Adjugate matrix

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,

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

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 M_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,

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

• 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,

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

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),

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

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 hence that adj(adj(A)) = A.

3 × 3 generic matrix

Consider the 3×3 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,

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

which is to say,

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

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, amounts to

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

The adjugate is the transpose of this cofactor matrix. Thus, for instance, the (3,2) entry of the adjugate is the (2,3) cofactor of A. (In this particular, non-generic, example, however, C happens to be its own transpose, so adj(A) = C, here.)

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.

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

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 finds, 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,

If A is a n×n matrix with n ≥ 2, then $\det\big(\mathrm{adj}(\mathbf{A})\big) = \det(\mathbf{A})^{n-1},$ and

If A is an invertible n×n matrix, then $\mathrm{adj}(\mathrm{adj}(\mathbf{A})) = \det(\mathbf{A})^{n - 2}\mathbf{A} ~,$

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

$\mathrm{adj}_k(\mathbf{A})=\det(\mathbf{A})^{\frac{(n-1)^k-(-1)^k}n}\mathbf{A}^{(-1)^k}~,$

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

Inverses

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

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

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 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

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

and equation (*) above implies

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

Similarly, the resolvent of A is

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

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

Characteristic polynomial

If

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

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

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

where

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

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,

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

Cayley–Hamilton formula

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

$\mathrm{adj}(\mathbf{A}) = \sum_{s=0}^{n-1}\mathbf{A}^{s}\sum_{k_1,k_2,\ldots ,k_{n-1}}\prod_{l=1}^{n-1} \frac{(-1)^{k_l+1}}{l^{k_l}k_{l}!}\mathrm{tr}(\mathbf{A}^l)^{k_l},$

where n is the dimension of A, and the sum is taken over s and all sequences of k_l ≥ 0 satisfying the linear Diophantine equation

$s+\sum_{l=1}^{n-1}lk_{l} = n - 1.$

For the 2×2 case this gives

$\mathrm{adj}(\mathbf{A})=\mathbf{I}_2\mathrm{tr}\mathbf{A}- \mathbf{A}.$

For the 3×3 case this gives

$\mathrm{adj}(\mathbf{A})=\frac{1}{2}\left( (\mathrm{tr}\mathbf{A})^{2}-\mathrm{tr}\mathbf{A}^{2}\right)\mathbf{I}_3 -\mathbf{A}\mathrm{tr}\mathbf{A}+\mathbf{A}^{2}.$

For the 4×4 case this gives

$\mathrm{adj}(\mathbf{A})=\frac{1}{6}\left( (\mathrm{tr}\mathbf{A})^{3}-3\mathrm{tr}\mathbf{A}\mathrm{tr}\mathbf{A}^{2}+2\mathrm{tr}\mathbf{A}^{3}\right)\mathbf{I}_4 -\frac{1}{2}\mathbf{A}\left( (\mathrm{tr}\mathbf{A})^{2}-\mathrm{tr}\mathbf{A}^{2}\right) +\mathbf{A}^{2}\mathrm{tr}\mathbf{A}-\mathbf{A}^{3}.$

The same results follow directly from the terminating step of the fast Faddeev–LeVerrier algorithm.

See also

• Cayley–Hamilton theorem
• Cramer's rule
• Trace diagram
• Jacobi's formula
• Faddeev–LeVerrier algorithm

References

1. F.R. Gantmacher, The Theory of Matrices v I (Chelsea Publishing, NY, 1960) ISBN 0-8218-1376-5 , pp 76-89
2. 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.
3. Householder, Alston S. (2006). The Theory of Matrices in Numerical Analysis. Dover Books on Mathematics. ISBN 0486449726. pp166-168

• Roger A. Horn and Charles R. Johnson (1991), Topics in Matrix Analysis. Cambridge University Press, ISBN 978-0-521-46713-1

External links

• Matrix Reference Manual
• Online matrix calculator (determinant, track, inverse, adjoint, transpose) Compute Adjugate matrix up to order 8
• "adjugate of { { a, b, c }, { d, e, f }, { g, h, i } }". Wolfram Alpha.