adjugate matrix

The adjugate of {{math|A}} is the transpose of the cofactor matrix {{math|C}} of {{math|A}},

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash mathbf\{C\}^\backslash mathsf\{T\}.$

In more detail, suppose {{math|R}} is a (unital) commutative ring and {{math|A}} is an {{math|n × n}} matrix with entries from {{math|R}}. The {{math|(i, j)}}-minor of {{math|A}}, denoted {{math|M_ij}}, is the determinant of the {{math|(n − 1) × (n − 1)}} matrix that results from deleting row {{mvar|i}} and column {{mvar|j}} of {{math|A}}. The cofactor matrix of {{math|A}} is the {{math|n × n}} matrix {{math|C}} whose {{math|(i, j)}} entry is the {{math|(i, j)}} cofactor of {{math|A}}, which is the {{math|(i, j)}}-minor times a sign factor:

:$\backslash mathbf\{C\}\; =\; \backslash left((-1)^\{i+j\}\; \backslash mathbf\{M\}\_\{ij\}\backslash right)\_\{1\; \backslash le\; i,\; j\; \backslash le\; n\}.$

The adjugate of {{math|A}} is the transpose of {{math|C}}, that is, the {{math|n × n}} matrix whose {{math|(i, j)}} entry is the {{math|(j,i)}} cofactor of {{math|A}},

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash mathbf\{C\}^\backslash mathsf\{T\}\; =\; \backslash left((-1)^\{i+j\}\; \backslash mathbf\{M\}\_\{ji\}\backslash right)\_\{1\; \backslash le\; i,\; j\; \backslash le\; n\}.$

The adjugate is defined so that the product of {{math|A}} with its adjugate yields a diagonal matrix whose diagonal entries are the determinant {{math|det(A)}}. That is,

:$\backslash mathbf\{A\}\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})\; \backslash mathbf\{A\}\; =\; \backslash det(\backslash mathbf\{A\})\; \backslash mathbf\{I\},$

where {{math|I}} is the {{math|n × n}} identity matrix. This is a consequence of the Laplace expansion of the determinant.

The above formula implies one of the fundamental results in matrix algebra, that {{math|A}} is invertible if and only if {{math|det(A)}} is an invertible element of {{math|R}}. When this holds, the equation above yields

:$\backslash begin\{align\}$

\operatorname{adj}(\mathbf{A}) &= \det(\mathbf{A}) \mathbf{A}^{-1}, \\

\mathbf{A}^{-1} &= \det(\mathbf{A})^{-1} \operatorname{adj}(\mathbf{A}).

\end{align}

Since the determinant of a 0 × 0 matrix is 1, the adjugate of any 1 × 1 matrix (complex scalar) is $\backslash mathbf\{I\}\; =\; \backslash begin\{bmatrix\}\; 1\; \backslash end\{bmatrix\}$. Observe that $\backslash mathbf\{A\}\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})\backslash mathbf\{A\}\; =\; (\backslash det\; \backslash mathbf\{A\})\; \backslash mathbf\; \{I\}.$

The adjugate of the 2 × 2 matrix

:

is

:

By direct computation,

:

In this case, it is also true that {{math|det}}({{math|adj}}(A)) = {{math|det}}(A) and hence that {{math|adj}}({{math|adj}}(A)) = A.

Consider a 3 × 3 matrix

:$\backslash mathbf\{A\}\; =\; \backslash begin\{bmatrix\}$

a_{1} & a_{2} & a_{3} \\

b_{1} & b_{2} & b_{3} \\

c_{1} & c_{2} & c_{3}

\end{bmatrix}.

Its cofactor matrix is

:$\backslash mathbf\{C\}\; =\; \backslash begin\{bmatrix\}$

+\begin{vmatrix} b_{2} & b_{3} \\ c_{2} & c_{3} \end{vmatrix} &

-\begin{vmatrix} b_{1} & b_{3} \\ c_{1} & c_{3} \end{vmatrix} &

+\begin{vmatrix} b_{1} & b_{2} \\ c_{1} & c_{2} \end{vmatrix} \\

\\

-\begin{vmatrix} a_{2} & a_{3} \\ c_{2} & c_{3} \end{vmatrix} &

+\begin{vmatrix} a_{1} & a_{3} \\ c_{1} & c_{3} \end{vmatrix} &

-\begin{vmatrix} a_{1} & a_{2} \\ c_{1} & c_{2} \end{vmatrix} \\

\\

+\begin{vmatrix} a_{2} & a_{3} \\ b_{2} & b_{3} \end{vmatrix} &

-\begin{vmatrix} a_{1} & a_{3} \\ b_{1} & b_{3} \end{vmatrix} &

+\begin{vmatrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{vmatrix}

\end{bmatrix},

where

:

= \det\!\begin{bmatrix} a & b \\ c & d \end{bmatrix} .

Its adjugate is the transpose of its cofactor matrix,

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash mathbf\{C\}^\backslash mathsf\{T\}\; =\; \backslash begin\{bmatrix\}$

+\begin{vmatrix} b_{2} & b_{3} \\ c_{2} & c_{3} \end{vmatrix} &

-\begin{vmatrix} a_{2} & a_{3} \\ c_{2} & c_{3} \end{vmatrix} &

+\begin{vmatrix} a_{2} & a_{3} \\ b_{2} & b_{3} \end{vmatrix} \\

& & \\

-\begin{vmatrix} b_{1} & b_{3} \\ c_{1} & c_{3} \end{vmatrix} &

+\begin{vmatrix} a_{1} & a_{3} \\ c_{1} & c_{3} \end{vmatrix} &

-\begin{vmatrix} a_{1} & a_{3} \\ b_{1} & b_{3} \end{vmatrix} \\

& & \\

+\begin{vmatrix} b_{1} & b_{2} \\ c_{1} & c_{2} \end{vmatrix} &

-\begin{vmatrix} a_{1} & a_{2} \\ c_{1} & c_{2} \end{vmatrix} &

+\begin{vmatrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{vmatrix}

\end{bmatrix}.

As a specific example, we have

:$\backslash operatorname\{adj\}\backslash !\backslash begin\{bmatrix\}$

-3 & 2 & -5 \\

-1 & 0 & -2 \\

3 & -4 & 1

\end{bmatrix} = \begin{bmatrix}

-8 & 18 & -4 \\

-5 & 12 & -1 \\

4 & -6 & 2

\end{bmatrix}.

It is easy to check the adjugate is the inverse times the determinant, {{math|−6}}.

The {{math|−1}} in the second row, third column of the adjugate was computed as follows. The (2,3) entry of the adjugate is the (3,2) cofactor of A. This cofactor is computed using the submatrix obtained by deleting the third row and second column of the original matrix A,

:

The (3,2) cofactor is a sign times the determinant of this submatrix:

:

and this is the (2,3) entry of the adjugate.

For any {{math|n × n}} matrix {{math|A}}, elementary computations show that adjugates have the following properties:

• $\backslash operatorname\{adj\}(\backslash mathbf\{I\})\; =\; \backslash mathbf\{I\}$, where $\backslash mathbf\{I\}$ is the identity matrix.
• $\backslash operatorname\{adj\}(\backslash mathbf\{0\})\; =\; \backslash mathbf\{0\}$, where $\backslash mathbf\{0\}$ is the zero matrix, except that if $n=1$ then $\backslash operatorname\{adj\}(\backslash mathbf\{0\})\; =\; \backslash mathbf\{I\}$.
• $\backslash operatorname\{adj\}(c\; \backslash mathbf\{A\})\; =\; c^\{n\; -\; 1\}\backslash operatorname\{adj\}(\backslash mathbf\{A\})$ for any scalar {{mvar|c}}.
• $\backslash operatorname\{adj\}(\backslash mathbf\{A\}^\backslash mathsf\{T\})\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})^\backslash mathsf\{T\}$.
• $\backslash det(\backslash operatorname\{adj\}(\backslash mathbf\{A\}))\; =\; (\backslash det\; \backslash mathbf\{A\})^\{n-1\}$.
• If {{math|A}} is invertible, then $\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; (\backslash det\; \backslash mathbf\{A\})\; \backslash mathbf\{A\}^\{-1\}$. It follows that:
• {{math|adj(A)}} is invertible with inverse {{math|(det A)⁻¹A}}.
• {{math|1=adj(A⁻¹) = adj(A)⁻¹}}.
• {{math|adj(A)}} is entrywise polynomial in {{math|A}}. In particular, over the real or complex numbers, the adjugate is a smooth function of the entries of {{math|A}}.

Over the complex numbers,

• $\backslash operatorname\{adj\}(\backslash overline\backslash mathbf\{A\})\; =\; \backslash overline\{\backslash operatorname\{adj\}(\backslash mathbf\{A\})\}$, where the bar denotes complex conjugation.
• $\backslash operatorname\{adj\}(\backslash mathbf\{A\}^*)\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})^*$, where the asterisk denotes conjugate transpose.

Suppose that {{math|B}} is another {{math|n × n}} matrix. Then

:$\backslash operatorname\{adj\}(\backslash mathbf\{AB\})\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{B\})\backslash operatorname\{adj\}(\backslash mathbf\{A\}).$

This can be proved in three ways. One way, valid for any commutative ring, is a direct computation using the Cauchy–Binet formula. The second way, valid for the real or complex numbers, is to first observe that for invertible matrices {{math|A}} and {{math|B}},

:$\backslash operatorname\{adj\}(\backslash mathbf\{B\})\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; (\backslash det\; \backslash mathbf\{B\})\backslash mathbf\{B\}^\{-1\}(\backslash det\; \backslash mathbf\{A\})\backslash mathbf\{A\}^\{-1\}\; =\; (\backslash det\; \backslash mathbf\{AB\})(\backslash mathbf\{AB\})^\{-1\}\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{AB\}).$

Because every non-invertible matrix is the limit of invertible matrices, continuity of the adjugate then implies that the formula remains true when one of {{math|A}} or {{math|B}} is not invertible.

A corollary of the previous formula is that, for any non-negative integer {{mvar|k}},

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\}^k)\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})^k.$

If {{math|A}} is invertible, then the above formula also holds for negative {{mvar|k}}.

From the identity

:$(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})\backslash operatorname\{adj\}(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})\backslash mathbf\{B\}\; =\; \backslash det(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})\backslash mathbf\{B\}\; =\; \backslash mathbf\{B\}\backslash operatorname\{adj\}(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\}),$

we deduce

:$\backslash mathbf\{A\}\backslash operatorname\{adj\}(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})\backslash mathbf\{B\}\; =\; \backslash mathbf\{B\}\backslash operatorname\{adj\}(\backslash mathbf\{A\}\; +\; \backslash mathbf\{B\})\backslash mathbf\{A\}.$

Suppose that {{math|A}} commutes with {{math|B}}. Multiplying the identity {{math|1=AB = BA}} on the left and right by {{math|adj(A)}} proves that

:$\backslash det(\backslash mathbf\{A\})\backslash operatorname\{adj\}(\backslash mathbf\{A\})\backslash mathbf\{B\}\; =\; \backslash det(\backslash mathbf\{A\})\backslash mathbf\{B\}\backslash operatorname\{adj\}(\backslash mathbf\{A\}).$

If {{math|A}} is invertible, this implies that {{math|adj(A)}} also commutes with {{math|B}}. Over the real or complex numbers, continuity implies that {{math|adj(A)}} commutes with {{math|B}} even when {{math|A}} is not invertible.

Finally, there is a more general proof than the second proof, which only requires that an n × n matrix has entries over a field with at least 2n + 1 elements (e.g. a 5 × 5 matrix over the integers modulo 11). {{math|det(A+tI)}} is a polynomial in t with degree at most n, so it has at most n roots. Note that the ijth entry of {{math|adj((A+tI)(B))}} is a polynomial of at most order n, and likewise for {{math|adj(A+tI)adj(B)}}. These two polynomials at the ijth entry agree on at least n + 1 points, as we have at least n + 1 elements of the field where {{math|A+tI}} is invertible, and we have proven the identity for invertible matrices. Polynomials of degree n which agree on n + 1 points must be identical (subtract them from each other and you have n + 1 roots for a polynomial of degree at most n – a contradiction unless their difference is identically zero). As the two polynomials are identical, they take the same value for every value of t. Thus, they take the same value when t = 0.

Using the above properties and other elementary computations, it is straightforward to show that if {{math|A}} has one of the following properties, then {{math|adjA}} does as well:

• upper triangular,
• lower triangular,
• diagonal,
• orthogonal,
• unitary,
• symmetric,
• Hermitian,
• normal.

If {{math|A}} is skew-symmetric, then {{math|adj(A)}} is skew-symmetric for even n and symmetric for odd n. Similarly, if {{math|A}} is skew-Hermitian, then {{math|adj(A)}} is skew-Hermitian for even n and Hermitian for odd n.

If {{math|A}} is invertible, then, as noted above, there is a formula for {{math|adj(A)}} in terms of the determinant and inverse of {{math|A}}. When {{math|A}} is not invertible, the adjugate satisfies different but closely related formulas.

• If {{math|1=rk(A) ≤ n − 2}}, then {{math|1=adj(A) = 0}}.
• If {{math|1=rk(A) = n − 1}}, then {{math|1=rk(adj(A)) = 1}}. (Some minor is non-zero, so {{math|adj(A)}} is non-zero and hence has rank at least one; the identity {{math|1=adj(A)A = 0}} implies that the dimension of the nullspace of {{math|adj(A)}} is at least {{math|n − 1}}, so its rank is at most one.) It follows that {{math|1=adj(A) = αxy^T}}, where {{math|α}} is a scalar and {{math|x}} and {{math|y}} are vectors such that {{math|1=Ax = 0}} and {{math|1=A^T y = 0}}.

{{see also|Cramer's rule}}

Partition {{math|A}} into column vectors:

:

Let {{math|b}} be a column vector of size {{math|n}}. Fix {{math|1 ≤ i ≤ n}} and consider the matrix formed by replacing column {{math|i}} of {{math|A}} by {{math|b}}:

:

Laplace expand the determinant of this matrix along column {{mvar|i}}. The result is entry {{mvar|i}} of the product {{math|adj(A)b}}. Collecting these determinants for the different possible {{mvar|i}} yields an equality of column vectors

:$\backslash left(\backslash det(\backslash mathbf\{A\}\; \backslash stackrel\{i\}\{\backslash leftarrow\}\; \backslash mathbf\{b\})\backslash right)\_\{i=1\}^n\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\})\backslash mathbf\{b\}.$

This formula has the following concrete consequence. Consider the linear system of equations

:$\backslash mathbf\{A\}\backslash mathbf\{x\}\; =\; \backslash mathbf\{b\}.$

Assume that {{math|A}} is non-singular. Multiplying this system on the left by {{math|adj(A)}} and dividing by the determinant yields

:$\backslash mathbf\{x\}\; =\; \backslash frac\{\backslash operatorname\{adj\}(\backslash mathbf\{A\})\backslash mathbf\{b\}\}\{\backslash det\; \backslash mathbf\{A\}\}.$

Applying the previous formula to this situation yields Cramer's rule,

:$x\_i\; =\; \backslash frac\{\backslash det(\backslash mathbf\{A\}\; \backslash stackrel\{i\}\{\backslash leftarrow\}\; \backslash mathbf\{b\})\}\{\backslash det\; \backslash mathbf\{A\}\},$

where {{math|x_i}} is the {{mvar|i}}th entry of {{math|x}}.

Let the characteristic polynomial of {{math|A}} be

:$p(s)\; =\; \backslash det(s\backslash mathbf\{I\}\; -\; \backslash mathbf\{A\})\; =\; \backslash sum\_\{i=0\}^n\; p\_i\; s^i\; \backslash in\; R[s].$

The first divided difference of {{math|p}} is a symmetric polynomial of degree {{math|n − 1}},

:

Multiply {{math|sI − A}} by its adjugate. Since {{math|1=p(A) = 0}} by the Cayley–Hamilton theorem, some elementary manipulations reveal

:$\backslash operatorname\{adj\}(s\backslash mathbf\{I\}\; -\; \backslash mathbf\{A\})\; =\; \backslash Delta\; p(s\backslash mathbf\{I\},\; \backslash mathbf\{A\}).$

In particular, the resolvent of {{math|A}} is defined to be

:$R(z;\; \backslash mathbf\{A\})\; =\; (z\backslash mathbf\{I\}\; -\; \backslash mathbf\{A\})^\{-1\},$

and by the above formula, this is equal to

:$R(z;\; \backslash mathbf\{A\})\; =\; \backslash frac\{\backslash Delta\; p(z\backslash mathbf\{I\},\; \backslash mathbf\{A\})\}\{p(z)\}.$

{{main|Jacobi's formula}}

The adjugate also appears in Jacobi's formula for the derivative of the determinant. If {{math|A(t)}} is continuously differentiable, then

:$\backslash frac\{d(\backslash det\; \backslash mathbf\{A\})\}\{dt\}(t)\; =\; \backslash operatorname\{tr\}\backslash left(\backslash operatorname\{adj\}(\backslash mathbf\{A\}(t))\; \backslash mathbf\{A\}\text{'}(t)\backslash right).$

It follows that the total derivative of the determinant is the transpose of the adjugate:

:$d(\backslash det\; \backslash mathbf\{A\})\_\{\backslash mathbf\{A\}\_0\}\; =\; \backslash operatorname\{adj\}(\backslash mathbf\{A\}\_0)^\{\backslash mathsf\{T\}\}.$

{{main|Cayley–Hamilton theorem}}

Let {{math|p_A(t)}} be the characteristic polynomial of {{math|A}}. The Cayley–Hamilton theorem states that

:$p\_\{\backslash mathbf\{A\}\}(\backslash mathbf\{A\})\; =\; \backslash mathbf\{0\}.$

Separating the constant term and multiplying the equation by {{math|adj(A)}} gives an expression for the adjugate that depends only on {{math|A}} and the coefficients of {{math|p_A(t)}}. These coefficients can be explicitly represented in terms of traces of powers of {{math|A}} using complete exponential Bell polynomials. The resulting formula is

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})\; =\; \backslash sum\_\{s=0\}^\{n-1\}\; \backslash mathbf\{A\}^\{s\}\; \backslash sum\_\{k\_1,\; k\_2,\; \backslash ldots,\; k\_\{n-1\}\}\; \backslash prod\_\{\backslash ell=1\}^\{n-1\}\; \backslash frac\{(-1)^\{k\_\backslash ell+1\}\}\{\backslash ell^\{k\_\backslash ell\}k\_\{\backslash ell\}!\}\backslash operatorname\{tr\}(\backslash mathbf\{A\}^\backslash ell)^\{k\_\backslash ell\},$

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

:$s+\backslash sum\_\{\backslash ell=1\}^\{n-1\}\backslash ell\; k\_\backslash ell\; =\; n\; -\; 1.$

For the 2 × 2 case, this gives

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})=\backslash mathbf\{I\}\_2(\backslash operatorname\{tr\}\backslash mathbf\{A\})\; -\; \backslash mathbf\{A\}.$

For the 3 × 3 case, this gives

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})=\backslash frac\{1\}\{2\}\backslash mathbf\{I\}\_3\backslash !\backslash left(\; (\backslash operatorname\{tr\}\backslash mathbf\{A\})^2-\backslash operatorname\{tr\}\backslash mathbf\{A\}^2\backslash right)\; -\; \backslash mathbf\{A\}(\backslash operatorname\{tr\}\backslash mathbf\{A\})\; +\; \backslash mathbf\{A\}^2\; .$

For the 4 × 4 case, this gives

:$\backslash operatorname\{adj\}(\backslash mathbf\{A\})=$

\frac{1}{6}\mathbf{I}_4\!\left(

(\operatorname{tr}\mathbf{A})^3

- 3\operatorname{tr}\mathbf{A}\operatorname{tr}\mathbf{A}^2

+ 2\operatorname{tr}\mathbf{A}^{3}

\right)

- \frac{1}{2}\mathbf{A}\!\left( (\operatorname{tr}\mathbf{A})^2 - \operatorname{tr}\mathbf{A}^2\right)

+ \mathbf{A}^2(\operatorname{tr}\mathbf{A})

- \mathbf{A}^3.

The same formula follows directly from the terminating step of the Faddeev–LeVerrier algorithm, which efficiently determines the characteristic polynomial of {{math|A}}.

In general, adjugate matrix of arbitrary dimension N matrix can be computed by Einstein's convention.

:$(\backslash operatorname\{adj\}(\backslash mathbf\{A\}))\_\{i\_N\}^\{j\_N\}\; =\; \backslash frac\{1\}\{(N-1)!\}\; \backslash epsilon\_\{i\_1\; i\_2\; \backslash ldots\; i\_N\}\; \backslash epsilon^\{j\_1\; j\_2\; \backslash ldots\; j\_N\}\; A\_\{j\_1\}^\{i\_1\}\; A\_\{j\_2\}^\{i\_2\}\; \backslash ldots\; A\_\{j\_\{N-1\}\}^\{i\_\{N-1\}\}$

The adjugate can be viewed in abstract terms using exterior algebras. Let {{math|V}} be an {{math|n}}-dimensional vector space. The exterior product defines a bilinear pairing

$$V\; \backslash times\; \backslash wedge^\{n-1\}\; V\; \backslash to\; \backslash wedge^n\; V.$$

Abstractly, $\backslash wedge^n\; V$ is isomorphic to {{math|R}}, and under any such isomorphism the exterior product is a perfect pairing. That is, it yields an isomorphism

$$\backslash phi\; \backslash colon\; V\backslash \; \backslash xrightarrow\{\backslash cong\}\backslash \; \backslash operatorname\{Hom\}(\backslash wedge^\{n-1\}\; V,\; \backslash wedge^n\; V).$$

This isomorphism sends each {{math|v ∈ V}} to the map $\backslash phi\_\{\backslash mathbf\{v\}\}$ defined by

$$\backslash phi\_\backslash mathbf\{v\}(\backslash alpha)\; =\; \backslash mathbf\{v\}\; \backslash wedge\; \backslash alpha.$$

Suppose that {{math|T : V → V}} is a linear transformation. Pullback by the {{math|(n − 1)}}th exterior power of {{math|T}} induces a morphism of {{math|Hom}} spaces. The adjugate of {{math|T}} is the composite

$$V\backslash \; \backslash xrightarrow\{\backslash phi\}\backslash \; \backslash operatorname\{Hom\}(\backslash wedge^\{n-1\}\; V,\; \backslash wedge^n\; V)\backslash \; \backslash xrightarrow\{(\backslash wedge^\{n-1\}\; T)^*\}\backslash \; \backslash operatorname\{Hom\}(\backslash wedge^\{n-1\}\; V,\; \backslash wedge^n\; V)\backslash \; \backslash xrightarrow\{\backslash phi^\{-1\}\}\backslash \; V.$$

If {{math|1=V = Rⁿ}} is endowed with its canonical basis {{math|e₁, ..., e_n}}, and if the matrix of {{math|T}} in this basis is {{math|A}}, then the adjugate of {{math|T}} is the adjugate of {{math|A}}. To see why, give $\backslash wedge^\{n-1\}\; \backslash mathbf\{R\}^n$ the basis

$$\backslash \{\backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash hat\backslash mathbf\{e\}\_k\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n\backslash \}\_\{k=1\}^n.$$

Fix a basis vector {{math|e_i}} of {{math|Rⁿ}}. The image of {{math|e_i}} under $\backslash phi$ is determined by where it sends basis vectors:

$$\backslash phi\_\{\backslash mathbf\{e\}\_i\}(\backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash hat\backslash mathbf\{e\}\_k\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n)$$

= \begin{cases} (-1)^{i-1} \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n, &\text{if}\ k = i, \\ 0 &\text{otherwise.} \end{cases}

On basis vectors, the {{math|(n − 1)}}st exterior power of {{math|T}} is

$$\backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash hat\backslash mathbf\{e\}\_j\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n\; \backslash mapsto\; \backslash sum\_\{k=1\}^n\; (\backslash det\; A\_\{jk\})\; \backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash hat\backslash mathbf\{e\}\_k\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n.$$

Each of these terms maps to zero under $\backslash phi\_\{\backslash mathbf\{e\}\_i\}$ except the {{math|1=k = i}} term. Therefore, the pullback of $\backslash phi\_\{\backslash mathbf\{e\}\_i\}$ is the linear transformation for which

$$\backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash hat\backslash mathbf\{e\}\_j\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n\; \backslash mapsto\; (-1)^\{i-1\}\; (\backslash det\; A\_\{ji\})\; \backslash mathbf\{e\}\_1\; \backslash wedge\; \backslash dots\; \backslash wedge\; \backslash mathbf\{e\}\_n.$$

That is, it equals

$$\backslash sum\_\{j=1\}^n\; (-1)^\{i+j\}\; (\backslash det\; A\_\{ji\})\backslash phi\_\{\backslash mathbf\{e\}\_j\}.$$

Applying the inverse of $\backslash phi$ shows that the adjugate of {{math|T}} is the linear transformation for which

$$\backslash mathbf\{e\}\_i\; \backslash mapsto\; \backslash sum\_\{j=1\}^n\; (-1)^\{i+j\}(\backslash det\; A\_\{ji\})\backslash mathbf\{e\}\_j.$$

Consequently, its matrix representation is the adjugate of {{math|A}}.

If {{math|V}} is endowed with an inner product and a volume form, then the map {{math|φ}} can be decomposed further. In this case, {{math|φ}} can be understood as the composite of the Hodge star operator and dualization. Specifically, if {{math|ω}} is the volume form, then it, together with the inner product, determines an isomorphism

$$\backslash omega^\backslash vee\; \backslash colon\; \backslash wedge^n\; V\; \backslash to\; \backslash mathbf\{R\}.$$

This induces an isomorphism

$$\backslash operatorname\{Hom\}(\backslash wedge^\{n-1\}\; \backslash mathbf\{R\}^n,\; \backslash wedge^n\; \backslash mathbf\{R\}^n)\; \backslash cong\; \backslash wedge^\{n-1\}\; (\backslash mathbf\{R\}^n)^\backslash vee.$$

A vector {{math|v}} in {{math|Rⁿ}} corresponds to the linear functional

$$(\backslash alpha\; \backslash mapsto\; \backslash omega^\backslash vee(\backslash mathbf\{v\}\; \backslash wedge\; \backslash alpha))\; \backslash in\; \backslash wedge^\{n-1\}\; (\backslash mathbf\{R\}^n)^\backslash vee.$$

By the definition of the Hodge star operator, this linear functional is dual to {{math|*v}}. That is, {{math|ω^∨∘ φ}} equals {{math|v ↦ *v^∨}}.

Let {{math|A}} be an {{math|n × n}} matrix, and fix {{math|r ≥ 0}}. The {{math|r}}th higher adjugate of {{math|A}} is an $\backslash binom\{n\}\{r\}\; \backslash !\backslash times\backslash !\; \backslash binom\{n\}\{r\}$ matrix, denoted {{math|adj_r A}}, whose entries are indexed by size {{math|r}} subsets {{math|I}} and {{math|J}} of {{math|{1, ..., m}}} {{Citation needed|date=November 2023}}. Let {{math|I{{i sup|c}}}} and {{math|J{{i sup|c}}}} denote the complements of {{math|I}} and {{math|J}}, respectively. Also let $\backslash mathbf\{A\}\_\{I^c,\; J^c\}$ denote the submatrix of {{math|A}} containing those rows and columns whose indices are in {{math|I{{i sup|c}}}} and {{math|J{{i sup|c}}}}, respectively. Then the {{math|(I, J)}} entry of {{math|adj_r A}} is

:$(-1)^\{\backslash sigma(I)\; +\; \backslash sigma(J)\}\backslash det\; \backslash mathbf\{A\}\_\{J^c,\; I^c\},$

where {{math|σ(I)}} and {{math|σ(J)}} are the sum of the elements of {{math|I}} and {{math|J}}, respectively.

Basic properties of higher adjugates include {{Citation needed|date=November 2023}}:

• {{math|1=adj₀(A) = det A}}.
• {{math|1=adj₁(A) = adj A}}.
• {{math|1=adj_n(A) = 1}}.
• {{math|1=adj_r(BA) = adj_r(A) adj_r(B)}}.
• $\backslash operatorname\{adj\}\_r(\backslash mathbf\{A\})C\_r(\backslash mathbf\{A\})\; =\; C\_r(\backslash mathbf\{A\})\backslash operatorname\{adj\}\_r(\backslash mathbf\{A\})\; =\; (\backslash det\; \backslash mathbf\{A\})I\_\{\backslash binom\{n\}\{r\}\}$, where {{math|C_r(A)}} denotes the {{math|r}}th compound matrix.

Higher adjugates may be defined in abstract algebraic terms in a similar fashion to the usual adjugate, substituting $\backslash wedge^r\; V$ and $\backslash wedge^\{n-r\}\; V$ for $V$ and $\backslash wedge^\{n-1\}\; V$, respectively.

Iteratively taking the adjugate of an invertible matrix A {{mvar|k}} times yields

:$\backslash overbrace\{\backslash operatorname\{adj\}\backslash dotsm\backslash operatorname\{adj\}\}^k(\backslash mathbf\{A\})=\backslash det(\backslash mathbf\{A\})^\{\backslash frac\{(n-1)^k-(-1)^k\}n\}\backslash mathbf\{A\}^\{(-1)^k\},$

:$\backslash det(\backslash overbrace\{\backslash operatorname\{adj\}\backslash dotsm\backslash operatorname\{adj\}\}^k(\backslash mathbf\{A\}))=\backslash det(\backslash mathbf\{A\})^\{(n-1)^k\}.$

For example,

:$\backslash operatorname\{adj\}(\backslash operatorname\{adj\}(\backslash mathbf\{A\}))\; =\; \backslash det(\backslash mathbf\{A\})^\{n\; -\; 2\}\; \backslash mathbf\{A\}.$

:$\backslash det(\backslash operatorname\{adj\}(\backslash operatorname\{adj\}(\backslash mathbf\{A\})))\; =\; \backslash det(\backslash mathbf\{A\})^\{(n\; -\; 1)^2\}.$

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

{{Reflist}}

• Roger A. Horn and Charles R. Johnson (2013), Matrix Analysis, Second Edition. Cambridge University Press, {{ISBN|978-0-521-54823-6}}
• Roger A. Horn and Charles R. Johnson (1991), Topics in Matrix Analysis. Cambridge University Press, {{ISBN|978-0-521-46713-1}}

• [http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/property.html#adjoint Matrix Reference Manual]
• [http://www.elektro-energetika.cz/calculations/matreg.php?language=english Online matrix calculator (determinant, track, inverse, adjoint, transpose)] Compute Adjugate matrix up to order 8
• {{cite web|url=http://www.wolframalpha.com/input/?i=adjugate+of+{+{+a%2C+b%2C+c+}%2C+{+d%2C+e%2C+f+}%2C+{+g%2C+h%2C+i+}+}|archive-url=|last=|first=|date=|title=Adjugate of { { a, b, c }, { d, e, f }, { g, h, i } }|archive-date=|access-date=|work=Wolfram Alpha}}

{{Matrix classes}}

Category:Matrix theory

Category:Linear algebra