Attachment matrix

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In modern mathematical terminology, it is called the matrix attached to the transposed conjugate matrix.

Given a square matrix A, its adjoint matrix or cofactor matrix cof(A) is the result of substituting each term a ij of A by the cofactor aij of A. The term adjoint matrix adj(A) is often confusing, since in many classical treatises on linear algebra it corresponds to the transposed cofactor matrix, however, in other texts, it corresponds to the cofactor matrix, since they call the cofactor adjoint in the same way and hence it is adjoint. In addition, the adj() symbol is also used interchangeably with cof() for the calculation on the elements of an array, thus making the confusion more extensive each time.

The main interest of the attached matrix is that it allows us to calculate the inverse of a matrix, since the relation is fulfilled:

A− − 1=1detAadj(A){displaystyle mathbf {A} ^{-1}={frac {1}{det mathbf {A}}}{;{mbox{adj}}}{mathbf {A}}}}}

where adj(A) corresponds to the transposed cofactor matrix, that is,

adj(A)=cof (A)T=CT{displaystyle mathrm {adj} (mathbf {A})=operatorname {cof} (mathbf {A})^{T}=mathbf {C} ^{T},}.

However, for matrices with large dimensions, this type of calculation is more expensive, in terms of operations, than other methods such as the Gaussian elimination method.

Definition and calculation formulas

Given a matrix A{displaystyle scriptstyle mathbf {A} } your attachment matrix is the only matrix B{displaystyle scriptstyle mathbf {B} } such that:

ABT=BTA=(detA)I{displaystyle mathbf {A} mathbf {B} ^{T}=mathbf {B} ^{T}mathbf {A} =(det mathbf {A})mathbf {I} }

This definition does not directly calculate the attachment matrix (or cofactors) so it is commonly defined also the attachment matrix by the following explicit formula. Given the explicit components of the matrix: (aij)=A한 한 Mn× × n{displaystyle (a_{ij})=mathbf {A} in M_{ntimes n}} for each i and j the matrix is defined A~ ~ (i,j){displaystyle {tilde {mathbf {A}}(i,j)} as the order matrix (n− − 1){displaystyle scriptstyle (n-1)} obtained from A{displaystyle mathbf {A} } removing the row i- and the column j- Sixty. And the amount is defined:

dij=(− − 1)i+jdetA~ ~ (i,j){displaystyle d_{ij}=(-1)^{i+j}det {tilde {mathbf {A}}}(i,j)}

And these are precisely the components of the attachment matrix (or cofactors), that is, cof(A)=(dij){displaystyle {mbox{cof}}(mathbf {A})=(d_{ij})},

2 x 2 matrices

Given a 2 x 2 matrix:

A=(A11A12A21A22){displaystyle mathbf {A} ={begin{pmatrix}A_{11} aliena_{12A_{21}{21}{22}end{pmatrix}}}}}}}}}}

Its adjoint matrix is given by:

adj(A)=CT=(A22− − A21− − A12A11)T=(A22− − A12− − A21A11){displaystyle {mbox{adj}}(mathbf {A})=mathbf {C} ^{T}={begin{pmatrix}A_{22}{mathbf}}{mathbf {C}}{mathbf}{mathbf}}{mathbf}}{mathbf}{mathbf}}}{mathbf}}}{mbox}{mbox}{cbf}}{mbox}{bf}}}{mbox}{mbox}{bf}}}{mbox}{mbox}{mbox}{mbox{bf}}{chmbox}}{mbox}{mbox}{mbox}{mbox}{mbox{mbox}{mbox}{mbox}{mbox}{mbox}}{mbox}{

where C is the cofactor matrix.

3 x 3 matrices

Given a 3 x 3 matrix:

A=(A11A12A13A21A22A23A31A32A33){displaystyle mathbf {A} ={begin{pmatrix}A_{11}{13}{12}{A_{13}A_{21}{21}{A_{22}{A_{23}{A_{23}A_{31}{31}{32}{end{pmatrix}}}}}}}

Its cofactor matrix is given by:

cof(A)=(+日本語A22A23A32A33日本語− − 日本語A21A23A31A33日本語+日本語A21A22A31A32日本語− − 日本語A12A13A32A33日本語+日本語A11A13A31A33日本語− − 日本語A11A12A31A32日本語+日本語A12A13A22A23日本語− − 日本語A11A13A21A23日本語+日本語A11A12A21A22日本語)=(A22A33− − A23A32A23A31− − A21A33A21A32− − A22A31A32A13− − A33A12A33A11− − A31A13A31A12− − A32A11A12A23− − A13A22A13A21− − A11A23A11A22− − A12A21)♪ I don't know ♪

and therefore the transpose of the cofactor matrix is the Adjoint matrix:

adj(A)=(+日本語A22A23A32A33日本語− − 日本語A21A23A31A33日本語+日本語A21A22A31A32日本語− − 日本語A12A13A32A33日本語+日本語A11A13A31A33日本語− − 日本語A11A12A31A32日本語+日本語A12A13A22A23日本語− − 日本語A11A13A21A23日本語+日本語A11A12A21A22日本語)T♪ I don't know ♪

For 3x3 matrices the following formula can also be used:

[chuckles]adj(A)]ij=12ε ε mniε ε pqjampanq{displaystyle [{mbox{adj}}(mathbf {A})]_{ij}{frac {1}{2}};epsilon _{mni};epsilon _{pqj};a_{mp};a_{nq}}}

Example

An example would be the following:

adj (2101− − 1102− − 1)=(− − 1111− − 2− − 22− − 4− − 3){displaystyle operatorname {adj} left({begin{array}{rrr}{rr}2}{1 hypo1}{1}{1}{1}{}{end{array}}}}}{left(}{begin{array}{rrrr}}-1 stranger11111{1}{1{1{1}{1{1}{1}{1{1}{1}{1}{1{1}{1{1}{1}{1}{1}{1}{1}{1}{1}{1{1}{1}{1}{1}{1}{1}{1}{1}{1}{1}{1}{1}{1}{11}{1}{1}{1}{

N x n matrices

For matrices with large n, the computational cost of calculating adjoints is large, so if the objective is to calculate the inverse of a matrix, other calculation algorithms that do not involve first compute the adjoint matrix. To calculate the adjoint matrix in the general case, the following formula can be used:

[chuckles]adj(A)]ij=1(n− − 1)!ε ε i1...... in− − 1iε ε j1...... jn− − 1jai1j1ai2j2...... ain− − 1jn− − 1{displaystyle}{mbox{adj}(mathbf {A}}}}{{frac {1}{(n-1)}}}}{i_n1}{i}{n1}}{n1}}{n1}}{n1}{n1}}{n1}{n1}{n1}{i}{n1}{s}{i}{s}{s}{s}}{s}{s}{s}{s}{j

Properties

Given a matrix A=(aij)한 한 Mn× × n{displaystyle mathbf {A} =(a_{ij})in M_{ntimes n}}} Defining B=(bij)=adj(A){displaystyle mathbf {B} =(b_{ij})={mbox{adj}}(A} it can be proved that the bij{displaystyle b_{ij},} can be written as a sum of degree monkeys n components aij{displaystyle a_{ij},}. That makes it as n increase the calculation of the attached matrix by application of direct formulas is complicated, becoming computationally very expensive.

If we consider the operation of looking for the attached matrix as a function: adj:Mn× × n→ → n× × n{displaystyle {mbox{adj}}:M_{ntimes n}to _{ntimes n} turns out that function is continuous. This can be seen from the continuity of the determining function. In addition, there are other interesting properties:

  • adj(AT)=adj(A)T{displaystyle {mbox{adj}}(mathbf {A} ^{T}={mbox{adj}}}(mathbf {A})^{T}}}}
  • adj(AB)=adj(B)adj(A){displaystyle {mbox{adj}}(mathbf {A} mathbf {B})={mbox{adj}}(mathbf {B}){mbox{adj}}}(mathbf {A}}}}}}
  • adj(I)=I{displaystyle {mbox{adj}}(mathbf {I})=mathbf {I} }
  • Aadj(AT)=adj(A)AT=det(A)I{displaystyle mathbf {A} ,mathrm {adj} (mathbf {A} ^{T})=mathrm {adj} (mathbf {A}),mathbf {A} ^{T}=det(mathbf {A},mathbf {I}}}}}}}}}} for A한 한 Mn× × n{displaystyle mathbf {A} in M_{ntimes n}}}.
  • adj(λ λ A)=λ λ n− − 1adj(A){displaystyle {mbox{adj}}(lambda mathbf {A})=lambda ^{n-1}{mbox{adj}}(mathbf {A})} for A한 한 Mn× × n{displaystyle mathbf {A} in M_{ntimes n}}}.
  • adj(adj(A))=det(A)n− − 2A{displaystyle {mbox{adj}}({mbox{adj}}}(mathbf {A})=det(mathbf {A})^{n-2}mathbf {A} }} for A한 한 Mn× × n{displaystyle mathbf {A} in M_{ntimes n}}}.
  • det(A)=tr(Aadj(A))/n{displaystyle det(mathbf {A})={mbox{tr}}(mathbf {A} {mbox{adj}}(mathbf {A})})/n} for A한 한 Mn× × n{displaystyle mathbf {A} in M_{ntimes n}}}.
  • det(adj(A))=det(A)n− − 1{displaystyle det {big}mathrm {adj} (mathbf {A}}{big)}=det(mathbf {A})^{n-1},}.

If p(t) = det(AtI) is the characteristic polynomial of A and we define the polynomial q(t) = (p(0) − p(t))/t, then:

adj(A)=q(A)=− − (p1I+p2A+p3A2+ +pnAn− − 1){displaystyle mathrm {adj} (mathbf {A})=q(mathbf {A})=-(p_{1}mathbf {I} +p_{2}mathbf {A} +p_{3}mathbf {A} ^{2}{2} +p_{n1}{n}{n}{n}{n}{n

Where pj{displaystyle p_{j},} are the coefficients of p(t(c):

p(t)=p0+p1t+p2t2+ pntn.{displaystyle p(t)=p_{0}+p_{1}t+p_{2}t^{2}+cdots p_{n}t^{n}. !

The adjoint function also appears in the formula for the derivative of the determinant:

det(A+H)− − det(A)=tr(adj(A)H)+or( H ){displaystyle det(mathbf {A+H})-det(mathbf {A})={mbox{tr}}({mbox{adj}}}}(mathbf {A}) mathbf {H})+o(associatedmathbf {H}organbf}}

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