n阶行列式第一行-b1 b1 0···0 0第二行0 -b2 b2···0 0第n行0 0 0···-bn bn最后一行全是1

2025-06-21 18:21:07
推荐回答(2个)
回答1:

-b1 b1 0···0 0
0 -b2 b2···0 0

0 0 0···-bn bn
1 1 1 ... 1 1

所有列加到第1列, 再按第1列展开
D = (-1)^(n+1+1) * (n+1) *b1b2...bn
= (-1)^n * (n+1) *b1b2...bn

回答2:

不知道,太深奥了