阶的形式．
解：将行列式按第一行展开，得Dn?2Dn?1?Dn?2. 即
Dn?Dn?1?Dn?1?Dn?2.
∴Dn?Dn?1?Dn?1?Dn?2???D2?D1?3?2?1. ∴Dn?1?Dn?1???1?1???1?Dn??n?1?
????n?1??2?n?1.
5.2 逐行相加减和套用范德蒙德行列式 例20 计算行列式
1D?1?sin?1sin?1?sin2?1sin2?1?sin3?111?sin?2sin?2?sin?22sin2?2?sin3?211?sin?3sin?3?sin?23sin2?3?sin3?311?sin?4解：从第
sin?4?sin?24sin2?4?sin3?4一行开始，依次用上一行的??1?倍加到下一行，进行逐行相加，得
1sin?1D?sin2?1sin3?11sin?1D?sin2?1sin3?11sin?2sin2?2sin3?21sin?2sin2?2sin3?21sin?3sin2?3sin3?31sin?3sin2?3sin3?31sin?4.
sin2?4sin3?41sin?4???sin?i?sin?j?.
sin2?41?j?i?4sin3?4再由范德蒙德行列式，得
5.3 构造法和套用范德蒙德行列式
20
1x1x12例21 求行列式Dn??x1n?2x1n1x22x2??????1xn2xn. ?n?2n?2x2?xnnx2nxn解：虽然Dn不是范德蒙德行列式，但可以考虑构造n?1阶的范德蒙德行列式来间接求出Dn的值．
构造n?1阶的范德蒙德行列式，得
1x1f?x??x12?x1n?2x1n?1x1n1x22x2????1xn2xn1xx2?xn?2xn?1xn.
?n?2x2n?1x2nx2?n?2?xnn?1?xn?nxn将f?x?按第n?1列展开，得
f?x??A1,n?1?A2,n?1x???An,n?1xn?1?An?1,n?1xn，
其中，xn?1的系数为
An,n?1???1?又根据范德蒙德行列式的结果知
n??n?1?Dn??Dn.
f?x???x?x1??x?x2???x?xn?由上式可求得xn?11?j?i?n??xi?xj?.
的系数为
??x1?x2?xn?故有：Dn??x1?x2???xn?1?j?i?n??xi?xj?.
1?j?i?n??xi?xj?.
21