Adjugate matrix

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In linear algebra, the adjugate, classical adjoint, or adjunct of a square matrix is the transpose of the 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[edit]

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 Aij, is the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j 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} = (-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),

\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[edit]

2 × 2 generic matrix[edit]

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[edit]

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

3 × 3 numeric matrix[edit]

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[edit]

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 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,

\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 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[edit]

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 ij 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[edit]

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})= p_0 ~ \mathbf{A}^{-1}  ~,

where pj are the coefficients of p(t),

 p(t) = p_0 + p_1 t + p_2 t^2 + \cdots + p_{n} t^{n}~.

(NB. The p(t) defined here differs by an over-all factor of (−)n from the conventional definition linked and the one utilized in the previous subsection.)

Jacobi's formula[edit]

Main article: 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[edit]

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[edit]

References[edit]

  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[edit]