Editing Minor (linear algebra) (section)

===Inverse of a matrix===
{{main|Invertible matrix}}

One can write down the inverse of an [[invertible matrix]] by computing its cofactors by using [[Cramer's rule]], as follows. The matrix formed by all of the cofactors of a square matrix {{math|'''A'''}} is called the '''cofactor matrix''' (also called the '''matrix of cofactors''' or, sometimes, ''comatrix''):

<math display=block>\mathbf C = \begin{bmatrix}
  C_{11}  & C_{12} & \cdots &   C_{1n} \\
  C_{21}  & C_{22} & \cdots &   C_{2n} \\
   \vdots & \vdots & \ddots & \vdots \\
  C_{n1}  & C_{n2} & \cdots &  C_{nn}
\end{bmatrix} </math>

Then the inverse of {{math|'''A'''}} is the transpose of the cofactor matrix times the reciprocal of the determinant of {{math|'''A'''}}:

<math display=block>\mathbf A^{-1} = \frac{1}{\operatorname{det}(\mathbf A)} \mathbf C^\mathsf{T}.</math>

The transpose of the cofactor matrix is called the [[adjugate]] matrix (also called the ''classical adjoint'') of {{math|'''A'''}}.

The above formula can be generalized as follows: Let 
<math display=block>\begin{align}
  I &= 1 \le i_1 < i_2 < \ldots < i_k \le n, \\[2pt]
  J &= 1 \le j_1 < j_2 < \ldots < j_k \le n,
\end{align}</math>
be ordered sequences (in natural order) of indexes (here {{math|'''A'''}} is an {{math|''n'' × ''n''}} matrix). Then<ref name="Prasolov1994">{{cite book|author=Viktor Vasil_evich Prasolov|title=Problems and Theorems in Linear Algebra|url=https://books.google.com/books?id=b4yKAwAAQBAJ&pg=PR15|date=13 June 1994|publisher=American Mathematical Soc.|isbn=978-0-8218-0236-6|pages=15–}}</ref>

<math display=block>[\mathbf A^{-1}]_{I,J} = \pm\frac{[\mathbf A]_{J',I'}}{\det \mathbf A},</math>

where {{math|''I′'', ''J′''}} denote the ordered sequences of indices (the indices are in natural order of magnitude, as above) complementary to {{math|''I'', ''J''}}, so that every index {{math|1, ..., ''n''}} appears exactly once in either {{mvar|I}} or {{mvar|I'}}, but not in both (similarly for the  {{mvar|J}} and {{mvar|J'}}) and {{math|['''A''']<sub>''I'', ''J''</sub>}} denotes the determinant of the submatrix of {{math|'''A'''}} formed by choosing the rows of the index set {{mvar|I}} and columns of index set {{mvar|J}}. Also, <math>[\mathbf A]_{I,J} = \det \bigl( (A_{i_p, j_q})_{p,q = 1, \ldots, k} \bigr).</math> A simple proof can be given using wedge product. Indeed,

<math display=block>\bigl[ \mathbf A^{-1} \bigr]_{I,J} (e_1\wedge\ldots \wedge e_n) = \pm(\mathbf A^{-1}e_{j_1})\wedge \ldots \wedge(\mathbf A^{-1}e_{j_k})\wedge e_{i'_1}\wedge\ldots \wedge e_{i'_{n-k}}, </math>

where <math>e_1, \ldots, e_n</math> are the basis vectors. Acting by {{math|'''A'''}} on both sides, one gets

<math display=block>\begin{align}
  &\ \bigl[\mathbf A^{-1} \bigr]_{I,J} \det \mathbf A (e_1\wedge\ldots \wedge e_n) \\[2pt]
  =&\ \pm (e_{j_1})\wedge \ldots \wedge(e_{j_k})\wedge (\mathbf A e_{i'_1})\wedge\ldots \wedge (\mathbf A e_{i'_{n-k}}) \\[2pt]
  =&\ \pm [\mathbf A]_{J',I'}(e_1\wedge\ldots \wedge e_n).
\end{align}</math>

The sign can be worked out to be 
<math display=block>(-1)^\wedge \!\!\left( \sum_{s=1}^{k} i_s - \sum_{s=1}^{k} j_s \right),</math>
so the sign is determined by the sums of elements in {{mvar|I}} and {{mvar|J}}.