這個(gè)問題有點(diǎn)麻煩
(數(shù)與數(shù)之間我用逗號(hào)隔開)
首先將每一列的數(shù)都家到第一列上
原式=
n(n+1)/2,2,3,4,...,n-1,n
n(n+1)/2,1,2,3,...,n-2,n-1
n(n+1)/2,n,1,2,...,n-3,n-2
.
n(n+1)/2,4,5,6,...,1,2
n(n+1)/2,3,4,5,...,n,1
第一列提出來n(n+1)/2
原式=n(n+1)/2*
1,2,3,4,...,n-1,n
1,1,2,3,...,n-2,n-1
1,n,1,2,...,n-3,n-2
.
1,4,5,6,...,1,2
1,3,4,5,...,n,1
從最后一行開始到第二行每一行都減去自己上面的一行
原式=n(n+1)/2*
1,2,3,4,...,n-1,n
0,-1,-1,-1,...,-1,-1
0,n-1,-1,-1,...,-1,-1
.
0,-1,-1,-1,...,-1,-1
0,-1,-1,-1,...,n-1,-1
從最后一列到第3列,每一列都減去自己左面一列得
原式=n(n+1)/2*
1,2,3,4,...,n-1,n
0,-1,-1,-1,...,-1,-1
0,n,0,0,...,0,0
.
0,0,0,0,...,0,0
0,0,0,0,...,n,0
按照第一列展開得
原式=n(n+1)/2*
-1,-1,-1,...,-1,-1
n,0,0,...,0,0
0,n,0,...,0,0
.
0,0,0,...,0,0
0,0,0,...,n,0
按最后一列展開得
原式=n(n+1)/2*（-1）^(1+n-1)*(-1)*
n,0,0,...,0
0,n,0,...,0
.
0,0,0,...,0
0,0,0,...,n
=(-1)^(n+1)*n(n+1)/2*n^(n-2)