n(n?1)?1)?(?1)22?(?1)n?(n?1)???1?n?1?n(n[?(i?j)]?(?1)?i?j?1n?1?[(i?j)]
?i?j?1?n?1?(i?j)
?i?j?1
an0bn??a1b(4) D12n?0c1d
1??cn0dnan?10bn?10??按第一行a1b1展开a0nc1d0?1
??cn?10dn?100?0dn0an?10bn?1??a1b1?(?1)2n?1b0c0n1d1
??cn?1dn?1cn00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??(aidi?bici)D2
i?2 11
而 Da1b12?cb1c1
1d?a1d1?1n得 D2n??(aidi?bici)
i?1(5)aij?i?j
0123?n?11012?n?2D?det(a2101?n?3nij)?3210?n?4
??????n?1n?2n?3n?4?0?1111?1?1?111?1r1?r2?1?1?11?1c2?c1,c3?c1r2?r3,??1?1?1?1?1c
4?c1,???????n?1n?2n?3n?4?0?1000?0?1?200?0?1?2?20?0?1?2?2?2?0=(?1)n?1(n?1)2n?2??????n?12n?32n?42n?5?n?11?a11?1(6)D11?a2?1c1?c2,c2?c3n?????cc
3?4,?11?1?an 12
a1?a2000a2?a3000a3???000000001111按最后一列
展开（由下往上）?a4????????000??an?1an?11000?0?an1?ana100?00?a2a20?000?a3a3?00(1?an)(a1a2?an?1)?00?a4?00??????000??an?2an?2000?00a100?00?a2a20?00?0?a3a3?00?????????
000??an?1an?1000?0?an?a2a20?000?a3a3?0000?a4?00??????
000??an?1an?1000?0?an?(1?an)(a1a2?an?1)?a1a2?an?3an?2an???a2a3?an
?(an11a2?an)(1??)
i?1ai0000 ?0?an13
3?59.设D=
21110?5?13132?4，D的(i,j)元的代数余子式记作Aij，求?1?3A31+3A32-2A33+2A34 .
3?5解：A31+3A32-2A33+2A34=
11113?5?123?4?223?3c4?c3r2?r13?1211113?5?1110 ?2030?111?111r?r按最后一列按第一列21?2?404?1?213?2r3?r1展开展开0?111?534?1c1?c23?1=8?3?24 ?8?(?1)8?110110.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,??3x1?x2?2x3?11x4?0;?1,?5x1?6x2??0,?x1?5x2?6x3?(2)?x2?5x3?6x4?0, ?x3?5x4?6x5?0,??x4?5x5?1.?111112?1401?23解 (1)D? ?2?3?1?50?5?3?7312110?2?181111 14
11111111?01?2301?2300?138?00?1?54??142
00?51400014251115111D?22?1405091??2?3?1?5??2?3?1?5 01211012111?5?1?91?5?1?9?0509?012110?13?3?230509
012110?13?3?231?5?1?91?5?1?9?01211?0121100?10?4600?138??142002312000014215111511D1?2?140?7?232?2?2?1?5?0?12?3?7 302110?15?1815111511?0?1320?132002311?00?1?19??284
003931000?2841151D12?243?2?3?2?5??426
3101115