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 [所有列减第1列]
-1 0 ...0 -n
-1 0 ...-n 0
......
-1 -n ...0 0
-1 0 ...0 0
[这是斜上三角行列式]
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+n^(n-1)]/2