which the map A scales n-dimensional volume, and the sign of det A records whether A preserves or reverses orientation.
A matrix is invertible exactly when det A ≠ 0, so the determinant is the single scalar test for singularity, linear independence of columns, and unique solvability of Ax = b.
The product rule det(AB) = det A · det B makes the determinant a homomorphism from GL(n) to nonzero scalars, which forces both transpose invariance and similarity invariance det(P⁻¹AP) = det A.
Practical routes are the Leibniz sum, the Laplace cofactor expansion, and — for anything large — Gaussian elimination reading the product of pivots in O(n³) rather than the O(n!) Leibniz cost.
The determinant equals the product of the eigenvalues and is the constant term of the characteristic polynomial, linking it to the trace, the adjugate, Cramer's rule, and the Jacobian of a change of variables.
Key idea: the determinant is the unique multilinear, alternating, normalized scalar function of a matrix's columns; invertibility, signed-volume scaling, the product rule, Cramer's rule, and the eigenvalue product are all short consequences of those three axioms.

The core insight

The determinant is the unique multilinear, alternating, normalized scalar function of a matrix's columns; everything else — invertibility, signed-volume scaling, the product rule, Cramer's rule, and the eigenvalue product — follows from those three axioms.

Where this lives today

The determinant is quietly load-bearing across computational physics and machine learning, almost always through its logarithm rather than the raw number, which for large matrices overflows or underflows. The log-determinant of a covariance matrix sets the normalization of every multivariate Gaussian, so training a Gaussian process or fitting a probabilistic model means differentiating log det Σ millions of times. In modern generative models, normalizing flows transform one probability density into another and pay for it with exactly the Jacobian determinant of equation (8) — the whole architecture is engineered so that determinant stays cheap to compute. In quantum many-body physics the Slater determinant builds an antisymmetric wavefunction for identical fermions, encoding the Pauli exclusion principle directly in the vanishing of a determinant with two equal columns; and lattice quantum chromodynamics spends most of its supercomputer time evaluating the fermion determinant that results from integrating out Grassmann fields. General relativity carries the metric determinant g under a square root in every covariant volume element √(−g) d⁴x, the curved-spacetime descendant of the spherical Jacobian above. And the humble sign of the determinant still classifies the orthogonal group into proper rotations and reflections, a distinction that underlies parity in particle physics and the orientability of manifolds.

The physical picture

Lay the columns of a matrix A end-to-origin as edge vectors of a box. In two dimensions two columns span a parallelogram; in three, three columns span a parallelepiped; in n dimensions they span an n-parallelepiped. The determinant is the signed volume of that box. This one image explains the axioms directly. Shear the box — add a multiple of one edge to another — and its base and height are unchanged, so the volume, and the determinant, do not move: that is column-replacement invariance. Stretch a single edge by λ and the volume scales by λ: that is linearity in one column. Flatten the box until its edges become linearly dependent and the volume, and the determinant, drop to zero: that is how det A = 0 detects singularity. The first figure makes "columns span a box, det is its signed area" a single concrete picture with real numbers.

Figure

The columns of a matrix span a parallelogram whose signed area is the determinant — here a₁ = (3, 1) and a₂ = (1, 2) give det = 3·2 − 1·1 = 5

The sign of the determinant is orientation. Swapping two columns exchanges two axes and reverses handedness, which is precisely why the axioms demand a sign flip; a reflection carries det = −1, a proper rotation det = +1. And the multiplicativity that makes the determinant basis-independent is just the statement that volume scaling compounds: apply B then A, and the two volume factors multiply. The second figure isolates the three elementary column moves acting on the humble unit square, turning each axiom into a number you can read off.

Figure

Four unit squares under different matrices — the identity keeps area 1, a shear keeps area 1 with the determinant unchanged, a diagonal stretch gives area 6, and dependent columns collapse the square to a line with det = 0

The core equations

The most explicit definition writes the determinant as a signed sum over all orderings of the columns. Each term picks exactly one entry from every row and every column; the sign is the parity of the permutation that made the selection. The sum has n! terms, which is why the Leibniz formula is best read as a definition rather than a working algorithm.

\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n}A_{i\,\sigma(i)} \qquad (1)

where: $S_n$ — the symmetric group of all $n!$ permutations of $\{1,\dots,n\}$ ; $\sigma$ — a permutation, with $\sigma(i)$ the image of index $i$ ; $\operatorname{sgn}(\sigma)$ — $+1$ for even and $-1$ for odd permutations, the parity of the number of transpositions; $A_{i\,\sigma(i)}$ — the entry in row $i$ , column $\sigma(i)$ . Reading: multiply one entry from each row so that no two share a column, attach the permutation's sign, and add over all such choices. Fixing one row and recursing gives the formula best suited to hand computation and to matrices with many zeros. Expanding along row $i$ turns an $n\times n$ determinant into $n$ determinants of size $n-1$ , each weighted by a checkerboard sign.

\det A=\sum_{j=1}^{n}(-1)^{i+j}A_{ij}\,M_{ij} \qquad (2)

where: $M_{ij}$ — the $(i,j)$ minor, the determinant of $A$ with row $i$ and column $j$ deleted; $C_{ij}=(-1)^{i+j}M_{ij}$ — the $(i,j)$ cofactor. Reading: pick any convenient row or column, and the more zeros it holds the fewer minors you must evaluate. The single most important algebraic property is that the determinant respects matrix multiplication. It converts a statement about a composite map into the product of two numbers, and it is the source of nearly every other identity below.

\det(AB)=\det A\,\det B \qquad (3)

where: $A,B$ — any $n\times n$ matrices over the same field. Reading: composing maps multiplies their volume factors, so the determinant is a homomorphism from the group of invertible matrices to the nonzero scalars. Because the Leibniz sum is symmetric under exchanging the roles of rows and columns, the determinant is blind to transposition. Every property stated for columns therefore holds verbatim for rows.

\det A^{\mathsf T}=\det A \qquad (4)

where: $A^{\mathsf T}$ — the transpose of $A$ . Reading: multilinearity and antisymmetry in the columns automatically give the same in the rows, so "expand along a row" and "expand along a column" are equally valid. Setting $B=A^{-1}$ in the product rule immediately inverts the determinant, and the cofactors assemble into an explicit formula for the inverse itself.

\det(A^{-1})=\frac{1}{\det A},\qquad A^{-1}=\frac{1}{\det A}\,\operatorname{adj}(A) \qquad (5)

where: $\operatorname{adj}(A)$ — the adjugate, the transpose of the cofactor matrix, $(\operatorname{adj}A)_{ij}=C_{ji}$ ; the inverse exists precisely when $\det A\neq0$ . Reading: the adjugate formula is the direct route from cofactors to the inverse, and it is exact but expensive for large $n$ . Applying the product rule to a change of basis makes the two conjugating factors cancel, so the determinant is a genuine property of the underlying linear operator, not of the coordinates chosen to write it.

\det(P^{-1}AP)=\det A \qquad (6)

where: $P$ — any invertible change-of-basis matrix. Reading: since $\det(P^{-1})\det(P)=1$ , the determinant survives every similarity transformation, which is what lets us speak of "the determinant of a linear map". Basis independence lets the determinant read the eigenvalues, which live in the operator, not the basis. The characteristic polynomial packages both the determinant and the trace as its extreme coefficients.

p(\lambda)=\det(A-\lambda I),\qquad \det A=\prod_{k=1}^{n}\lambda_k \qquad (7)

where: $\lambda_k$ — the eigenvalues of $A$ counted with algebraic multiplicity, the roots of $p(\lambda)$ ; $I$ — the $n\times n$ identity. Reading: the constant term $p(0)=\det A$ is the product of the eigenvalues, while the trace $\operatorname{tr}A=\sum_k\lambda_k$ is their sum; a single zero eigenvalue forces $\det A=0$ . Finally, the geometric meaning of the determinant as a volume factor is exactly what makes it the correct weight when rescaling a volume element from one coordinate system to another.

\mathrm{d}^n y=\left|\det\frac{\partial y}{\partial x}\right|\,\mathrm{d}^n x \qquad (8)

where: $\partial y/\partial x$ — the Jacobian matrix of the map $x\mapsto y$ ; its absolute determinant is the local factor by which volume elements stretch. Reading: this is why $\mathrm{d}x\,\mathrm{d}y = r\,\mathrm{d}r\,\mathrm{d}\theta$ in the plane and $r^2\sin\theta$ appears in spherical coordinates.

Derivation AThe Leibniz formula forced by three axioms

The goal is to show that the three defining demands — linearity in each column, a sign change whenever two columns are swapped, and the value one on the identity — do not merely constrain the determinant but pin it down completely, and that unpacking them mechanically produces the explicit n!-term Leibniz sum. We expand every column in the standard basis, let multilinearity distribute the product across all basis choices, use antisymmetry to discard every choice with a repeated index, and read off the permutation signs from the surviving terms.

Step 1. Write each column of $A$ in the standard basis, $a_j=\sum_i A_{ij}\,e_i$ , so the matrix is presented purely as combinations of the fixed vectors $e_i$ . Why: the axioms know how to act on the $e_i$ — they are exactly the arguments on which multilinearity and antisymmetry are stated.

Step 2. Push the determinant through every column by multilinearity, distributing over each independent basis choice:

\det A=\sum_{i_1,\dots,i_n}A_{i_1 1}A_{i_2 2}\cdots A_{i_n n}\,\det\!\big(e_{i_1},e_{i_2},\dots,e_{i_n}\big).

Why: linearity in each column separately turns one determinant of sums into a sum of determinants of pure basis vectors, with the matrix entries pulled out front as ordinary scalars.

Step 3. Discard every term in which two indices coincide. Antisymmetry makes $\det$ vanish whenever two of its arguments are equal, so only the choices with all $i_1,\dots,i_n$ distinct — that is, the permutations $\sigma$ of $\{1,\dots,n\}$ — survive. Why: this is where the $n^n$ raw terms collapse to just the $n!$ meaningful ones; a repeated basis vector means a flattened box of zero volume.

Step 4. Evaluate each survivor. Reordering $(e_{\sigma(1)},\dots,e_{\sigma(n)})$ back to standard order costs one sign per transposition, so $\det(e_{\sigma(1)},\dots,e_{\sigma(n)})=\operatorname{sgn}(\sigma)\det I=\operatorname{sgn}(\sigma)$ , giving $\det A=\sum_\sigma\operatorname{sgn}(\sigma)\prod_i A_{i\,\sigma(i)}$ . Why: the normalization $\det I=1$ fixes the last free constant, so the axioms produce not just a formula but the unique determinant. *Verified: for $n=2$ the sum has the two permutations, identity and swap, giving $A_{11}A_{22}-A_{12}A_{21}=ad-bc$ ; for the $3\times3$ case it reproduces the familiar six-term rule.*

Go deeper · Derivation B · 1 minWhy determinants multiply

Derivation BWhy determinants multiply

The goal is to prove the product rule det(AB) = det A · det B without ever expanding a single sum, using only the fact that the determinant is the unique multilinear alternating normalized form. We hold A fixed, view det(AB) as a function of the columns of B, check that it inherits both defining properties, invoke uniqueness to conclude it must be a constant multiple of det B, and then pin the constant by evaluating at the identity.

Step 1. Fix $A$ and define $D(B)=\det(AB)$ , regarded as a function of the columns $b_1,\dots,b_n$ of $B$ . Note that column $j$ of $AB$ is exactly $Ab_j$ . Why: recasting a statement about the product $AB$ as a statement about the columns of $B$ lets the column axioms do all the work.

Step 2. Show $D$ is multilinear. Since $Ab_j$ is linear in $b_j$ and $\det$ is linear in each of its column arguments, the composition $D$ is linear in each $b_j$ separately. Why: linearity survives composition with a linear map, so no new computation is needed.

Step 3. Show $D$ is alternating. If two columns of $B$ coincide, $b_i=b_j$ , then $Ab_i=Ab_j$ , so $AB$ has two equal columns and $\det(AB)=0$ . Why: the alternating property of $\det$ transfers straight through, because $A$ maps equal inputs to equal outputs.

Step 4. Invoke uniqueness. Any multilinear alternating function of $n$ vectors in $n$ -dimensional space is a constant multiple of the determinant, so $D(B)=c\,\det B$ for some scalar $c$ . Evaluate at $B=I$ : $D(I)=\det(AI)=\det A$ , while $\det I=1$ , so $c=\det A$ and $\det(AB)=\det A\,\det B$ . Why: the uniqueness theorem is the whole lever — it converts "shares both properties" into "equals det, up to one number", and the identity fixes that number. *Verified: setting $B=A^{-1}$ gives $\det(A^{-1})=1/\det A$ , and $\det(P^{-1}AP)=\det(P^{-1})\det A\det P=\det A$ , recovering equations (5) and (6) as one-line corollaries.*

Go deeper · Derivation C · 1 minCramer's rule from the column axioms

Derivation CCramer's rule from the column axioms

The goal is to derive the closed-form solution of a square linear system Ax = b directly from multilinearity and antisymmetry, without inverting the matrix, showing that each unknown is a ratio of two determinants. We replace one column of A with the right-hand side, substitute the expansion of b in the columns of A, split the determinant by linearity, and let antisymmetry annihilate every term but one.

Step 1. Let $A$ have columns $a_1,\dots,a_n$ and suppose $Ax=b$ , so $b=\sum_k x_k a_k$ . Form $A_j$ , the matrix $A$ with its $j$ -th column replaced by $b$ , and consider $\det A_j=\det[a_1,\dots,b,\dots,a_n]$ . Why: placing $b$ in slot $j$ sets up a single determinant whose value we can force to reveal $x_j$ alone.

Step 2. Substitute the expansion of $b$ into slot $j$ and split by multilinearity:

\det A_j=\det\!\Big[a_1,\dots,\textstyle\sum_k x_k a_k,\dots,a_n\Big]=\sum_k x_k\,\det[a_1,\dots,a_k\text{ in slot }j,\dots,a_n].

Why: linearity in the $j$ -th column pulls each unknown $x_k$ out in front of its own determinant.

Step 3. Kill the cross terms. For every $k\neq j$ the column $a_k$ now appears twice — once in its home slot $k$ , once in slot $j$ — so antisymmetry makes that determinant zero. Only $k=j$ survives, and there the matrix is just $A$ again: $\det A_j=x_j\,\det A$ . Why: antisymmetry is the filter that isolates the single unknown, exactly as a repeated column flattens the box.

Step 4. Divide to obtain $x_j=\det A_j/\det A$ , valid whenever $\det A\neq0$ . Why: a nonzero determinant is both the invertibility condition and the denominator, so Cramer's rule and "unique solution exists" are the same statement. *Verified: for $2\times2$ with $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $b=(e,f)^{\mathsf T}$ , the rule gives $x_1=(ed-bf)/(ad-bc)$ , the standard elimination answer.*

Limiting cases

Special structure collapses the general n! sum into something you can evaluate at sight, and each case is a direct reading of the axioms.

Case	Result	Meaning
Triangular or diagonal	$\det A=\prod_{i=1}^{n}A_{ii}$	Only the identity permutation avoids a zero below or above the diagonal, so the determinant is the product of the diagonal entries — the basis of the Gaussian-elimination method.
Scalar multiple	$\det(cA)=c^{n}\det A$	Scaling the whole matrix scales all $n$ columns, contributing one factor of $c$ each — not a single factor.
Block triangular	$\det\begin{pmatrix}A&B\\0&D\end{pmatrix}=\det A\,\det D$	A zero off-diagonal block lets the determinant factor over the square diagonal blocks, for any conformable $A$ and $D$ .
Orthogonal or unitary	$\lvert\det U\rvert=1$	Volume-preserving maps have determinant on the unit circle, so $\det=\pm1$ for real orthogonal matrices, distinguishing rotations from reflections.
Two-by-two closed form	$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$	The smallest nontrivial case, equal to the signed area of the parallelogram spanned by the columns.

Common trapThe determinant adds like a linear map

The most common error is to treat the determinant as if it were linear in the matrix as a whole — to write $\det(A+B)=\det A+\det B$ , or $\det(cA)=c\,\det A$ . Both are false. The multilinearity in the definition refers to each column separately, holding every other column fixed; it says nothing about adding two full matrices or scaling all columns at once. Concretely, $\det(cA)=c^n\det A$ because all $n$ columns get scaled, and $\det(I+I)=\det(2I)=2^n$ while $\det I+\det I=2$ , so the two sides disagree already for $n\ge2$ . There is no useful formula for $\det(A+B)$ in general; the determinant interacts cleanly with products, not with sums.

Worked examples

1 · A 3×3 determinant two ways, and what it tells you (confidence builder)

Compute the determinant of the matrix below by cofactor expansion and again by row reduction, confirm the two agree, and state whether the matrix is invertible.

A=\begin{pmatrix}2&1&1\\[2pt]1&3&2\\[2pt]1&0&0\end{pmatrix}

Step 1. Expand along the third row, which carries two zeros so only one minor survives. The entries are $(1,0,0)$ , so $\det A=1\cdot(-1)^{3+1}M_{31}$ , where $M_{31}=\det\begin{pmatrix}1&1\\3&2\end{pmatrix}=1\cdot2-1\cdot3=-1$ . Hence $\det A=1\cdot(+1)\cdot(-1)=\mathbf{-1}$ .

Step 2. Cross-check by expanding along the first row instead, with signs $+,-,+$ : the minors are $M_{11}=\det\begin{pmatrix}3&2\\0&0\end{pmatrix}=0$ , $M_{12}=\det\begin{pmatrix}1&2\\1&0\end{pmatrix}=-2$ , and $M_{13}=\det\begin{pmatrix}1&3\\1&0\end{pmatrix}=-3$ . Then $\det A=2\cdot0-1\cdot(-2)+1\cdot(-3)=0+2-3=\mathbf{-1}$ , agreeing with Step 1.

Step 3. Now reduce to triangular form, tracking that adding a multiple of one row to another leaves the determinant unchanged. Using the first pivot, $R_2\to R_2-\tfrac12R_1=(0,\ \tfrac52,\ \tfrac32)$ and $R_3\to R_3-\tfrac12R_1=(0,\ -\tfrac12,\ -\tfrac12)$ ; then $R_3\to R_3+\tfrac15R_2=(0,\ 0,\ -\tfrac15)$ . The pivots are $2,\ \tfrac52,\ -\tfrac15$ , and their product is $2\cdot\tfrac52\cdot(-\tfrac15)=\mathbf{-1}$ — the same value, obtained in $O(n^3)$ work rather than by summing permutations.

Step 4. Interpret. Since $\det A=-1\neq0$ , the matrix is invertible, its columns are linearly independent and form a basis of $\mathbb{R}^3$ , and the system $Ax=b$ has a unique solution for every $b$ . The product of the three eigenvalues equals $-1$ , and the negative sign means $A$ reverses orientation. det A = −1; A is invertible and orientation-reversing.

2 · Solving a linear system by Cramer's rule

Solve the system below using ratios of determinants, and verify the solution.

2x+y=5,\qquad x+3y=10

Step 1. The coefficient matrix is $A=\begin{pmatrix}2&1\\1&3\end{pmatrix}$ with $\det A=2\cdot3-1\cdot1=\mathbf{5}$ . Because this is nonzero, the system has a unique solution and Cramer's rule applies.

Step 2. Replace the first column by the right-hand side $(5,10)^{\mathsf T}$ : $A_1=\begin{pmatrix}5&1\\10&3\end{pmatrix}$ , $\det A_1=5\cdot3-1\cdot10=5$ , so $x=\det A_1/\det A=5/5=\mathbf{1}$ .

Step 3. Replace the second column instead: $A_2=\begin{pmatrix}2&5\\1&10\end{pmatrix}$ , $\det A_2=2\cdot10-5\cdot1=15$ , so $y=\det A_2/\det A=15/5=\mathbf{3}$ .

Step 4. Verify by substitution: $2(1)+3=5$ ✓ and $1+3(3)=10$ ✓. Cramer's rule is elegant and exact for small systems, but note each additional unknown costs another full determinant, so for large systems Gaussian elimination is far cheaper. x = 1, y = 3.

3 · The Jacobian of spherical coordinates

Compute the determinant of the Jacobian for the spherical map x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ, and confirm the standard volume element.

Step 1. Assemble the Jacobian matrix $J=\partial(x,y,z)/\partial(r,\theta,\varphi)$ :

J=\begin{pmatrix}\sin\theta\cos\varphi & r\cos\theta\cos\varphi & -r\sin\theta\sin\varphi\\[2pt]\sin\theta\sin\varphi & r\cos\theta\sin\varphi & r\sin\theta\cos\varphi\\[2pt]\cos\theta & -r\sin\theta & 0\end{pmatrix}

Step 2. Expand along the third row, since its last entry is zero. The cofactor of $\cos\theta$ evaluates to $r^2\sin\theta\cos\theta$ (using $\cos^2\varphi+\sin^2\varphi=1$ ), and the cofactor of $-r\sin\theta$ evaluates to $-r\sin^2\theta$ , so $\det J=\cos\theta\,(r^2\sin\theta\cos\theta)+(-r\sin\theta)(-r\sin^2\theta)$ .

Step 3. Combine: $\det J=r^2\sin\theta\cos^2\theta+r^2\sin^3\theta=r^2\sin\theta\,(\cos^2\theta+\sin^2\theta)=\mathbf{r^2\sin\theta}$ .

Step 4. By equation (8) the volume element is $\mathrm{d}^3\mathbf{r}=|\det J|\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi$ , the familiar factor that appears in every integral over a sphere and in the normalization of hydrogen wavefunctions. det J = r² sinθ.

Physics of the portrait

The determinant is one of the oldest ideas in linear algebra and one that had to be discovered several times before it acquired a name and a theory. Its modern form — a single multiplicative, basis-independent number — is largely the work of Augustin-Louis Cauchy.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · the first systematic theory of determinants and the multiplication theorem

Determinant-like calculations are older than the concept itself: Leibniz manipulated the coefficients of linear systems in the 1680s, and Seki Takakazu did the same independently in Japan in the same decade, while Cramer's rule (1750) and Laplace's cofactor expansion (1772) each produced pieces of the machinery without a unifying idea. It was Augustin-Louis Cauchy who, in an 1812 memoir published in 1815, gathered these fragments into a coherent subject. He fixed the modern meaning of the word déterminant, organized the theory around what we now call minors and cofactors, and — most consequentially — proved the multiplication theorem det(AB) = det A · det B in full generality, the identity from which basis independence, the eigenvalue product, and the whole notion of a determinant of an operator descend. Cauchy was a founder of rigorous analysis, giving the field its modern definitions of limit, continuity, and convergence, and he brought the same demand for proof to determinants, replacing lists of formulas with theorems. The thread he tied off was later extended by Jacobi, whose 1841 study of functional determinants gave physics the Jacobian, and by Cayley and Sylvester, who in the 1850s embedded determinants in the algebra of matrices where they now live.

History

1683

— Leibniz and Seki Takakazu independently reduce linear systems using determinant-like elimination

1750

— Cramer publishes the rule solving a system by ratios of determinants

1772

— Laplace gives the cofactor expansion along a row or column

1812

— Cauchy proves the multiplication theorem det(AB) = det A det B and fixes the modern name

1841

— Jacobi develops the functional determinant, the Jacobian of a change of variables

Going further

Texts. Stone & Goldbart, Mathematics for Physics, §A.1 (linear algebra) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 2 · Hassani, Mathematical Physics, Ch. 3–4 (finite-dimensional vector spaces and operators) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 8 · Axler, Linear Algebra Done Right, Ch. 10 · Strang, Linear Algebra and Its Applications, Ch. 4.

Provenance. The three axioms and the uniqueness theorem, and the Leibniz and Laplace formulas, are standard (Axler Ch. 10; Hassani Ch. 3; Arfken–Weber Ch. 2). The product rule, transpose invariance, and Cramer's rule are derived above from uniqueness (Stone & Goldbart §A.1). The eigenvalue product and characteristic-polynomial link (Byron & Fuller Ch. 4) and the Jacobian volume-scaling law (Strang Ch. 4) are the geometric consequences. A verifier should confirm: the $2\times2$ value $ad-bc$ and the six-term $3\times3$ rule from equation (1); the worked $3\times3$ determinant $-1$ by both cofactor expansion and pivot product; the Cramer solution $x=1,\ y=3$ against direct substitution; and the spherical Jacobian $\det J=r^2\sin\theta$ . The numerical checks ( $5^4=625$ , the eigenvalue product $2\cdot3\cdot0=0$ , and $\det(2I)=2^n$ ) are elementary recomputations.

Prerequisites. Linear Operators and Matrix Representations · Linear Independence, Basis, and Dimension · Vector Spaces and Subspaces. Leads to. Eigenvalues, Eigenvectors, and the Characteristic Equation · Change of Basis and Similarity Transformations · Rank, Nullity, and the Fundamental Theorem of Linear Maps · Diagonalization and the Spectral Decomposition.

Ask the tutor: Ask me to walk through any determinant by hand — cofactor expansion, row reduction to pivots, or a Jacobian for your favorite coordinate system — and I will show every sign and step.

Why is det(A + B) not equal to det A + det B?

Because the determinant is linear in each column separately, holding the others fixed, not linear in the whole matrix. Adding two matrices adds all their columns simultaneously, which is a different operation from the one-column-at-a-time linearity the axioms grant. A quick counterexample settles it: take A = B = I in two dimensions. Then det(A + B) = det(2I) = 4, while det A + det B = 1 + 1 = 2. There is no general formula for the determinant of a sum; the determinant plays well with matrix products, through det(AB) = det A · det B, but not with matrix sums. Whenever you are tempted to split a determinant across a sum, stop and reach for a product or a row operation instead.

Rows or columns — does it matter which the axioms use?

It does not, and that is a theorem rather than a convention. Because the Leibniz sum is symmetric under exchanging the roles of rows and columns, det Aᵀ = det A. Every property stated for columns therefore holds verbatim for rows: the determinant is multilinear and alternating in the rows too, adding a multiple of one row to another leaves it unchanged, and you may expand a cofactor along any row or any column you like. In practice you exploit this freedom constantly — you pick whichever row or column has the most zeros to minimize the number of minors, and you use row operations during Gaussian elimination even though the geometric picture was drawn with columns. The two viewpoints are genuinely interchangeable.

Why does a zero determinant mean the matrix is not invertible?

Geometrically, det A = 0 says the box spanned by the columns has collapsed to zero volume, which happens exactly when the columns are linearly dependent — they fail to span the full space. A map that squashes n-dimensional space into something lower-dimensional cannot be undone, because many input vectors get sent to the same output, so no inverse exists. Algebraically the same fact appears in equation (5): the inverse formula A⁻¹ = adj(A)/det A divides by det A, which is only legal when det A ≠ 0. And through equation (7), det A = 0 means at least one eigenvalue is zero, so the matrix has a nontrivial null space. All three statements — collapsed volume, dependent columns, zero eigenvalue — are the same singularity seen from three angles.

How is the determinant actually computed for a large matrix?

Never by the Leibniz formula, whose n! terms make even a 20×20 determinant astronomically expensive. Instead you row-reduce the matrix to triangular form by Gaussian elimination and read off the product of the pivots, tracking a sign for each row swap and a factor for each row scaling. This costs on the order of n³ arithmetic operations, which is entirely practical for matrices with thousands of rows. For very large or ill-conditioned problems, numerical libraries use an LU or QR factorization and, crucially, accumulate the logarithm of the determinant rather than the determinant itself, because the raw value overflows or underflows floating point almost immediately. The cofactor expansion sits between these extremes: fine by hand for a 3×3 with zeros, hopeless for anything large.

What does the sign of the determinant tell you?

The sign records orientation. A positive determinant means the map preserves handedness — a right-handed frame stays right-handed — while a negative determinant means it flips handedness, as a mirror reflection does. This is why swapping two columns, which exchanges two axes, must flip the sign. The distinction is not cosmetic: for orthogonal matrices, det = +1 picks out the proper rotations (the group SO(3)) and det = −1 picks out the improper ones that include a reflection, and this split is exactly the difference between a physical rotation you can perform and one you can only reach by turning an object inside out. The magnitude |det A| gives the volume scale factor; the sign gives the orientation, and you usually need both.

How are the determinant and the trace related?

They are the two ends of the same object, the characteristic polynomial p(λ) = det(A − λI). Its roots are the eigenvalues, and the elementary symmetric functions of those eigenvalues appear as its coefficients. The determinant is the product of all the eigenvalues — the constant term p(0) — while the trace is their sum, the coefficient one step down from the top. So the determinant and trace are the most extreme of a whole ladder of invariants, all of them basis-independent because the characteristic polynomial is. In physics this pairing is everywhere: for a 2×2 system the trace and determinant alone decide stability (a stable node needs negative trace and positive determinant), and for quantum Hamiltonians the trace gives the sum of energy levels while the determinant gives their product.