Question

N阶行列式计算求高手，跪求。

|1234....n|
|x123...n-1|
|xx12.. n-2|
............
....
|xxx... 1|

Answer 1

1 2 3 4 ... n
x 1 2 3 ... n-1
x x 1 2 ... n-2
x x x 1 ... n-3
... ... ...
x x x x ... 1

ri - r(i+1), i=1,2,...,n-1
1-x 1 1 1 ... 1
0 1-x 1 1 ... 1
0 0 1-x 1 ... 1
0 0 0 1-x ... 1
... ... ...
0 0 0 0 ...1-x 1
x x x x ... x 1

当x=1时, 按第1列展开得行列式 = (-1)^(n+1).

当x≠1时, 目标是把行列式化成下三角.
ci+(1/(x-1))c1, i=2,3,...,n [用a11 将第1行左边消0]
最后一行变成:
x, x^2/(x-1), x^2/(x-1), x^2/(x-1), ..., , x^2/(x-1),(2x-1)/(x-1)

同样的方法处理第2行: ci+(1/(x-1))c2, i=2,3,...,n
最后一行变成:
x, x^2/(x-1), x^3/(x-1), x^3/(x-1), ..., , x^3/(x-1),???

问题转化为: 最后一行最后一列位置的元素ann的算法
若找到这个规律, 此行列式就解了
呵呵 就到这里吧