则Dn按第1列展开:
?1xDn?xDn?1?an(?1)n?1?10?1?1????00?x00?xDn?1?an?右边 ??1所以，对于n阶行列式命题成立.
6.设n阶行列式D?det(aij),把D上下翻转、或逆时针旋转90?、或依副对角线翻转，依次得
an1?anna1n?annann?a1nD1???，D2??? ，D3???，
a11?a1na11?an1an1?a11证明D1?D2?(?1)
证明 ?D?det(aij)
n(n?1)2D,D3?D.
an1?D1??a11a11?a1na11?a1n?anna21?a2na?annann?? ??(?1)n?1n1?(?1)n?1(?1)n?2an1???a1n??a21?a2na31?a3na11?a1nn(n?1)n?1n?21?2???(n?2)?(n?1)?(?1)(?1)?(?1)???(?1)D?(?1)2D
an1?annn(n?1)2同理可证D2?(?1)a11?an1n(n?1)n(n?1)T???(?1)2D?(?1)2D a1n?annn(n?1)2D3?(?1)
n(n?1)2D2?(?1)(?1)n(n?1)2D?(?1)n(n?1)D?D
7.计算下列各行列式（Dk为k阶行列式）：
6
a(1)Dn?1?,其中对角线上元素都是a，未写出的元素都是0；
1xa(2)Dn??aax?aa????aa; ?x?(a?n)n?(a?n)n?1??; 提示：利用范德蒙德行列式的结果． ?a?n?1(3) Dn?1an(a?1)nan?1(a?1)n?1???aa?111an?(4) D2n?0bn0a1b1c1d1?0??0; dncn(5)Dn?det(aij),其中aij?i?j;
1?a1111?a2(6)Dn???11解
?1?1,其中a1a2?an?0.
???1?ana00(1) Dn??010a0?0000a?00??????000?a0100 ?0a00a?0000?0?????000?a7
0a按最后一行展开n?1(?1)0?0
1a0 0?(?1)2n?a??a(n?1)(n?1)0(n?1)?(n?1)(再按第一行展开)
a?(?1)n?1?(?1)n?a(n?2)(n?2)(2)将第一行乘(?1)分别加到其余各行，得
?an?an?an?2?an?2(a2?1)
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
n(n?1)2Dn?1?(?1)n?1?i?j?1?[(a?i?1)?(a?j?1)] ?[?(i?j)]?(?1)n(n?1)2?(?1)n(n?1)2?(?1)n?(n?1)???1?n?1?i?j?1n?1?i?j?1?[(i?j)]
?
n?1?i?j?1?(i?j)
8
an?(4)D2n?00?a1b1?c1d10an?1?bn
?dn0?a1b1?c1d10??dn?100dncn?1cn00??(?1)2n?1cnbn?100an?1?bn0?0?a1b1c1d1?bn?10dn?10按第一行展开an0cn?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??(adii?2ni?bici)D2
而 D2?a1b1?a1d1?b1c1
c1d1得
D2n??(aidi?bici)
i?1n(5)aij?i?j
012310122101Dn?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 9
?1000?1?200c2?c1,c3?c1?1?2?20?1?2?2?2c4?c1,?????n?12n?32n?42n?5?0?0?0=(?1)n?1(n?1)2n?2
?0???n?1a100?a2a200?a3a300?a4???000000?001?001?001?001 ??????an?1an?11?0?an1?an?000?000?000?000 ??????an?2an?20?00?an?00?00?00
?????an?1an?1?0?an1?a1111?a2(6)Dn???11?1c1?c2,c2?c3?1??c3?c4,??1?ana100?a2a200?a3a3按最后一列(1?an)(a1a2?an?1)?00?a4展开（由下往上）???000000a100?a2a200?a3a3????000000?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?1n1) ai
8.用克莱姆法则解下列方程组：
?1,?5x1?6x2x?x?x?x?5,?1234?x?5x?6x?0,123?x?2x?x?4x??2,??1?234(1)?(2)?x2?5x3?6x4?0, ?2x1?3x2?x3?5x4??2,?x3?5x4?6x5?0,???3x1?x2?2x3?11x4?0;?x4?5x5?1.?
10