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→2 × 2 generic matrix: the adjugate matrix should be a transpose of the cofactor matrix. |
Delimited matrices with square brackets to visually distinguish from parentheses everywhere used for grouping |
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=== 1 × 1 generic matrix ===
The adjugate of any non-zero 1×1 matrix (complex scalar) is <math>\mathbf{I} =
=== 2 × 2 generic matrix ===
The adjugate of the 2×2 matrix
:<math>\mathbf{A} = \begin{
</math>
is
:<math>
\operatorname{adj}(\mathbf{A}) = \begin{
</math>
By direct computation,
:<math>\mathbf{A} \operatorname{adj}(\mathbf{A}) = \begin{
In this case, it is also true that det(adj('''A''')) = det('''A''') and hence that adj(adj('''A''')) = '''A'''.
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Consider a 3×3 matrix
:<math>
\mathbf{A} = \begin{
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{
</math>
Its cofactor matrix is
:<math>
\mathbf{C} = \begin{
+\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{
</math>
where
:<math>
\begin{vmatrix} a_{im} & a_{in} \\ a_{jm} & a_{jn} \end{vmatrix}
=\det\begin{
</math>
Its adjugate is the transpose of its cofactor matrix,
:<math>
\operatorname{adj}(\mathbf{A}) = \mathbf{C}^\mathsf{T} = \begin{
+\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} &
-\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} &
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-\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{
</math>
=== 3 × 3 numeric matrix ===
As a specific example, we have
:<math>\operatorname{adj} \begin{
-3 & 2 & -5 \\
-1 & 0 & -2 \\
3 & -4 & 1
\end{
-8 & 18 & -4 \\
-5 & 12 & -1 \\
4 & -6 & 2
\end{
</math>
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''',
:<math>\begin{
The (3,2) cofactor is a sign times the determinant of this submatrix:
:<math>(-1)^{3+2}\operatorname{det}\begin{
and this is the (2,3) entry of the adjugate.
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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''') = ''α'''''xy'''<sup>T</sup>}}, where {{math|''α''}} is a scalar and {{math|'''x'''}} and {{math|'''y'''}} are vectors such that {{math|1='''Ax''' = '''0'''}} and {{math|1='''A'''<sup>T</sup> '''y''' = '''0'''}}.
=== Column substitution and Cramer's rule ===
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Partition {{math|'''A'''}} into column vectors:
:<math>\mathbf{A} =
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'''}}:
:<math>(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\ \stackrel{\text{def}}{=}\ \begin{
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
:<math>\left(\det(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\right)_{i=1}^n = \operatorname{adj}(\mathbf{A})\mathbf{b}.</math>
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:<math>V\ \xrightarrow{\phi}\ \operatorname{Hom}(\wedge^{n-1} V, \wedge^n V)\ \xrightarrow{(\wedge^{n-1} T)^*}\ \operatorname{Hom}(\wedge^{n-1} V, \wedge^n V)\ \xrightarrow{\phi^{-1}}\ V.</math>
If {{math|1=''V'' = '''R'''<sup>''n''</sup>}} is endowed with its coordinate basis {{math|'''e'''<sub>1</sub>,
:<math>\{\mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_k \wedge \dots \wedge \mathbf{e}_n\}_{k=1}^n.</math>
Fix a basis vector {{math|'''e'''<sub>''i''</sub>}} of {{math|'''R'''<sup>''n''</sup>}}. The image of {{math|'''e'''<sub>''i''</sub>}} under <math>\phi</math> is determined by where it sends basis vectors:
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== Higher adjugates ==
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 <math display="inline">
:<math>(-1)^{\sigma(I) + \sigma(J)}\det \mathbf{A}_{J^c, I^c},</math>
where {{math|σ(''I'')}} and {{math|σ(''J'')}} are the sum of the elements of {{math|''I''}} and {{math|''J''}}, respectively.
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