Template:No footnotes In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of the cofactor matrix.

The adjugate has sometimes been called the "adjoint", but today the "adjoint" of a matrix normally refers to its corresponding adjoint operator, which is its conjugate transpose.

## Definition

The adjugate of A is the transpose of the cofactor matrix C of A:

${\displaystyle {\mathrm {adj} }({\mathbf {A} })={\mathbf {C} }^{\mathsf {T}}}$.

In more detail: suppose R is a commutative ring and A is an n×n matrix with entries from R.

• The (i,j) minor of A, denoted Aij, is the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j of A.
${\displaystyle \mathbf {C} _{ij}=(-1)^{i+j}\mathbf {M} _{ij}\,}$.

where ${\displaystyle \mathbf {M} _{ij}}$ is the (i,j) minor of A.

• The adjugate of A is the transpose of C, that is, the n×n matrix whose (i,j) entry is the (j,i) cofactor of A:
${\displaystyle {\mathrm {adj} }({\mathbf {A} })_{ij}={\mathbf {C} }_{ji}\,}$.

The adjugate is defined as it is so that the product of A and its adjugate yields a diagonal matrix whose diagonal entries are det(A):

${\displaystyle {\mathbf {A} }\,{\mathrm {adj} }({\mathbf {A} })=\det({\mathbf {A} })\,{\mathbf {I} }\,}$.

A is invertible if and only if det(A) is an invertible element of R, and in that case the equation above yields:

${\displaystyle {\mathrm {adj} }({\mathbf {A} })=\det({\mathbf {A} }){\mathbf {A} }^{-1}\,}$,
${\displaystyle {\mathbf {A} }^{-1}={\frac {1}{\det({\mathbf {A} })}}\,{\mathrm {adj} }({\mathbf {A} })\,}$.

## Examples

### 2 × 2 generic matrix

The adjugate of the 2 × 2 matrix

${\displaystyle {\mathbf {A} }={\begin{pmatrix}{a}&{b}\\{c}&{d}\end{pmatrix}}}$

is

${\displaystyle \operatorname {adj} ({\mathbf {A} })={\begin{pmatrix}\,\,\,{d}&\!\!{-b}\\{-c}&{a}\end{pmatrix}}}$.

### 3 × 3 generic matrix

Consider the ${\displaystyle 3\times 3}$ matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}}}$

Its adjugate is the transpose of the cofactor matrix

${\displaystyle {\mathbf {C} }={\begin{pmatrix}+\left|{\begin{matrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{matrix}}\right|&-\left|{\begin{matrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{matrix}}\right|&+\left|{\begin{matrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{matrix}}\right|\\&&\\-\left|{\begin{matrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{matrix}}\right|&+\left|{\begin{matrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{matrix}}\right|&-\left|{\begin{matrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{matrix}}\right|\\&&\\+\left|{\begin{matrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{matrix}}\right|&-\left|{\begin{matrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{matrix}}\right|&+\left|{\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}}\right|\end{pmatrix}}={\begin{pmatrix}+\left|{\begin{matrix}5&6\\8&9\end{matrix}}\right|&-\left|{\begin{matrix}4&6\\7&9\end{matrix}}\right|&+\left|{\begin{matrix}4&5\\7&8\end{matrix}}\right|\\&&\\-\left|{\begin{matrix}2&3\\8&9\end{matrix}}\right|&+\left|{\begin{matrix}1&3\\7&9\end{matrix}}\right|&-\left|{\begin{matrix}1&2\\7&8\end{matrix}}\right|\\&&\\+\left|{\begin{matrix}2&3\\5&6\end{matrix}}\right|&-\left|{\begin{matrix}1&3\\4&6\end{matrix}}\right|&+\left|{\begin{matrix}1&2\\4&5\end{matrix}}\right|\end{pmatrix}}}$

So that we have

${\displaystyle \operatorname {adj} ({\mathbf {A} })={\begin{pmatrix}+\left|{\begin{matrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{matrix}}\right|&-\left|{\begin{matrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{matrix}}\right|&+\left|{\begin{matrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{matrix}}\right|\\&&\\-\left|{\begin{matrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{matrix}}\right|&+\left|{\begin{matrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{matrix}}\right|&-\left|{\begin{matrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{matrix}}\right|\\&&\\+\left|{\begin{matrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{matrix}}\right|&-\left|{\begin{matrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{matrix}}\right|&+\left|{\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}}\right|\end{pmatrix}}={\begin{pmatrix}+\left|{\begin{matrix}5&6\\8&9\end{matrix}}\right|&-\left|{\begin{matrix}2&3\\8&9\end{matrix}}\right|&+\left|{\begin{matrix}2&3\\5&6\end{matrix}}\right|\\&&\\-\left|{\begin{matrix}4&6\\7&9\end{matrix}}\right|&+\left|{\begin{matrix}1&3\\7&9\end{matrix}}\right|&-\left|{\begin{matrix}1&3\\4&6\end{matrix}}\right|\\&&\\+\left|{\begin{matrix}4&5\\7&8\end{matrix}}\right|&-\left|{\begin{matrix}1&2\\7&8\end{matrix}}\right|&+\left|{\begin{matrix}1&2\\4&5\end{matrix}}\right|\end{pmatrix}}}$

where

${\displaystyle \left|{\begin{matrix}a_{im}&a_{in}\\\,\,a_{jm}&a_{jn}\end{matrix}}\right|=\det \left({\begin{matrix}a_{im}&a_{in}\\\,\,a_{jm}&a_{jn}\end{matrix}}\right)}$.

Therefore C, the matrix of cofactors for A, is

${\displaystyle {\mathbf {C} }={\begin{pmatrix}-3&6&-3\\6&-12&6\\-3&6&-3\end{pmatrix}}}$

The adjugate is the transpose of the cofactor matrix. Thus, for instance, the (3,2) entry of the adjugate is the (2,3) cofactor of A. (In this example, C happens to be its own transpose, so adj(A) = C.)

### 3 × 3 numeric matrix

As a specific example, we have

${\displaystyle \operatorname {adj} {\begin{pmatrix}\!-3&\,2&\!-5\\\!-1&\,0&\!-2\\\,3&\!-4&\,1\end{pmatrix}}={\begin{pmatrix}\!-8&\,18&\!-4\\\!-5&\!12&\,-1\\\,4&\!-6&\,2\end{pmatrix}}}$.

The −6 in the third row, second column of the adjugate was computed as follows:

${\displaystyle (-1)^{2+3}\;\operatorname {det} {\begin{pmatrix}\!-3&\,2\\\,3&\!-4\end{pmatrix}}=-((-3)(-4)-(3)(2))=-6.}$

Again, the (3,2) entry of the adjugate is the (2,3) cofactor of A. Thus, the submatrix

${\displaystyle {\begin{pmatrix}\!-3&\,\!2\\\,\!3&\!-4\end{pmatrix}}}$

was obtained by deleting the second row and third column of the original matrix A.

## Properties

${\displaystyle \mathrm {adj} (\mathbf {I} )=\mathbf {I} ,}$
${\displaystyle {\mathrm {adj} }({\mathbf {AB} })={\mathrm {adj} }({\mathbf {B} })\,{\mathrm {adj} }({\mathbf {A} }),}$
${\displaystyle {\mathrm {adj} }(c{\mathbf {A} })=c^{n-1}{\mathrm {adj} }({\mathbf {A} })}$

for n×n matrices A and B. The second line follows from equations adj(B)adj(A) = det(B)B−1 det(A)A−1 = det(AB)(AB)−1. Substituting in the second line B = Am − 1 and performing the recursion, one gets for all integer m

${\displaystyle {\mathrm {adj} }({\mathbf {A} }^{m})={\mathrm {adj} }({\mathbf {A} })^{m}.}$

${\displaystyle {\mathrm {adj} }({\mathbf {A} }^{\mathsf {T}})={\mathrm {adj} }({\mathbf {A} })^{\mathsf {T}}.}$

Furthermore,

${\displaystyle \det {\big (}{\mathrm {adj} }({\mathbf {A} }){\big )}=\det({\mathbf {A} })^{n-1},}$
${\displaystyle {\mathrm {adj} }({\mathrm {adj} }({\mathbf {A} }))=\det({\mathbf {A} })^{n-2}{\mathbf {A} }}$

so if n = 2 and A is invertible, then det(adj(A)) = det(A) and adj(adj(A)) = A.

Taking the adjugate ${\displaystyle k}$ times of an invertible matrix ${\displaystyle \mathbf {A} }$ yields:

${\displaystyle \mathrm {adj} _{k}(\mathbf {A} )=\det(\mathbf {A} )^{\frac {(n-1)^{k}-(-1)^{k}}{n}}\mathbf {A} ^{(-1)^{k}}}$
${\displaystyle \det {\big (}\mathrm {adj} _{k}(\mathbf {A} ){\big )}=\det(\mathbf {A} )^{(n-1)^{k}}}$

### Inverses

As a consequence of Laplace's formula for the determinant of an n×n matrix A, we have

${\displaystyle {\mathbf {A} }\,{\mathrm {adj} }({\mathbf {A} })={\mathrm {adj} }({\mathbf {A} })\,{\mathbf {A} }=\det({\mathbf {A} })\,{\mathbf {I} }_{n}\qquad (*)}$

where ${\displaystyle {\mathbf {I} }_{n}}$ is the n×n identity matrix. Indeed, the (i,i) entry of the product A adj(A) is the scalar product of row i of A with row i of the cofactor matrix C, which is simply the Laplace formula for det(A) expanded by row i. Moreover, for ij the (i,j) entry of the product is the scalar product of row i of A with row j of C, which is the Laplace formula for the determinant of a matrix whose i and j rows are equal and is therefore zero.

From this formula follows one of the most important results in matrix algebra: A matrix A over a commutative ring R is invertible if and only if det(A) is invertible in R.

For if A is an invertible matrix then

${\displaystyle 1=\det({\mathbf {I} }_{n})=\det({\mathbf {A} }{\mathbf {A} }^{-1})=\det({\mathbf {A} })\det({\mathbf {A} }^{-1}),}$

and equation (*) above shows that

${\displaystyle {\mathbf {A} }^{-1}=\det({\mathbf {A} })^{-1}\,{\mathrm {adj} }({\mathbf {A} }).}$

### Characteristic polynomial

If p(t) = det(A − t I) is the characteristic polynomial of A and we define the polynomial q(t) = (p(0) − p(t))/t, then

${\displaystyle {\mathrm {adj} }({\mathbf {A} })=q({\mathbf {A} })=-(p_{1}{\mathbf {I} }+p_{2}{\mathbf {A} }+p_{3}{\mathbf {A} }^{2}+\cdots +p_{n}{\mathbf {A} }^{n-1}),}$

where ${\displaystyle p_{j}}$ are the coefficients of p(t),

${\displaystyle p(t)=p_{0}+p_{1}t+p_{2}t^{2}+\cdots +p_{n}t^{n}.}$

### Jacobi's formula

The adjugate also appears in Jacobi's formula for the derivative of the determinant:

${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }\alpha }}\det(A)=\operatorname {tr} \left(\operatorname {adj} (A){\frac {{\mathrm {d} }A}{{\mathrm {d} }\alpha }}\right).}$

### Cayley–Hamilton formula

Cayley–Hamilton theorem allows the adjugate of A to be represented in terms of traces and powers of A:

${\displaystyle \mathrm {adj} (\mathbf {A} )=\sum _{s=0}^{n-1}\mathbf {A} ^{s}\sum _{k_{1},k_{2},\ldots ,k_{n-1}}\prod _{l=1}^{n-1}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}\mathrm {tr} (\mathbf {A} ^{l})^{k_{l}},}$

where n is the dimension of A, and the sum is taken over s and all sequences of kl ≥ 0 satisfying the linear Diophantine equation

${\displaystyle s+\sum _{l=1}^{n-1}lk_{l}=n-1.}$

For the 2×2 case this gives

${\displaystyle \mathrm {adj} (\mathbf {A} )=\mathrm {tr} \mathbf {A} -\mathbf {A} .}$

For the 3×3 case this gives

${\displaystyle \mathrm {adj} (\mathbf {A} )={\frac {1}{2}}\left((\mathrm {tr} \mathbf {A} )^{2}-\mathrm {tr} \mathbf {A} ^{2}\right)-\mathbf {A} \mathrm {tr} \mathbf {A} +\mathbf {A} ^{2}.}$

For the 4×4 case this gives

${\displaystyle \mathrm {adj} (\mathbf {A} )={\frac {1}{6}}\left((\mathrm {tr} \mathbf {A} )^{3}-3\mathrm {tr} \mathbf {A} \mathrm {tr} \mathbf {A} ^{2}+2\mathrm {tr} \mathbf {A} ^{3}\right)-{\frac {1}{2}}\mathbf {A} \left((\mathrm {tr} \mathbf {A} )^{2}-\mathrm {tr} \mathbf {A} ^{2}\right)+\mathbf {A} ^{2}\mathrm {tr} \mathbf {A} -\mathbf {A} ^{3}.}$