Adjugate matrix

In linear algebra, the adjugate, classical adjoint, or 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.

\mathbf{C}_{ij} = (-1)^{i+j} \mathbf{M}_{ij} \,.
\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 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 = Am − 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}

so if n = 2 and A is invertible, then det(adj(A)) = det(A) and adj(adj(A)) = A.

Taking the adjugate k times of an invertible matrix \mathbf{A} 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

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).

Cayley–Hamilton formula

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 kl ≥ 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}.

See also

References

    External links