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In linear algebra, the adjugate or classical adjoint of a square matrix is a matrix that 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
$\mathbf{C}_{ij} = (-1)^{i+j} \mathbf{M}_{ij}. \,$
• 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:

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

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

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

## Examples

### 2 × 2 generic matrix

The adjugate of the 2 × 2 matrix

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

is

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

### 3 × 3 generic matrix

Consider the $3\times 3$ 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 the transpose of the cofactor matrix

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

So that we have

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

where

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

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.

### 3 × 3 numeric matrix

As a specific example, we have

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

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

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

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

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

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

$\mathbf{A}\, \mathrm{adj}(\mathbf{A}) = \mathrm{adj}(\mathbf{A})\, \mathbf{A} = \det(\mathbf{A})\, \mathbf I_n \qquad (*)$

where $\mathbf I_n$ 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

$1 = \det(\mathbf I_n) = \det(\mathbf{A} \mathbf{A}^{-1}) = \det(\mathbf{A}) \det(\mathbf{A}^{-1}),$

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

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

## Properties

$\mathrm{adj}(\mathbf{I}) = \mathbf{I}\,$
$\mathrm{adj}(\mathbf{AB}) = \mathrm{adj}(\mathbf{B})\,\mathrm{adj}(\mathbf{A})\,$

for all n×n matrices A and B.

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

Furthermore

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

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

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

where pj are the coefficients of p(t),

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

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

## References

• Strang, Gilbert (1988). "Section 4.4: Applications of determinants". Linear Algebra and its Applications (3rd ed.). Harcourt Brace Jovanovich. pp. 231–232. ISBN 0-15-551005-3.

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