an?1?按第一行展开an0?cn?100an?10?a1c10?0?bn?1b1d1?dn?1000?
0dnbn?1?a1c1b1d1?dn?1000?(?1)2n?1bn0?cn?1cn
都按最后一行展开andnD2n?2?bncnD2n?2
由此得递推公式：
D2n?(andn?bncn)D2n?2
n即 D2n?而 D2??(ai?2idi?bici)D2
a1c1nb1d1?a1d1?b1c1
得 D2n?(5)aij?i?j
01Dn?det(aij)?23?n?11012??(adii?1i?bici)
2101?n?33210?n?4??????n?1n?2n?3n?4?011
n?2?1?1r1?r2r2?r3,?1?1?1?1?n?200?2?2?2n?411?a2?100a3???????a111?1?1?n?3000?2?2n?5????0000??an?100a2?a30?00111?1?n?4??????11?1?an0000?an?1?an00??????0000?n?1111c2?c1,c3?c11?0c4?c1,??1?1?n?1
?1?1?1?1?n?10?2?2?2?2n?31?a11?10a2?a30?00=(?1)n?1(n?1)2n?2
(6)Dn?c1?c2,c2?c3c3?c4,?
a1?a200?001111?11?an???????0000??an?200000?an?200000?0?an?a4?00按最后一列展开（由下往上）
?a20a3?a4?00(1?an)(a1a2?an?1)?0?00
12
a1?a2?0?00?a200?00a20a2?a3?0000a3?000a3?a4?00n????????????1ai000??an?10000??an?10000?an?1?an000?an?1?an???
?a30?00
?(1?an)(a1a2?an?1)?a1a2?an?3an?2an???a2a3?an
?(a1a2?an)(1??)
i?1
8.用克莱姆法则解下列方程组：
?x1?x2?x3?x4?5,??x1?2x2?x3?4x4??2,(1)? ?2x1?3x2?x3?5x4??2,?3x?x?2x?11x?0;234?1?1,?5x1?6x2?x?5x2?6x3?0,?1?(2)?x2?5x3?6x4?0, ?x3?5x4?6x5?0,??x4?5x5?1.?112?311?1?1214?511?100011?5?21?2?3?113?78解 (1)D?123
13
1?0005D1??2?201?0001?0001D2?1231?0001D3?1235110012?31?55?131?51005?2?20131?2?13?51?1?12?10?32?12?10231?1?121211315?2?201381414?511?99?2311?911?4612014?5111?00014?511??1000110051?2?1015?31?515?13?51005?713?5414210?12?120?3?12?101?2?3?112?19?284??284 19?511?9119?23?9113814213?78??142??142
?0?201
?0001
?000
10005?100?12?1513?10
?10012?312339??426
14
112?31D1D1?1?12?1,5?2?20x2?D2D?2,x3?D3D?3,x4?D4D??1 D4?123?142
?x1?
51651000651000651000655按最后一行展开651006510006?5D??6D??
(2)D?0005D??100?5(5D???6D???)?6D???19D???30D???
?65D????114D?????65?19?114?5?665
?的余子式,D???,D????类推) (D?为行列式D中a11的余子式,D??为D?中a1110D1?00165100406510006510006546按第一列展开065100650006D??510
?D??6?19D????30?????6?1507
51D2?0005?10100016510651000065100065按第二列展开1?000651006510065?5100065100650006
6?5?6??65?1080??1145 5315