1个回答
展开全部
Dn+1 经 n(n+1)/2 次换行后,及 n(n+1)/2 次换列后,矩阵化为:
Vandermonde(范德蒙) 行列式 V :
V=
1 1 ... 1
a-n a-n+1 ... a
..........................
(a-n)^n (a-n+1)^n a^n
∴ Dn+1 = (-1)^[n(n+1)/2 + n(n+1)/2 ]*V = V
记: ak=a-n+k-1 ,则: aj-ai= j-i
V = ∏(1≤i<j≤n+1) (aj-ai) = ∏(1≤i<j≤n+1) (j-i)
= ∏(2≤j≤n+1) {∏(1≤i≤j-1) (j-i)
= ∏(2≤j≤n+1) (j-1)!
= 1!*2!*3!*...*n!
∴ Dn+1 = 1!*2!*3!*...*n!
Vandermonde(范德蒙) 行列式 V :
V=
1 1 ... 1
a-n a-n+1 ... a
..........................
(a-n)^n (a-n+1)^n a^n
∴ Dn+1 = (-1)^[n(n+1)/2 + n(n+1)/2 ]*V = V
记: ak=a-n+k-1 ,则: aj-ai= j-i
V = ∏(1≤i<j≤n+1) (aj-ai) = ∏(1≤i<j≤n+1) (j-i)
= ∏(2≤j≤n+1) {∏(1≤i≤j-1) (j-i)
= ∏(2≤j≤n+1) (j-1)!
= 1!*2!*3!*...*n!
∴ Dn+1 = 1!*2!*3!*...*n!
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询