# 拉普拉斯展开

（重定向自拉普拉斯展开式

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

## 公式

B = (bij)是一个n × n矩阵。B关于第i行第j列的余子式Mij是指B中去掉第i行第j列后得到的n−1阶子矩阵的行列式。有时可以简称为B的（ij余子式B的（ij代数余子式Cij是指B的（ij）余子式Mij与(−1)i + j的乘积：Cij = (−1)i + j Mij

{\displaystyle {\begin{aligned}|B|&{}=b_{i1}C_{i1}+b_{i2}C_{i2}+\cdots +b_{in}C_{in}\\&{}=b_{1j}C_{1j}+b_{2j}C_{2j}+\cdots +b_{nj}C_{nj}.\end{aligned}}}

## 例子

${\displaystyle B={\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}}}$

${\displaystyle |B|=1\cdot {\begin{vmatrix}5&6\\8&9\end{vmatrix}}-2\cdot {\begin{vmatrix}4&6\\7&9\end{vmatrix}}+3\cdot {\begin{vmatrix}4&5\\7&8\end{vmatrix}}}$
${\displaystyle {}=1\cdot (-3)-2\cdot (-6)+3\cdot (-3)=0}$

${\displaystyle |B|=-2\cdot {\begin{vmatrix}4&6\\7&9\end{vmatrix}}+5\cdot {\begin{vmatrix}1&3\\7&9\end{vmatrix}}-8\cdot {\begin{vmatrix}1&3\\4&6\end{vmatrix}}}$
${\displaystyle {}=-2\cdot (-6)+5\cdot (-12)-8\cdot (-6)=0}$

## 证明

B是一个n × n的矩阵，ij ∈ {1, 2, ..., n}。为了明确起见，将${\displaystyle M_{ij}}$的系数记为${\displaystyle (a_{st})}$，其中1 ≤ s,t ≤ n − 1.

${\displaystyle \operatorname {sgn} \tau \,b_{1,\tau (1)}\cdots b_{i,j}\cdots b_{n,\tau (n)}=\operatorname {sgn} \tau \,b_{ij}a_{1,\sigma (1)}\cdots a_{n-1,\sigma (n-1)}}$

${\displaystyle \tau \,=(n,n-1,\ldots ,i)\,\sigma '\,(j,j+1,\ldots ,n)}$

${\displaystyle \operatorname {sgn} \tau \,=(-1)^{2n-(i+j)}\operatorname {sgn} \sigma '\,=(-1)^{i+j}\operatorname {sgn} \sigma }$

 ${\displaystyle \sum _{\tau \in S_{n}:\tau (i)=j}}$ ${\displaystyle \operatorname {sgn} \tau \,b_{1,\tau (1)}\cdots b_{n,\tau (n)}}$ ${\displaystyle =\sum _{\sigma \in S_{n-1}}(-1)^{i+j}\operatorname {sgn} \sigma \,b_{ij}a_{1,\sigma (1)}\cdots a_{n-1,\sigma (n-1)}}$ ${\displaystyle =\ b_{ij}(-1)^{i+j}|M_{ij}|,}$