The Adjugate and Cofactor Matrix

Suppose you need the inverse of a matrix. Suppose you only have access to determinants and a lot of patience. Is there a formula for $A^{-1}$ built purely out of smaller determinants of $A$ ? Could this work with no row reduction at all? The answer is yes. It is one of the most elegant identities in elementary linear algebra. The adjugate matrix, built from cofactors, satisfies $A \cdot \operatorname{adj}(A) = \det(A) I$ . This holds for every square matrix $A$ , invertible or not. This identity matters for an important reason. It converts a global, seemingly indivisible invariant — the determinant — into a matrix built entirely from local data. That local data is the $(n-1)\times(n-1)$ minors of $A$ . It is the bridge between Computing Determinants by Cofactor Expansion and the existence of a closed-form inverse. This is one expressible entirely in terms of determinants, rather than an algorithmic row-reduction procedure. This closed form is indispensable in several situations. You may need to differentiate a matrix inverse symbolically. You may need to bound entries of $A^{-1}$ . Or you may need to reason about inverses over rings where division is not generally available. The big picture is this. The cofactor expansion of the determinant computes a single number. But it secretly contains within it all the information needed to invert the matrix. Every cofactor $C_{ij}$ was originally introduced just as a bookkeeping device. Its purpose was expanding $\det(A)$ along a row. But it turns out to be exactly the right ingredient — after a transpose — to build $A^{-1}$ . This works whenever $\det(A) \ne 0$ . This article proves the adjugate identity in full generality. This holds over any commutative ring, in particular any field. It derives the formula $A^{-1} = \frac{1}{\det A}\operatorname{adj}(A)$ . It works two nontrivial numerical examples. It shows exactly why invertibility is essential to extracting the inverse from the identity. It closes by pointing toward Cramer's Rule. This is this same identity, read one column at a time.

Key takeaways

The cofactor $C_{ij}$ of $A$ is $(-1)^{i+j}$ times the determinant of $A$ with row $i$ and column $j$ deleted.
The adjugate $\operatorname{adj}(A)$ is the transpose of the cofactor matrix: $\operatorname{adj}(A)_{ij} = C_{ji}$ .
The central identity holds for every square matrix, singular or not: $A\,\operatorname{adj}(A) = \operatorname{adj}(A)\,A = \det(A)\,I$ .
When $\det(A) \ne 0$ , this identity instantly gives the closed-form inverse $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$ .
The proof needs no row reduction — only properties of determinants and a clever "repeated row" trick.
Key idea: the determinant's own cofactor expansion machinery is exactly the data needed to build the inverse.

Where this lives today

The adjugate identity is far more than a computational curiosity. In differential geometry and control theory, Jacobi's formula for the derivative of a determinant, $\frac{d}{dt}\det(A(t)) = \operatorname{tr}\!\big(\operatorname{adj}(A(t))\, A'(t)\big)$ , is proved directly from the adjugate identity, and underlies sensitivity analysis of dynamical systems. In algebraic geometry and commutative algebra, the adjugate is used to show that a matrix over an integral domain is invertible over that same domain exactly when its determinant is a unit — the key step in proving results about unimodular matrices and lattices such as $\mathbb{Z}^n$ . In numerical analysis, the closed form clarifies (rather than replaces) practical inversion: it explains why inverting a near-singular matrix is numerically unstable, since entries of $A^{-1}$ blow up as $\det(A)\to 0$ while $\operatorname{adj}(A)$ stays bounded. In cryptography, modular arithmetic inverses of matrices (e.g., in Hill ciphers) are computed via the adjugate formula reduced modulo an integer, since row reduction with modular division is more error-prone than the cofactor formula. Finally, the theorem is the direct engine behind Cramer's Rule: solving $A\mathbf{x} = \mathbf{b}$ by writing $\mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{\det A}\operatorname{adj}(A)\mathbf{b}$ and recognizing each coordinate of the result as a ratio of determinants.

Ask the tutor: How does the adjugate identity turn into Cramer's Rule when solving $A\mathbf{x} = \mathbf{b}$ column by column?

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Intuition

The cofactor $C_{ij}$ arose originally as the coefficient of $a_{ij}$ in the cofactor expansion of $\det(A)$ along row $i$ :

\det(A) = \sum_{j=1}^{n} a_{ij} C_{ij}.

This is a diagonal fact — it tells you how row $i$ interacts with its own cofactors. The magic of the adjugate identity is an off-diagonal fact: if you instead pair row $i$ of $A$ with the cofactors $C_{kj}$ of a different row $k \ne i$ , the sum collapses to zero. Intuitively, that sum equals the cofactor expansion (along row $k$ ) of a matrix obtained from $A$ by replacing row $k$ with a duplicate copy of row $i$ — and a matrix with two identical rows always has determinant zero. So the full matrix product $A\,\operatorname{adj}(A)$ has $\det(A)$ on every diagonal entry (the "self" pairings) and $0$ everywhere off the diagonal (the "cross" pairings, killed by the repeated-row trick). That is precisely the identity $A\,\operatorname{adj}(A) = \det(A) I$ . Geometrically, $\det(A)$ measures the signed volume scaling factor of the linear map $A$ . The adjugate can be seen as measuring, coordinate by coordinate, how the map scales $(n-1)$ -dimensional volumes (via minors) with an orientation-tracking sign. When $A$ is invertible, dividing by the full $n$ -dimensional scaling factor $\det(A)$ recovers the map that undoes $A$ exactly.

Figure

2×2 case: cofactors swap and negate the off-diagonal entries

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(i,k)

Worked examples

Example 1: A $3\times 3$ matrix with no special symmetry. Let

A = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 4 & 1 \\ 5 & 2 & -2 \end{pmatrix}.

Step 1 — determinant. Expand along row 1:

\det(A) = 2\det\begin{pmatrix}4&1\\2&-2\end{pmatrix} - (-1)\det\begin{pmatrix}0&1\\5&-2\end{pmatrix} + 3\det\begin{pmatrix}0&4\\5&2\end{pmatrix}.

Compute each minor: $\det\begin{pmatrix}4&1\\2&-2\end{pmatrix} = (4)(-2) - (1)(2) = -8-2=-10$ ; $\det\begin{pmatrix}0&1\\5&-2\end{pmatrix} = 0 - 5 = -5$ ; $\det\begin{pmatrix}0&4\\5&2\end{pmatrix} = 0 - 20 = -20$ . So $\det(A) = 2(-10) + 1(-5) + 3(-20) = -20 - 5 - 60 = -85$ . Since $\det(A) = -85 \ne 0$ , $A$ is invertible. Step 2 — all nine cofactors. Recall $C_{ij} = (-1)^{i+j}\det(A_{ij})$ . Row 1 minors (delete row 1):

C_{11} = +\det\begin{pmatrix}4&1\\2&-2\end{pmatrix} = -10,\quad C_{12} = -\det\begin{pmatrix}0&1\\5&-2\end{pmatrix} = -(-5) = 5,\quad C_{13} = +\det\begin{pmatrix}0&4\\5&2\end{pmatrix} = -20.

Row 2 minors (delete row 2):

C_{21} = -\det\begin{pmatrix}-1&3\\2&-2\end{pmatrix} = -\big((-1)(-2)-3\cdot2\big) = -(2-6) = 4,

C_{22} = +\det\begin{pmatrix}2&3\\5&-2\end{pmatrix} = (2)(-2) - (3)(5) = -4-15=-19,

C_{23} = -\det\begin{pmatrix}2&-1\\5&2\end{pmatrix} = -\big(4-(-5)\big) = -9.

Row 3 minors (delete row 3):

C_{31} = +\det\begin{pmatrix}-1&3\\4&1\end{pmatrix} = (-1)(1)-3(4) = -1-12=-13,

C_{32} = -\det\begin{pmatrix}2&3\\0&1\end{pmatrix} = -(2-0) = -2,

C_{33} = +\det\begin{pmatrix}2&-1\\0&4\end{pmatrix} = 8-0=8.

Step 3 — sanity check via the diagonal identity. Row 1 cofactor expansion of $\det(A)$ should reproduce $-85$ :

a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13} = 2(-10) + (-1)(5) + 3(-20) = -20-5-60=-85. \checkmark

Also check an off-diagonal cancellation predicted by Step 3 of the proof, say row $1$ paired with row $2$ 's cofactors:

a_{11}C_{21}+a_{12}C_{22}+a_{13}C_{23} = 2(4)+(-1)(-19)+3(-9) = 8+19-27 = 0. \checkmark

This confirms the off-diagonal vanishing predicted by the theorem before we even build $\operatorname{adj}(A)$ . Step 4 — assemble the adjugate. The cofactor matrix is

C = \begin{pmatrix} -10 & 5 & -20 \\ 4 & -19 & -9 \\ -13 & -2 & 8 \end{pmatrix}, \qquad \operatorname{adj}(A) = C^{\mathsf T} = \begin{pmatrix} -10 & 4 & -13 \\ 5 & -19 & -2 \\ -20 & -9 & 8 \end{pmatrix}.

Step 5 — the inverse.

A^{-1} = \frac{1}{-85}\begin{pmatrix} -10 & 4 & -13 \\ 5 & -19 & -2 \\ -20 & -9 & 8 \end{pmatrix} = \begin{pmatrix} \tfrac{10}{85} & -\tfrac{4}{85} & \tfrac{13}{85} \\[2pt] -\tfrac{5}{85} & \tfrac{19}{85} & \tfrac{2}{85} \\[2pt] \tfrac{20}{85} & \tfrac{9}{85} & -\tfrac{8}{85} \end{pmatrix} = \begin{pmatrix} \tfrac{2}{17} & -\tfrac{4}{85} & \tfrac{13}{85} \\[2pt] -\tfrac{1}{17} & \tfrac{19}{85} & \tfrac{2}{85} \\[2pt] \tfrac{4}{17} & \tfrac{9}{85} & -\tfrac{8}{85} \end{pmatrix}.

Verification. Multiply $A A^{-1}$ entry $(1,1)$ : $2\cdot\tfrac{10}{85} + (-1)\cdot(-\tfrac{5}{85}) + 3\cdot\tfrac{20}{85} = \tfrac{20+5+60}{85} = \tfrac{85}{85}=1$ . Entry $(1,2)$ : $2\cdot(-\tfrac{4}{85}) + (-1)\cdot\tfrac{19}{85} + 3\cdot\tfrac{9}{85} = \tfrac{-8-19+27}{85}=0$ . These match $I$ , consistent with the theorem.

Example 2: A $2\times 2$ matrix over $\mathbb{Z}$ , used to show non-invertibility over $\mathbb{Z}$ but invertibility over $\mathbb{Q}$ . Let

A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} \in M_2(\mathbb{Z}).

Then $\det(A) = 3(2) - 5(1) = 6 - 5 = 1$ . The cofactors are $C_{11}=2$ , $C_{12}=-1$ , $C_{21}=-5$ , $C_{22}=3$ , so

\operatorname{adj}(A) = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}.

Check the identity directly:

A\cdot\operatorname{adj}(A) = \begin{pmatrix}3&5\\1&2\end{pmatrix}\begin{pmatrix}2&-5\\-1&3\end{pmatrix} = \begin{pmatrix} 3\cdot2+5\cdot(-1) & 3\cdot(-5)+5\cdot3 \\ 1\cdot2+2\cdot(-1) & 1\cdot(-5)+2\cdot3 \end{pmatrix} = \begin{pmatrix}1 & 0\\0&1\end{pmatrix} = \det(A) I.

Because $\det(A) = 1$ is a unit in $\mathbb{Z}$ , the formula $A^{-1} = \operatorname{adj}(A)$ holds with integer entries — a matrix with integer entries and determinant $\pm 1$ has an integer-entry inverse, a fact used constantly in the theory of unimodular matrices and lattice bases. Contrast this with

B = \begin{pmatrix} 3 & 5 \\ 1 & 4 \end{pmatrix}, \qquad \det(B) = 12 - 5 = 7,

which is invertible over $\mathbb{Q}$ (since $7 \ne 0$ ) with $B^{-1} = \tfrac{1}{7}\begin{pmatrix}4&-5\\-1&3\end{pmatrix}$ , but not invertible as an integer matrix, since $7$ is not a unit in $\mathbb{Z}$ : the adjugate identity $B\,\operatorname{adj}(B) = 7I$ still holds over $\mathbb{Z}$ , but dividing by $7$ leaves $\mathbb{Z}$ . This illustrates precisely how the theorem's ring-level identity is unconditional, while the inverse formula needs $\det(A)$ to be a unit in the ring of scalars being used.

A\,\operatorname{adj}(A) = \det(A) I

Scope, edge cases & common traps

Situation	What holds	What can fail
$A$ any square matrix over any commutative ring	$A\,\operatorname{adj}(A) = \operatorname{adj}(A)\,A = \det(A) I$ always	Nothing — this is unconditional
$\det(A)$ a unit of $R$ (e.g. $\det A \ne 0$ over a field)	$A^{-1} = \dfrac{1}{\det A}\operatorname{adj}(A)$ , and it's a two-sided inverse	—
$\det(A) = 0$	Identity still holds, both sides are the zero matrix	$A^{-1}$ does not exist; formula undefined
$A$ over $\mathbb{Z}$ , $\det(A) = \pm1$	$\operatorname{adj}(A) = \pm A^{-1}$ has integer entries	For other integer determinants, $A^{-1}$ exists over $\mathbb{Q}$ but not over $\mathbb{Z}$
$1\times 1$ matrix $A=(a)$	$\operatorname{adj}(A) = (1)$ by convention, and $A\operatorname{adj}(A) = (a) = \det(A) I$	None; degenerate but consistent
Computing $\operatorname{adj}(A)$ for large $n$	Correct in principle	Requires $n^2$ determinants of size $(n-1)$ ; row reduction is far cheaper for numerical inversion when $n$ is large

Common trapThe classic error

Students often state the adjugate identity as if it automatically produces an inverse, forgetting the quantifier "provided $\det(A) \ne 0$ " (or, over a general ring, "provided $\det(A)$ is a unit"). The identity $A\,\operatorname{adj}(A) = \det(A) I$ is unconditionally true — it says nothing about invertibility on its own. Do not write " $A^{-1} = \operatorname{adj}(A)/\det(A)$ " without first checking $\det(A) \ne 0$ ; and never claim the converse "if $A\operatorname{adj}(A) = 0$ then $A = 0$ " — the singular counterexample above shows $A \ne 0$ can still give $A\,\operatorname{adj}(A) = 0$ .

The mathematician behind it

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

Augustin-Louis Cauchy

1789–1857 · Paris, École Polytechnique and Sorbonne

Cauchy was one of the principal architects of rigorous analysis and one of the earliest systematic developers of determinant theory. Working in the aftermath of Leibniz, Cramer, and Vandermonde, Cauchy organized the algebra of determinants — multiplicativity, expansion by minors, and the adjugate-style relations among cofactors — into the coherent framework still used today. He also introduced much of the modern terminology and notation for determinants, treating them as objects worthy of study in their own right rather than mere byproducts of solving linear systems.

History

- 1750: Gabriel Cramer publishes what becomes known as Cramer's Rule, implicitly using ratios of determinants to solve linear systems.

- 1772: Pierre-Simon Laplace formalizes cofactor (minor) expansion for computing determinants of arbitrary size.

- 1812: Augustin-Louis Cauchy gives one of the first systematic treatments of determinants, proving the multiplication theorem

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

and organizing cofactor identities.

- 1841: Carl Gustav Jacobi's memoir on functional determinants ("Jacobians") popularizes determinant methods across analysis.

- Late 1800s: the term "adjugate" (and the older "adjoint") becomes standard in textbooks treating the cofactor matrix as the natural bridge to matrix inversion.

What exactly is the difference between a "minor," a "cofactor," and the "adjugate"?

A minor $M_{ij}$ is the determinant of the smaller matrix left after deleting row $i$ and column $j$ from $A$ ; it carries no sign. A cofactor $C_{ij} = (-1)^{i+j}M_{ij}$ attaches the checkerboard sign to that minor. The cofactor matrix collects all $C_{ij}$ into an $n\times n$ array in the natural $(i,j)$ position. The adjugate is that cofactor matrix transposed: $\operatorname{adj}(A)_{ij} = C_{ji}$ . The transpose is essential — without it, the adjugate identity $A\operatorname{adj}(A)=\det(A)I$ fails, since cofactor expansion pairs row $i$ of $A$ with cofactors from the same row, which land in column $i$ of the adjugate only after transposing.

Why is computing the inverse via the adjugate rarely used in practice for large matrices?

The adjugate formula requires computing $n^2$ cofactors, each itself an $(n-1)\times(n-1)$ determinant. Using cofactor expansion recursively, computing one $n\times n$ determinant costs on the order of $n!$ arithmetic operations, so building the full adjugate is vastly more expensive than Gaussian elimination, which inverts an $n\times n$ matrix in about $O(n^3)$ operations. For numerical work with large matrices, row reduction (or LU decomposition) is standard. The adjugate formula remains invaluable for small matrices, symbolic computation, theoretical proofs, and situations (like modular arithmetic) where division is delicate and an explicit closed form is preferable to an iterative algorithm.

Does the adjugate identity require the matrix to be over a field?

No. The identity $A\,\operatorname{adj}(A) = \operatorname{adj}(A)\,A = \det(A) I$ is proved using only the multilinear, alternating construction of the determinant via permutations, which makes sense over any commutative ring $R$ , and the fact that a matrix with a repeated row has determinant zero, which also holds over any commutative ring. Fields are only needed for the second step, dividing both sides by $\det(A)$ to isolate $A^{-1}$ — and even there, all you truly need is that $\det(A)$ is a unit of $R$ , not that $R$ is a field. This is why the identity applies equally well to integer matrices, polynomial matrices, and matrices over more exotic commutative rings.

How does the adjugate relate to Cramer's Rule?

Cramer's Rule is the adjugate identity applied one coordinate at a time to solve $A\mathbf{x}=\mathbf{b}$ . Writing $\mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{\det(A)}\operatorname{adj}(A)\mathbf{b}$ and expanding the matrix-vector product coordinate-wise shows that the $k$ -th entry of $\mathbf{x}$ equals $\frac{1}{\det(A)}$ times a sum of cofactors dotted with $\mathbf b$ , which is exactly the cofactor expansion of the determinant of the matrix $A_k(\mathbf b)$ formed by replacing column $k$ of $A$ with $\mathbf b$ . So Cramer's Rule is not a separate theorem requiring new machinery — it is literally the adjugate identity read column by column against a specific vector $\mathbf b$ .

Is there a version of the adjugate identity for matrices that are not square?

No — the adjugate is defined only for square matrices, because the definition of a cofactor $C_{ij}=(-1)^{i+j}\det(A_{ij})$ requires deleting one row and one column from $A$ to leave a well-defined smaller square matrix whose determinant makes sense, and this only produces a well-behaved $n\times n$ array when $A$ itself is $n\times n$ . For non-square matrices, related but distinct tools exist (such as the Moore–Penrose pseudoinverse, or minors used in the Cauchy–Binet formula for products of rectangular matrices), but they are genuinely different constructions, not adjugates.

What happens to the adjugate identity when

A

itself has rank less than

n-1

Something striking happens: if $\operatorname{rank}(A) \le n-2$ , then every $(n-1)\times(n-1)$ minor of $A$ is zero (because any $(n-1)\times(n-1)$ submatrix would need $n-1$ linearly independent rows or columns, impossible when the rank is at most $n-2$ ). Hence every cofactor $C_{ij}=0$ , so $\operatorname{adj}(A)$ is the zero matrix. The identity $A\operatorname{adj}(A)=\det(A)I$ becomes the trivial statement $A\cdot 0 = 0 \cdot I$ , which is consistent since $\det(A)=0$ automatically in this case too. Only when $\operatorname{rank}(A) = n-1$ exactly can $\operatorname{adj}(A)$ be a nonzero (but still singular) matrix.

Can the adjugate be used to find eigenvalues or characteristic polynomial coefficients?

Yes, indirectly. Since $A\,\operatorname{adj}(A) = \det(A) I$ , when $A$ is invertible, $\operatorname{adj}(A) = \det(A) A^{-1}$ , and $A^{-1}$ 's eigenvalues are reciprocals of $A$ 's eigenvalues (for invertible $A$ ). More precisely, the entries of $\operatorname{adj}(\lambda I - A)$ appear in the resolvent expansion, and Jacobi's formula connects $\operatorname{adj}(A)$ directly to derivatives of the characteristic polynomial $p_A(\lambda) = \det(\lambda I - A)$ , since $\operatorname{adj}(\lambda I - A)$ can be expressed via the coefficients of $p_A$ using the Cayley–Hamilton theorem. This link is used heavily in control theory to compute transfer functions symbolically.