Question

线性代数 下面行列式求简单方法？

1+a 1 1 1 a0 1 1 ....1
1 1-a 1 1 1 a1 0 ...0
1 1 1+b 1 1 0 a2 ...0
1 1 1 1-b . . . .
1 0 0 ....an

Answer 1

(1)
若b=0, 由行列式=0
当b≠0时,
ri-r1, i=2,3,4
1+a 1 1 1
-a -a 0 0
-a 0 b 0
-a 0 0 -b

c1-r2+(a/b)c3-(a/b)c4
a 1 1 1
0 -a 0 0
0 0 b 0
0 0 0 -b

行列式 = a^2b^2.
当b=0时,上式也成立.

(2)
c1-(1/ai)c(i+1), i=1,2,...,n
t 1 1 ... 1
0 a1 0 ... 0
0 0 a2 ... 0
......
0 0 0 ... an

t = a0-1/a1-1/a2-...-1/an
行列式 = (a0-1/a1-1/a2-...-1/an)a1a2...an