Question

求解，这个行列式的计算方法

Answer 1

解: c1+c2+...+cn [所有列加到第1列]
n(n+1)/2 2 3 ... n-1 n
n(n+1)/2 3 4 ... n 1
n(n+1)/2 4 5 ... 1 2
... ...
n(n+1)/2 n 1 ... n-3 n-2
n(n+1)/2 1 2 ... n-2 n-1

第1列提出公因子 n(n+1)/2, 然后
ri-r(i-1), i=n,n-1,...,2 [从最后一行开始,每一行减上一行]
1 2 3 ... n-1 n
0 1 1 ... 1 1-n
0 1 1 ... 1-n 1
... ...
0 1 1-n ... 1 1
0 1-n 1 ... 1 1

按第1列展开
1 1 ... 1 1-n
1 1 ... 1-n 1
... ...
1 1-n ... 1 1
1-n 1 ... 1 1

c1+c2+...+cn-1 [所有列加到第1列]
-1 1 ... 1 1-n
-1 1 ... 1-n 1
... ...
-1 1-n ... 1 1
-1 1 ... 1 1

ci+c1, i=2,3,...,n-1
-1 0 ... 0 -n
-1 0 ...-n 0
... ...
-1 -n ... 0 0
-1 0 ... 0 0

行列式 = n(n+1)/2 * (-1)^[(n-2)(n-1)/2]*(-1)^(n-1)*n^(n-2)
= (-1)^[n(n-1)/2] * n^(n-2) * n(n+1)/2.

= (-1)^[n(n-1)/2]*[n^n+n^(n-1)]/2.

Answer 2

Answer 3

Answer 4