计算n阶行列式Dn=.x1+1x1+2…x1+

2025-06-22 19:15:48
推荐回答(1个)
回答1:


①当n=2时,
D2

x1+1 x1+2
x2+1 x2+2
x1?x2
②当n>2时,
Dn
x1+1 x1+2 x1+(n?1) x1+n
x2+1 x2+2 x2+(n?1) x2+n
xn+1 xn+2 xn+(n?1) xn+n
cj?cj?1  (j=n,n?1,…2)
x1+1 1 1 1
x2+1 1 1 1
xn+1 1 1 1
=0.