a
?(?1)n?1?(?1)n?a(n?2)(n?2)?an?an?an?2?an?2(a2?1)
(2)将第一行乘(?1)分别加到其余各行，得
xaaa?xx?a0Dn?a?x0x?a???a?x00再将各列都加到第一列上，得
?a?0?0 ??0x?ax?(n?1)aaa0x?a0Dn?00x?a???000?a?0?0?[x?(n?1)a](x?a)n?1 ??0x?a(3) 从第n?1行开始，第n?1行经过n次相邻对换，换到第1行，第n行经(n?1)次对换换到第2行…，
经n?(n?1)???1?n(n?1)次行交换，得 2Dn?111a?1n(n?1)a?(?1)2??an?1(a?1)n?1an(a?1)n?1?a?n?? ?(a?n)n?1?(a?n)n此行列式为范德蒙德行列式
Dn?1?(?1)?(?1)n(n?1)2n?1?i?j?1n(n?1)2?[(a?i?1)?(a?j?1)]
n(n?1)2n?1?i?j?1?[?(i?j)]?(?1)?(?1)n?(n?1)???12?n?1?i?j?1?[(i?j)]
?
n?1?i?j?1?(i?j)
an?(4) D2n?00?a1b1?c1d10?bn
cndn6
an?1?按第一行展开0?a1b1?c1d10??bn?100?dn?100dn2n?10an?1??(?1)bn0?cn?1cn0?a1b1c1d1?0bn?10dn?10an0cn?10
都按最后一行展开 andnD2n?2?bncnD2n?2
由此得递推公式：
D2n?(andn?bncn)D2n?2 即 D2n?而 D2??(adii?2ni?bici)D2
a1b1?a1d1?b1c1
c1d1n得 (5)aij?i?j
D2n??(aidi?bici)
i?1012310122101Dn?det(aij)?3210????n?1n?2n?3n?4??????n?1n?2n?3r1?r2n?4r2?r3,??0?1111?1?111?1?1?11?1?1?1?1????n?1n?2n?3n?4??????111 1?0?1000?1?200c2?c1,c3?c1?1?2?20?1?2?2?2c4?c1,?????n?12n?32n?42n?5?0?0?0n?1n?2=(?1)(n?1)2
?0???n?1a100?a2a200?a3a300?a4???000000?001?001?001?001 ??????an?1an?11?0?an1?an1?a1111?a2(6)Dn???11?1c1?c2,c2?c3?1??c3?c4,??1?an7
a100?a2a200?a3a3按最后一列(1?an)(a1a2?an?1)?00?a4展开（由下往上）???000000a100?a2a200?a3a3????000000n?000?000?000?000 ??????an?2an?20?00?an?00?00?00
?????an?1an?1?0?an?00?00?00?????an?1an?1?0?an????a2a200?a3a300?a4???000000?(1?an)(a1a2?an?1)?a1a2?an?3an?2an???a2a3?an ?(a1a2?an)(1??i?11) ai
8.用克莱姆法则解下列方程组：
?1,?5x1?6x2?x1?x2?x3?x4?5,?x?5x?6x?0,123?x?2x?x?4x??2,??1?234(1)?x2?5x3?6x4?0, (2)??2x1?3x2?x3?5x4??2,?x3?5x4?6x5?0,???3x1?x2?2x3?11x4?0;?x4?5x5?1.?1112解 (1)D?2?33151?22D1??2?301111111111?1401?2301?2???1?50?5?3?700?132110?2?1800?51151?1405??1?5?2?32110110?1211?5905??50?13110111111301?23???142 800?1?5414000142?1?91?50901??3?23052110?13?1?9211
09?3?231?5?1?91?5?1?90121101211??142 ??00?10?4600?138002312000014215111511151?2?140?7?230?1D2???2?2?1?50?12?3?700302110?15?180013233912?1131150?100001132??284
?1?190?2848
11511112?2412D3???426 ; D4?2?3?2?52?33101131?x1?15?1?2?142
?1?220D1DDD?1,x2?2?2,x3?3?3,x4?4??1 DDDD51(2) D?00065100065100065100按最后一行5D??0展开6551006510065100?5D??6D?? 06?5(5D???6D???)?6D???19D???30D????65D????114D?????65?19?114?5?665
?的余子式(D?为行列式D中a11的余子式,D??为D?中a11,D???,D????类推)
10D1?00151D2?00065100100010651006510006510065100按第一列D??0展开6565100651
0651006500?D??64?19D????30?????64?1507 060065
0
5600
?156?5?63 0
0156
016
0按第二列05
?001展开600
5050016?
605501
??65?1080??1145
51D3?00065100100010065100065按第三列展开100051000651056015?
60150000650
1605600
?056?6150 0
0150166
?19?6?114?703
51D4?00051D5?0006510065100065100651010001006510006510001按第四列展开10?00100051005100651065100501?60506510065105600??5?6156??395 00156按最后一列展开06?D??1?211?212 519
?x1?1507;665x2??1145;665x3?703;665x4??395;665x4?212． 665??x1?x2?x3?0?9.问?,?取何值时,齐次线性方程组?x1??x2?x3?0有非零解？
?x?2?x?x?023?111解 D3?1?1?????， 齐次线性方程组有非零解，则D3?0
12?1即 ?????0 得 ??0或??1
不难验证，当??0或??1时,该齐次线性方程组确有非零解.
??(1??)x1?2x2?4x3?0?10.问?取何值时,齐次线性方程组?2x1?(3??)x2?x3?0 有非零解？
?x?x?(1??)x?023?1解
1???241???3??4D?23??1?21??1
111??101???(1??)3?(??3)?4(1??)?2(1??)(?3??)?(1??)3?2(1??)2???3
齐次线性方程组有非零解，则D?0 得 ??0,??2或??3
不难验证，当??0,??2或??3时，该齐次线性方程组确有非零解.
10