Question

高等代数 求行列式的值

Answer 1

a1 b b ... b
b-a1 a2-b 0 0 ... 0
b-a1 0 a3-b 0 ... 0
.......
b-a1 0 ... an-b
按最后一列展开得
D[n]=-b(b-a1)(a2-b)(a3-b)...(a[n-1]-b)+(an-b)D[n-1]
=(an-b)D[n-1]+b(a1-b)(a2-b)...(a[n-1]-b)
=(an-b)((a[n-1]-b)D[n-2]+b(a1-b)(a2-b)...(a[n-2]-b))
+b(a1-b)(a2-b)...(a[n-1]-b)
=(an-b)(a[n-1]-b)D[n-2]+b(a1-b)(a2-b)...(an-b)(1/(a[n-1]-b)+1/(an-b))
=...
=(an-b)(a[n-1]-b)...(a2-b)D1+b(a1-b)(a2-b)...(an-b)(∑1/(ak-b)) k从2求和到n
=(an-b)(a[n-1]-b)...(a2-b)a1+b(a1-b)(a2-b)...(an-b)(∑1/(ak-b)) k从2求和到n
=(an-b)(a[n-1]-b)...(a2-b)(a1-b)+b(a1-b)(a2-b)...(an-b)(∑1/(ak-b)) k从1求和到n
记a0=2b，则上式可以统一写成
=(a0-b)(a1-b)(a2-b)...(an-b)(∑1/(ak-b)) k从0求和到n
{注：这里的结果只是个形式记号，若ak-b=0，那么便把
(a0-b)(a1-b)(a2-b)...(an-b)中的ak-b和1/(ak-b)约掉即可)

Answer 2

这类行列式可化为箭形行列式
所有行减第1行
D =
a1 b b .... b
b-a1 a2-b 0 ... 0
b-a1 0 a3-b ... 0
... ...
b-a1 0 0 ... an-b
第1列提出 a1-b, 第2列提出 a2-b ...... 第n列提出 an-b
然后将所有列加到第1列
行列式化为上三角形式
D = (a1-b)(a2-b)...(an-b) [ a1/(a1-b) +b/(a2-b) + ...+b/(an-b)]
或 (a1-b)(a2-b)...(an-b) [ 1 + b/(a1-b) +b/(a2-b) + ...+b/(an-b)]

注意: 这需要 ai≠b