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:

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})={\mathbf{C}}^{\mathsf{T}}$.

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:

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}{)}_{ij}={\mathbf{C}}_{ji}$.

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):

$\mathbf{A}\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})=\det (\mathbf{A})\mathbf{I}$.

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

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})=\det (\mathbf{A}){\mathbf{A}}^{-1}$,

${\mathbf{A}}^{-1}=\frac{1}{\det (\mathbf{A})}\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})$.

Examples

2 × 2 generic matrix

The adjugate of the 2 × 2 matrix

$\mathbf{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

is

$\mathrm{adj}(\mathbf{A})=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$.

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

3 × 3 generic matrix

Consider the $3\times 3$ matrix

$\mathbf{A}=\left(\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$

Its adjugate is the transpose of the cofactor matrix

$\mathbf{C}=\left(\begin{array}{ccc}+\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& -\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|& +\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\\ & & \\ -\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{32}& a_{33}\end{array}\right|& +\left|\begin{array}{cc}{a}_{11}& a_{13}\\ a_{31}& a_{33}\end{array}\right|& -\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{31}& a_{32}\end{array}\right|\\ & & \\ +\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|& -\left|\begin{array}{cc}{a}_{11}& a_{13}\\ a_{21}& a_{23}\end{array}\right|& +\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\end{array}\right)=\left(\begin{array}{ccc}+\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|& -\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|& +\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|\\ & & \\ -\left|\begin{array}{cc}2& 3\\ 8& 9\end{array}\right|& +\left|\begin{array}{cc}1& 3\\ 7& 9\end{array}\right|& -\left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|\\ & & \\ +\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|& -\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|& +\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|\end{array}\right)$

So that we have

$\mathrm{adj}(\mathbf{A})=\left(\begin{array}{ccc}+\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|& -\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{32}& a_{33}\end{array}\right|& +\left|\begin{array}{cc}{a}_{12}& a_{13}\\ a_{22}& a_{23}\end{array}\right|\\ & & \\ -\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|& +\left|\begin{array}{cc}{a}_{11}& a_{13}\\ a_{31}& a_{33}\end{array}\right|& -\left|\begin{array}{cc}{a}_{11}& a_{13}\\ a_{21}& a_{23}\end{array}\right|\\ & & \\ +\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|& -\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{31}& a_{32}\end{array}\right|& +\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\end{array}\right)=\left(\begin{array}{ccc}+\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|& -\left|\begin{array}{cc}2& 3\\ 8& 9\end{array}\right|& +\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|\\ & & \\ -\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|& +\left|\begin{array}{cc}1& 3\\ 7& 9\end{array}\right|& -\left|\begin{array}{cc}1& 3\\ 4& 6\end{array}\right|\\ & & \\ +\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|& -\left|\begin{array}{cc}1& 2\\ 7& 8\end{array}\right|& +\left|\begin{array}{cc}1& 2\\ 4& 5\end{array}\right|\end{array}\right)$

where

$\left|\begin{array}{cc}{a}_{im}& a_{in}\\ a_{jm}& a_{jn}\end{array}\right|=\det \left(\begin{array}{cc}{a}_{im}& a_{in}\\ a_{jm}& a_{jn}\end{array}\right)$.

Therefore C, the matrix of cofactors for A, is

$\mathbf{C}=\left(\begin{array}{ccc}-3& 6& -3\\ 6& -12& 6\\ -3& 6& -3\end{array}\right)$

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

$\mathrm{adj}\left(\begin{array}{ccc}-3& 2& -5\\ -1& 0& -2\\ 3& -4& 1\end{array}\right)=\left(\begin{array}{ccc}-8& 18& -4\\ -5& 12& -1\\ 4& -6& 2\end{array}\right)$.

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

$(-1{)}^{2+3}\det \left(\begin{array}{cc}-3& 2\\ 3& -4\end{array}\right)=-((-3)(-4)-(3)(2))=-6.$

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

$\left(\begin{array}{cc}-3& 2\\ 3& -4\end{array}\right)$

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

Properties

The adjugate has the properties

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{I})=\mathbf{I},$

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}\mathbf{B})=\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{B})\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}),$

$\mathrm{a}\mathrm{d}\mathrm{j}(c\mathbf{A})=c^{n-1}\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})$

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

$\mathrm{a}\mathrm{d}\mathrm{j}({\mathbf{A}}^m)=\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}{)}^m.$

The adjugate preserves transposition:

$\mathrm{a}\mathrm{d}\mathrm{j}({\mathbf{A}}^{\mathsf{T}})=\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}{)}^{\mathsf{T}}.$

Furthermore,

$\det (\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}))=\det (\mathbf{A}{)}^{n-1},$

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}))=\det (\mathbf{A}{)}^{n-2}\mathbf{A}$

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

$\mathbf{A}\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})=\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})\mathbf{A}=\det (\mathbf{A}){\mathbf{I}}_n(*)$

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

${\mathbf{A}}^{-1}=\det (\mathbf{A}{)}^{-1}\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A}).$

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

$\mathrm{a}\mathrm{d}\mathrm{j}(\mathbf{A})=q(\mathbf{A})=-(p_1\mathbf{I}+p_2\mathbf{A}+p_3{\mathbf{A}}^2+\cdots +p_n{\mathbf{A}}^{n-1}),$

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:

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

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
• Template:Cite web