?a10 (4)011000?b01?c11
da1000r1?ar201?aba10 解 ?0100?b110?c11??????d01b0?011 ?c1d ?(?1)(?1)2?11??aba0c3?dc 21?abaad01?c11d??????01?c11?0cd
?(?1)(?1)3?21??ab11?adcdabcdabcdad 5 证明:
2a2ab1aa?b2 (1)1b21b(ab)3
;
证明
a2 21aaab?1bb221bc?c?2?c???12a2aabb??aa22b2b??a2 3?c11002a
?(?1)3?1abb??aa22b2b??a22a?(b?a)(b?a)a1b?2a
(2)axay?byay?bzaz?bxxyzaz??bzbxazax??bxbyaxay??bybz?(a3?b3)yzzxx;
y 证明
axay?byay?bzaz?bxaz??bzbxazax??bxbyaxay??bybz
(ab)31
xay?bzaz?bxyay?bzaz?bx ?ayaz?bxax?by?bzaz?bxax?by
zax?byay?bzxax?byay?bz ?axay?bzzyzaz?bx2yzazax??bxbyxy?b2zxxyaxay??by
bz ?axyzyzx3yzzxxy?b3zxxyy
z ?axyyzzxyz3xxy?b3yzzxx
zy?(axyz3?b3)y
zzxxya2(a?1)2(a?2)2(a?3)2b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2?0; d2(d?1)2(d?2)2(d?3)2证明 a2(a?1)2(a?2)2(a?3)2b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2(c4
c3d2(d?1)2(d?2)2(d?3)2a222?b22ac2b??112ab??3322ab??55(d222dc??1122dc??3322dc?c4c3?55 c3c23c2得)2c1得)
(3) c
c a22b ?c2d2
1a (4)a2a41bb2b42a?12b?12c?12d?11cc2c4222222?022
1d d2d4 (ab)(ac)(ad)(bc)(bd)(cd)(abcd);
1bb2b41cc2c41d d2d4 证明 1a a2a411110b?ac?ad?a? 0b(b?a)c(c?a)d(d?a) 0b2(b2?a2)c2(c2?a2)d2(d2?a2)111cd ?(b?a)(c?a)(d?a)b
222b(b?a)c(c?a)d(d?a)111c?bd?b ?(b?a)(c?a)(d?a)0
0c(c?b)(c?b?a)d(d?b)(d?b?a)1 ?(b?a)(c?a)(d?a)(c?b)(d?b)c(c?1b?a)d(d?b?a)
=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)
x0 (5)? ? ?0an
?1x? ? ?0an?10?1? ? ?0an?2? ? ?? ? ?? ? ?? ? ?? ? ?0000?? ? ?x?1a2x?a1xna1xn1
an1xan
证明 用数学归纳法证明 当n2时
命题成立
x?1?x2?ax?a D2?a122x?a1 假设对于(n1)阶行列式命题成立 Dn1
即
xn1
a1 xn 有
2
an2xan00 ? ? ? ?1
1
则Dn按第一列展开
?1 Dn?xDn?1?an(?1)n?1 ? x? ? 1 因此
0?1 ? ? ? 11
? ? ? ? ? ? ? ? ? ? ? ?
00 ? ? ? x
xD n1
anxna1xnan1xan
对于n阶行列式命题成立
6 设n阶行列式Ddet(aij), 把D上下翻转、或逆时针旋转90、或依副对角线翻转an1? ? ?ann D1?? ? ?? ? ?? ? ?a11? ? ?a1n 依次得
ann? ? ?a1n D3?? ? ?? ? ?? ? ?an1? ? ?a11a1n? ? ?ann D2?? ? ?? ? ?? ? ?
a11? ? ?an1
证明D1?D2?(?1)n(n?1)2D D3
D
证明 因为Ddet(aij) 所以
