Definitions

In linear algebra, the adjugate or classical adjoint of a square matrix is a matrix which plays a role similar to the inverse of a matrix; it can however be defined for any square matrix without the need to perform any divisions.

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

## Definition

Suppose R is a commutative ring and A is an n×n matrix with entries from R. The definition of the adjugate of A is a multi-step process:

• Define the (i,j) minor of A, denoted Mij, as the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j of A.
• Define the (i,j) cofactor of A as

$mathbf\left\{C\right\}_\left\{ij\right\} = \left(-1\right)^\left\{i+j\right\} mathbf\left\{M\right\}_\left\{ij\right\}. ,$

• Define the cofactor matrix of A, as the n×n matrix C whose (i,j) entry is the (i,j) cofactor of A.

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

$mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right) = mathbf\left\{C\right\}^T ,$.

That is, the adjugate of A is the n×n matrix whose (i,j) entry is the (j,i) cofactor of A:

$mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right)_\left\{ij\right\} = mathbf\left\{C\right\}_\left\{ji\right\} ,$.

## Examples

### 2x2 generic matrix

The adjugate of the $2times 2$ matrix

$mathbf\left\{A\right\} = begin\left\{pmatrix\right\} & & end\left\{pmatrix\right\}$
is
$operatorname\left\{adj\right\}\left(mathbf\left\{A\right\}\right) = begin\left\{pmatrix\right\} ,,, & !! & end\left\{pmatrix\right\}$.

### 3x3 generic matrix

Consider the $3times 3$ matrix


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
$operatorname\left\{adj\right\}\left(mathbf\left\{A\right\}\right) = begin\left\{pmatrix\right\}$
+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
$left| begin\left\{matrix\right\} A_\left\{im\right\} & A_\left\{in\right\} ,,A_\left\{jm\right\} & A_\left\{jn\right\} end\left\{matrix\right\} right|=$
detleft( begin{matrix} A_{im} & A_{in} ,,A_{jm} & A_{jn} end{matrix} right).

Note that 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.

### 3x3 numeric matrix

As a specific example, we have

$operatorname\left\{adj\right\}begin\left\{pmatrix\right\}$
!-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:

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

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

$begin\left\{pmatrix\right\}!-3&,2 ,3&!-4end\left\{pmatrix\right\}$
was obtained by deleting the second row and third column of the original matrix A.

## Applications

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

$mathbf\left\{A\right\}, mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right) = mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right), mathbf\left\{A\right\} = det\left(mathbf\left\{A\right\}\right), mathbf\left\{I\right\}qquad \left(*\right)$

where I 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

$1 = det\left(mathbf I\right) = det\left(mathbf\left\{A\right\} mathbf\left\{A\right\}^\left\{-1\right\}\right) = det\left(mathbf\left\{A\right\}\right) det\left(mathbf\left\{A\right\}^\left\{-1\right\}\right),$

and if det(A) is a unit then (*) above shows that

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

## Properties

$mathrm\left\{adj\right\}\left(mathbf\left\{I\right\}\right) = mathbf\left\{I\right\},$
$mathrm\left\{adj\right\}\left(mathbf\left\{AB\right\}\right) = mathrm\left\{adj\right\}\left(mathbf\left\{B\right\}\right),mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right),$
for all n×n matrices A and B.

$mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}^T\right) = mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right)^T,$.

Furthermore,

$detbig\left(mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right)big\right) = det\left(mathbf\left\{A\right\}\right)^\left\{n-1\right\},$.

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

$mathrm\left\{adj\right\}\left(mathbf\left\{A\right\}\right) = q\left(mathbf\left\{A\right\}\right) = -\left(p_1 mathbf\left\{I\right\} + p_2 mathbf\left\{A\right\} + p_3 mathbf\left\{A\right\}^2 + cdots + p_\left\{n\right\} mathbf\left\{A\right\}^\left\{n-1\right\}\right)$,

where $p_j$ are the coefficients of p(t),

$p\left(t\right) = p_0 + p_1 t + p_2 t^2 + cdots p_\left\{n\right\} t^\left\{n\right\}.$

The adjugate also appears in the formula of the derivative of the determinant.

## References

• Strang, Gilbert (1988). Linear Algebra and its Applications. Third edition, Harcourt Brace Jovanovich.