21?1(2)
12?1???; 11?2n?11?111?1解:原式=
n?12?11???=(n?1)12???? n?11?211?211?1 =(n?1)01?0????n?1. 00?1
x1?mx2?xn (3)Dn?x1x2?m?xn???x1x2?xn?m解:将第2列,L,第n列分别加到第一列,并提取第一列的
公因子,得
x1?x2???xn?mx2?xnDx1?x2???xn?mx2?m?xnn????x1?x2???xn?mx2?xn?m1x2?xn?(x1x2?m?1?x2???xxnn?m)???1x2?xn?m
10?0?(x1?m?01?x2???xn?m)???
10??m?(x1?x2???xn?m)(?m)n?1
b1b2b3?bn?1bn?a1a20?00 (4)Dn?0?a2a3?00 ?????000??an?1an(其中ai?0,i?1,2,?,n)
?a1a20?0 解: D0?a20?0n?(?1)1?nbn????000??an?1b1b2?bn?2bn?1?a1a2?00?an0?a2?00 ????00?an?2an?1?abn1a2?an?a?anDn?1 n ???a?nbi?1a2?an?????i?1ai??. 三、证明题
1.试证:如果n次多项式f(x)?a0?a1x???anxn对n+1个不同的x值都是零,则此多项式恒等于零.
(提示:用范德蒙行列式证明)
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