# 餘因子矩陣

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

## 範例

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

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

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

## 餘因子分解

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

$\ \det(A) = a_{1j}C_{1j} + a_{2j}C_{2j} + a_{3j}C_{3j} + ... + a_{nj}C_{nj}$
（對第 j 縱行的餘因子分解）
$\ \det(A) = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} + ... + a_{in}C_{in}$
（對第 i 橫列的餘因子分解）

## 古典伴隨矩陣

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

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

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

## 克萊姆法則

$\mathrm{cof}(A)^t A = A\mathrm{cof}(A)^t = \det(A) I_n$

$\det A \neq 0$ 時，$A$ 的逆矩陣由下式給出：

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

## 文獻

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