1.
解: D =
c1+c2+c3+c4
a+b+c+d b c d
a+b+c+d a d c
a+b+c+d d a b
a+b+c+d c b a
r2-r1,r3-r1,r4-r1
a+b+c+d b c d
0 a-b d-c c-d
0 d-b a-c b-d
0 c-b b-c a-d
c2+c3
a+b+c+d b+c c d
0 a-b-c+d d-c c-d
0 a-b-c+d a-c b-d
0 0 b-c a-d
r3-r2
a+b+c+d b+c c d
0 a-b-c+d d-c c-d
0 0 a-d b-c
0 0 b-c a-d
c3+c4
a+b+c+d b+c c d
0 a-b-c+d d-c c-d
0 0 a+b-c-d b-c
0 0 a+b-c-d a-d
r4-r3
a+b+c+d b+c c d
0 a-b-c+d d-c c-d
0 0 a+b-c-d b-c
0 0 0 a-b+c-d
行列式 = (a+b+c+d)(a-b-c+d)(a+b-c-d)(a-b+c-d).
2.
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+n^(n-1)]/2
