# 余因子矩阵本文重定向自 余因子矩阵

${\displaystyle \mathbf {A} ={\begin{bmatrix}1&2\\3&4\end{bmatrix}}}$

## 范例

${\displaystyle B={\begin{bmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\\\end{bmatrix}}}$

${\displaystyle M_{23}={\begin{vmatrix}b_{11}&b_{12}&\Box \\\Box &\Box &\Box \\b_{31}&b_{32}&\Box \\\end{vmatrix}}}$ 给出 ${\displaystyle M_{23}={\begin{vmatrix}b_{11}&b_{12}\\b_{31}&b_{32}\\\end{vmatrix}}=b_{11}b_{32}-b_{31}b_{12}}$

${\displaystyle \ C_{23}=(-1)^{2+3}(M_{23})}$
${\displaystyle \ C_{23}=(-1)^{5}(b_{11}b_{32}-b_{31}b_{12})}$
${\displaystyle \ C_{23}=b_{31}b_{12}-b_{11}b_{32}.}$

## 余因子分解

${\displaystyle A={\begin{bmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\cdots &a_{nn}\end{bmatrix}}}$

${\displaystyle \ \det(A)=a_{1j}C_{1j}+a_{2j}C_{2j}+a_{3j}C_{3j}+...+a_{nj}C_{nj}}$
（对第 j 纵行的余因子分解）
${\displaystyle \ \det(A)=a_{i1}C_{i1}+a_{i2}C_{i2}+a_{i3}C_{i3}+...+a_{in}C_{in}}$
（对第 i 横列的余因子分解）

## 古典伴随矩阵

${\displaystyle A^{-1}={\frac {\mathrm {adj} (A)}{\det(A)}}}$

${\displaystyle {\begin{bmatrix}C_{11}&C_{12}&\cdots &C_{1n}\\C_{21}&C_{22}&\cdots &C_{2n}\\\vdots &\vdots &\ddots &\vdots \\C_{n1}&C_{n2}&\cdots &C_{nn}\end{bmatrix}}}$

${\displaystyle \mathrm {adj} (A)={\begin{bmatrix}C_{11}&C_{21}&\cdots &C_{n1}\\C_{12}&C_{22}&\cdots &C_{n2}\\\vdots &\vdots &\ddots &\vdots \\C_{1n}&C_{2n}&\cdots &C_{nn}\end{bmatrix}}}$

## 克莱姆法则

${\displaystyle \mathrm {cof} (A)^{t}A=A\mathrm {cof} (A)^{t}=\det(A)I_{n}}$

${\displaystyle \det A\neq 0}$ 时，${\displaystyle A}$ 的逆矩阵由下式给出：

${\displaystyle A^{-1}={\dfrac {\mathrm {cof} (A)^{t}}{\det A}}}$

## 文献

• Anton, Howard and Chris, Rorres, Elementary Linear Algebra, 9th edition (2005), John Wiley and Sons. ISBN 0-471-66959-8