xay?bzzy分别再分2ayaz?bxx?0?0?bzzax?byyxzaz?bxxax?by yay?bz分别再分xyzyzxa3yzx?b3zxy zxyxyzxyzxyz?a3yzx?b3yzx(?1)2?右边 zxyzxya2a2?(2a?1)(a?2)2(a?3)2b2左边?b2?(2b?1)(b?2)2(b?3)2c2c2?(2c?1)(c?2)2(c?3)2 d2d2?(2d?1)(d?2)2(d?3)2a22a?14a?46a?9c22?c1b2b?14b?46b?9c23?c1c2c?14c?46c?9 c4?c1d22d?14d?46d?9a2a4a?46a?9a214a?4按第二列b4b?46b?914b?4分成二项2b2c2c4c?46c?9?b2c214c?4d2d4d?46d?9d214d?4第一项c3?4c2a2a49a214a6ac4?6c2b2b49b2第二项c3?4c2c2c49?14b6bc214c6c?0c4?9c2d2d49d214d6d1000左边?ab?ac?ad?aa2b2?a2c2?a2d2?a2 a4b4?a4c4?a4d4?a4b?ac?ad?a=b2?a2c2?a2d2?a2 b2(b2?a2)c2(c2?a2)d2(d2?a2)6a?96b?96c?96d?96
(3) (4)
111c?ad?a =(b?a)(c?a)(d?a)b?ab2(b?a)c2(c?a)d2(d?a)=(b?a)(c?a)(d?a)?
100 b?ac?bd?bb2(b?a)c2(c?a)?b2(b?a)d2(d?a)?b2(b?a)=(b?a)(c?a)(d?a)(c?b)(d?b)?
11 2222(c?bc?b)?a(c?b)(d?bd?b)?a(d?b)=(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d)
(5) 用数学归纳法证明
x?1当n?1时,D1??a1x?a0,命题成立.
a0a1假设对于n阶行列式命题成立，即
Dn?1?an?1xn?1?an?2xn?2???a1x?a0,
则Dn按第n+1列展开:
x?1?000x?00Dn?(?1)(?1)2n?1Dn?1?an(?1)2n?2?????00?0x
?Dn?1?anxn?右边
所以，对于n?1阶行列式命题成立.
7.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?annD1??a11a1n?annann?a1n?， D2??? ，D3???， ?a1na11?an1an1?a11n(n?1)2D,D3?D. 证明D1?D2?(?1)证明 ?D?det(aij)
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an1?ann?D1????(?1)n?1a11?a1nan1?ann??a21?a2na11?a1n
a11?a1na21?a2nann?? ?(?1)n?1(?1)n?2an1??a31?a3na11?a1n?(?1)n?1(?1)n?2?(?1)??
an1?ann?(?1)1?2???(n?2)?(n?1)D?(?1)a?an1n(n?1)11同理可证D2?(?1)2n(n?1)2D
DT?(?1)n(n?1)2???(?1)a1n?annD2?(?1)n(n?1)2n(n?1)2D
D3?(?1)
n(n?1)2(?1)n(n?1)2D?(?1)n(n?1)D?D
8.计算下列各行列式（Dk为k阶行列式）：
a(1)Dn?1?,其中对角线上元素都是a，未写出的元素都是0；
1axa?aax?a(2)Dn?;
????aa?x 8
an(a?1)n?(a?n)nan?1(a?1)n?1?(a?n)n?1(3) Dn?1?????; aa?1?a?n11?1提示：利用范德蒙德行列式的结果．
anbn?0?(4) Da1b12n?0c1d0;
1?0?cndn(5)Dn?det(aij),其中aij?i?j;
1?a11?1(6)D11?a2?1n?????,其中a1a2?an?0.
11?1?an解
a00?010a0?00(1) D00a?00按最后一行展开n???????
000?a0100?0a000?01a00?00a(?1)n?10a0?00?(?1)2n?a???????000?a0(n?1)?(n?1)(再按第一行展开)
a(n?1)(n?1)9
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)
xa?xDn?a?x?a?xax?a0?0a0??a00 ?x?ax?a???00再将各列都加到第一列上，得
x?(n?1)a0Dn?0?0ax?a0?0n?1a0?0
???0a00?x?a
x?a??[x?(n?1)a](x?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?交换，得
n(n?1)次行 21Dn?1?(?1)n(n?1)21a?1??1
a?an?1an?a?n??(a?1)n?1?(a?n)n?1(a?1)n?(a?n)n此行列式为范德蒙德行列式
Dn?1?(?1)n(n?1)2n?1?i?j?1?[(a?i?1)?(a?j?1)]
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