Adjugate matrix: Difference between revisions

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.

The adjugate of A is the transpose of the cofactor matrix C of A,

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

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,

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

• 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,

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

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),

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

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

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

The adjugate of the 2 × 2 matrix

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

is

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

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

Consider the 3×3 matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}}}$

Its adjugate is the transpose of the cofactor matrix,

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

which is to say,

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

where

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

Therefore C, the matrix of cofactors for A, amounts to

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

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.)

As a specific example, we have

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

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

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

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

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

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.

The adjugate has the properties

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

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

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

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,

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

The adjugate preserves transposition,

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

Furthermore,

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

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

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

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

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

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

where ${\displaystyle \mathbf {I} _{n}}$ 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

${\displaystyle 1=\det(\mathbf {I} _{n})=\det(\mathbf {A} \mathbf {A} ^{-1})=\det(\mathbf {A} )\det(\mathbf {A} ^{-1})~,}$

and equation (*) above implies

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

Similarly, the resolvent of A is

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

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

.

If

${\displaystyle p\colon R[t],\qquad p(t){\stackrel {\text{def}}{=}}\det(t\mathbf {I} -\mathbf {A} )=\sum _{i=0}^{n}p_{i}t^{i}~,}$

is the characteristic polynomial of ${\displaystyle \mathbf {A} \colon \mathrm {M} _{n}(R)~,}$ then

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

where ${\displaystyle \mathrm {\Delta } \!p\colon R[s,t]}$,

${\displaystyle \mathrm {\Delta } \!p(s,t){\stackrel {\text{def}}{=}}{\frac {p(s)-p(t)}{s-t}}=\sum _{j=0}^{n}\sum _{i=0}^{n-j-1}p_{i+j+1}s^{i}t^{j}}$

is the first divided difference of p, a symmetric polynomial of degree n−1.

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

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

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

${\displaystyle \mathrm {adj} (\mathbf {A} )=\sum _{s=0}^{n-1}\mathbf {A} ^{s}\sum _{k_{1},k_{2},\ldots ,k_{n-1}}\prod _{l=1}^{n-1}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}\mathrm {tr} (\mathbf {A} ^{l})^{k_{l}},}$

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

${\displaystyle s+\sum _{l=1}^{n-1}lk_{l}=n-1.}$

For the 2×2 case this gives

${\displaystyle \mathrm {adj} (\mathbf {A} )=\mathbf {I} _{2}\mathrm {tr} \mathbf {A} -\mathbf {A} .}$

For the 3×3 case this gives

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

For the 4×4 case this gives

${\displaystyle \mathrm {adj} (\mathbf {A} )={\frac {1}{6}}\left((\mathrm {tr} \mathbf {A} )^{3}-3\mathrm {tr} \mathbf {A} \mathrm {tr} \mathbf {A} ^{2}+2\mathrm {tr} \mathbf {A} ^{3}\right)\mathbf {I} _{4}-{\frac {1}{2}}\mathbf {A} \left((\mathrm {tr} \mathbf {A} )^{2}-\mathrm {tr} \mathbf {A} ^{2}\right)+\mathbf {A} ^{2}\mathrm {tr} \mathbf {A} -\mathbf {A} ^{3}.}$

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

• Trace diagram
• Jacobi's formula
• Faddeev–LeVerrier algorithm

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

• 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.