In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of its cofactor matrix.[1]

The adjugate[2] has sometimes been called the "adjoint",[3] 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,

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

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 Mij, is the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j of A.
• The cofactor matrix of A is the n×n matrix C whose (i, j) entry is the (i, j) cofactor of A,
${\displaystyle \mathbf {C} _{ij}=(-1)^{i+j}\mathbf {M} _{ij}~.}$
• 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,
${\displaystyle \operatorname {adj} (\mathbf {A} )_{ij}=\mathbf {C} _{ji}=(-1)^{i+j}\mathbf {M} _{ji}\,}$.

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

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

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

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

## Examples

### 1 × 1 generic matrix

The adjugate of any 1×1 matrix is ${\displaystyle \mathbf {I} =(1)}$.

### 2 × 2 generic matrix

The adjugate of the 2×2 matrix

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

is ${\displaystyle \operatorname {adj} (\mathbf {A} )={\begin{pmatrix}{d}&{-b}\\{-c}&{a}\end{pmatrix}}}$. It is seen that det(adj(A)) = det(A) and hence that adj(adj(A)) = A.

### 3 × 3 generic matrix

Consider a 3×3 matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}}$,

so its cofactor matrix is

${\displaystyle \mathbf {C} ={\begin{pmatrix}+{\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}}\\&&\\-{\begin{vmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{vmatrix}}\\&&\\+{\begin{vmatrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}\end{pmatrix}}}$

where

${\displaystyle {\begin{vmatrix}a_{im}&a_{in}\\a_{jm}&a_{jn}\end{vmatrix}}=\det {\begin{pmatrix}a_{im}&a_{in}\\a_{jm}&a_{jn}\end{pmatrix}}}$.

Its adjugate is the transpose of its cofactor matrix,

${\displaystyle \operatorname {adj} (\mathbf {A} )=\mathbf {C} ^{\mathsf {T}}={\begin{pmatrix}+{\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix}}\\&&\\-{\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{vmatrix}}\\&&\\+{\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}\end{pmatrix}}}$.

### 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.

It is easy to check the adjugate is the inverse times the determinant, −6.

## Properties

The adjugate has the properties

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

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 finds, for all integer m,

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

The adjugate preserves transposition,

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

Furthermore,

If A is a n×n matrix with n ≥ 2, then ${\displaystyle \det(\operatorname {adj} (\mathbf {A} ))=\det(\mathbf {A} )^{n-1},}$ and
If A is an invertible n×n matrix, then ${\displaystyle \operatorname {adj} (\operatorname {adj} (\mathbf {A} ))=\det(\mathbf {A} )^{n-2}\mathbf {A} ~,}$

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

Taking the adjugate of an invertible matrix A k times yields

${\displaystyle \operatorname {adj} _{k}(\mathbf {A} )=\det(\mathbf {A} )^{\frac {(n-1)^{k}-(-1)^{k}}{n}}\mathbf {A} ^{(-1)^{k}}~,}$
${\displaystyle \det(\operatorname {adj} _{k}(\mathbf {A} ))=\det(\mathbf {A} )^{(n-1)^{k}}~.}$

### Inverses

In consequence of Laplace's formula for the determinant of an n×n matrix A,

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

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 therefore vanishes.

From this formula follows one of the central 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 implies

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

Similarly, the resolvent of A is

${\displaystyle R(t;\mathbf {A} )\equiv {\frac {\mathbf {I} }{t\mathbf {\mathbf {I} } -\mathbf {A} }}={\frac {\operatorname {adj} (t\mathbf {I} -\mathbf {A} )}{p(t)}}~,}$

where p(t) is the characteristic polynomial of A.

### Characteristic polynomial

If

${\displaystyle p(t)~{\stackrel {\text{def}}{=}}~\det(t\mathbf {I} -\mathbf {A} )=\sum _{i=0}^{n}p_{i}t^{i}\in R[t],}$

is the characteristic polynomial of the matrix n-by-n matrix ${\displaystyle \mathbf {A} }$ with coefficients in the ring R, then

${\displaystyle \operatorname {adj} \,(s\mathbf {I} -\mathbf {A} )=\mathrm {\Delta } \!p(s,\mathbf {A} )~,}$

where

${\displaystyle \mathrm {\Delta } \!p(s,t)~{\stackrel {\text{def}}{=}}~{\frac {p(s)-p(t)}{s-t}}=\sum _{j=0}^{n-1}\sum _{i=0}^{n-j-1}p_{i+j+1}s^{i}t^{j}\in R[s,t]}$

is the first divided difference of p, a symmetric polynomial of degree n − 1.

### 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

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

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

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

${\displaystyle \operatorname {adj} (\mathbf {A} )=\mathbf {I} _{2}\operatorname {tr} \mathbf {A} -\mathbf {A} ~.}$

For the 3×3 case, this gives

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

For the 4×4 case, this gives

${\displaystyle \operatorname {adj} (\mathbf {A} )={\frac {1}{6}}\left((\operatorname {tr} \mathbf {A} )^{3}-3\operatorname {tr} \mathbf {A} \operatorname {tr} \mathbf {A} ^{2}+2\operatorname {tr} \mathbf {A} ^{3}\right)\mathbf {I} _{4}-{\frac {1}{2}}\mathbf {A} \left((\operatorname {tr} \mathbf {A} )^{2}-\operatorname {tr} \mathbf {A} ^{2}\right)+\mathbf {A} ^{2}\operatorname {tr} \mathbf {A} -\mathbf {A} ^{3}~..}$

The same ${\displaystyle \operatorname {adj} (\mathbf {A} )=(-)^{n-1}\sum _{s=0}^{n-1}c_{s+1}\mathbf {A} ^{s}}$ follows directly from the terminating step of the fast Faddeev–LeVerrier algorithm, where the coefficients determined above are those of the characteristic polynomial of A, namely, ${\displaystyle p(\lambda )=\sum _{k=0}^{n}c_{k}\lambda ^{k}}$.