Runge's phenomenon

Template:No footnotes In linear algebra, the adjugate or classical adjoint (occasionally referred to as 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:

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

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 A_ij, 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:

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

• 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:

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

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):

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

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

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

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

Examples

2 × 2 generic matrix

The adjugate of the 2 × 2 matrix

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

is

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

It is seen that det(adj(A)) = det(A) and adj(adj(A)) = A.

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

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

Therefore C, the matrix of cofactors for A, is

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

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

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

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.

Properties

The adjugate has the properties

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

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

$\mathrm {adj} (c\mathbf {A} )=c^{n-1}\mathrm {adj} (\mathbf {A} )$

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 = A^{m - 1} and performing the recursion, one gets for all integer m

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

The adjugate preserves transposition:

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

Furthermore,

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

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

and, if det(A) is a unit, then det(adj(A)) = det(A) and adj(adj(A)) = A.

Inverses

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 i ≠ j 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 equation (*) above shows that

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

See also Cramer's rule.

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

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

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

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

Jacobi's formula

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

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

See also

• Trace diagram
• [1]

References

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External links

• Matrix Reference Manual
• Online matrix calculator (determinant, track, inverse, adjoint, transpose) Compute Adjugate matrix up to order 8
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