Adjugate matrix

The adjugate of any non-zero 1×1 matrix (complex scalar) is ${\displaystyle \mathbf {I} =(1)}$. By convention, adj(0) = 0.

The adjugate of the 2×2 matrix

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

is

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

By direct computation,

${\displaystyle \mathbf {A} \operatorname {adj} (\mathbf {A} )={\begin{pmatrix}ad-bc&0\\0&ad-bc\end{pmatrix}}=(\det \mathbf {A} )\mathbf {I} .}$

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

Consider a 3×3 matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}.}$

Its cofactor matrix is

${\displaystyle \mathbf {C} ={\begin{pmatrix}+{\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}}\\&&\\-{\begin{vmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{vmatrix}}\\&&\\+{\begin{vmatrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}\end{pmatrix}},}$

where

${\displaystyle {\begin{vmatrix}a_{im}&a_{in}\\a_{jm}&a_{jn}\end{vmatrix}}=\det {\begin{pmatrix}a_{im}&a_{in}\\a_{jm}&a_{jn}\end{pmatrix}}}$.

Its adjugate is the transpose of its cofactor matrix,

${\displaystyle \operatorname {adj} (\mathbf {A} )=\mathbf {C} ^{\mathsf {T}}={\begin{pmatrix}+{\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix}}\\&&\\-{\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{13}\\a_{21}&a_{23}\end{vmatrix}}\\&&\\+{\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}}&-{\begin{vmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{vmatrix}}&+{\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}\end{pmatrix}}}$.

As a specific example, we have

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

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

The −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,

${\displaystyle {\begin{pmatrix}-3&-5\\-1&-2\end{pmatrix}}.}$

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

${\displaystyle (-1)^{3+2}\operatorname {det} {\begin{pmatrix}-3&-5\\-1&-2\end{pmatrix}}=-(-3\cdot -2--5\cdot -1)=-1,}$

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

Partition A into column vectors:

${\displaystyle \mathbf {A} =(\mathbf {a} _{1}\ \cdots \ \mathbf {a} _{n}).}$

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

${\displaystyle (\mathbf {A} {\stackrel {i}{\leftarrow }}\mathbf {b} )\ {\stackrel {\text{def}}{=}}\ {\begin{pmatrix}\mathbf {a} _{1}&\cdots &\mathbf {a} _{i-1}&\mathbf {b} &\mathbf {a} _{i+1}&\cdots &\mathbf {a} _{n}\end{pmatrix}}.}$

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

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

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

${\displaystyle \mathbf {A} \mathbf {x} =\mathbf {b} .}$

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

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

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

${\displaystyle x_{i}={\frac {\det(\mathbf {A} {\stackrel {i}{\leftarrow }}\mathbf {b} )}{\det \mathbf {A} }},}$

where x_i is the ith entry of x.

Let the characteristic polynomial of A be

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

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

${\displaystyle \Delta p(s,t)={\frac {p(s)-p(t)}{s-t}}=\sum _{0\leq j+k<n}p_{j+k+1}s^{j}t^{k}\in R[s,t].}$

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

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

In particular, the resolvent of A is defined to be

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

and by the above formula, this is equal to

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

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

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

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

${\displaystyle d(\det \mathbf {A} )_{\mathbf {A} _{0}}=\operatorname {adj} (\mathbf {A} _{0})^{\mathsf {T}}.}$

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

${\displaystyle p_{\mathbf {A} }(\mathbf {A} )=\mathbf {0} .}$

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

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

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

${\displaystyle s+\sum _{\ell =1}^{n-1}\ell k_{\ell }=n-1.}$

For the 2×2 case, this gives

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

For the 3×3 case, this gives

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

For the 4×4 case, this gives

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

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