解: Dn =
a b b ... b+0
b a b ... b+0
a a b ... a+0
... ...
a a a ... a+(b-a)
按第n列分拆, Dn =
a b b ... b
b a b ... b
a a b ... a
... ...
a a a ... a
+ (b-a)Dn-1
(所有列减第n列, 化为上三角行列式)
= (b-a)Dn-1 + a(a-b)^2(b-a)^(n-3)
= a(b-a)^(n-1) + (b-a)Dn-1
-- 迭代
= a(b-a)^(n-1) + (b-a)[a(b-a)^(n-2) + (b-a)Dn-2]
= 2a(b-a)^(n-1) + (b-a)^2Dn-2
= ...
= (n-4)a(b-a)^(n-1) + (b-a)^(n-4)D4
= (n-4)a(b-a)^(n-1) + (b-a)^(n-4)[-(b-a)^4]
= [(n-3)a-b](b-a)^(n-1)
注:
n>=4 时成立.
D3 = -b*(a - b)^2
D4 = -(a - b)^4
D5 = (2*a - b)*(a - b)^4
D6 = -(3*a - b)*(a - b)^5