线性代数课后习题答案总主编邹庭荣主编李仁所张洪谦第1章_行列式习题解

1习题11-1．计算下列行列式解一由三阶行列式定义得71350116330765311110335161709010154234.解二2331123361105105105361056317317018rrrrrr23325105105018018340560034rrrr.（2）解213241120112011201135001510151015601560007123400330033rrrrrr34120101512100330007rr.（3）dcba100110011001.2解342312100011101010110101001001rcrrbrraraaababcdbbbcdccddd2111(1)(1)1011010101aababcdaababcdcdcddd1221011101(1)(1)1011.rraabcdabcdaabcdabcdcdddabadcdabcd（4）2010411063143211111.解43433232211111111111111234012301231361001360013141020014100014rrrrrrrrrr4311110123100130001rr.（5）49362516362516925169416941.解43433232211491614916149164916253579357909162536579112222162536497911132222rrrrrrrrrr.3（6）222111abcabc.解222111()()()abccbcabaabc.1-2．计算行列式abcdbadccdabdcba.解12341111()rrrrabcdbadcbadcabcdcdabcdabdcbadcba4132211000()ccccccbabdacbabcdcdcadbcdcdbcad()abdacbabcddcadbccdbcad3221()000rrrrabdacbabcdabcdabcdabcd21()()(1)dacbabcdabcdabcdabcd()()()()()()()()().abcdabcdabcddacbabcdabcdabcdabcd1-3．计算n阶行列式4（1）n321332122211111.解1122111111111122201111123300111230001nnnnrrrrrrn.（2）14321432113213121321nnnnnnnn.解12123112312131113123111311(1)22341134123411341ncccnnnnnnnnnnnnnnnnnn2131112310100001200(1)20112001111nrrrrrrnnnnnn.5110001200(1)113021111(1)!(1).2nnnnn（3）21111121111211112．解21111111112111021111211012111111210112nD,按第一列展开成两个行列式得1111111112110211112101211111120112nD213111121221221111032200320003333333nnrrrrrrnnnnnnnnDDDD12212221333333512nnnn612213313333111132nnnn.1-4．证明：（1）322)(11122babbaababa.证322122222()()222()211111100100ccccaabbaababbaaabaabbabababaa左32222(321)3()210()(1)()100()ccaababaababaab右.（2）2221112222221111112cbacbacbabaaccbbaaccbbaaccb.证11111111111111112222222222222222bccaabbcaabccaabbccaabbcaabccaabbccaabbcaabccaab左=1111111122222222bcaacaabbcaacaabbcaacaab111111222222bcacabbcacabbcacab1112222abcabcabc右.7（3）321321321332321332321332321cccbbbaaacmcclckccbmbblbkbbamaalakaa.证1323123233122312323312231232331223clccmcakalaamaaakaaabkblbbmbbbkbbbckclccmccckccc左=12123123123ckcaaabbbccc右.（4）222244441111abcdabcdabcd()()()()()()()bacadacbdbdcabcd.证243322122224444222222222111111110=0()()()0()()()rarrarrarabcdbacadaabcdbbaccaddaabcdbbaccadda左222222222()()()()()()bacadabbaccaddabbaccadda222111()()()()()()bacadabcdbbaccadda21222111()()()()()()rarbacadabacadabbaccadda823121()2222111()()()00()()()()rbrrbarbacadacbdbcbcadbda2222()()()()()()()cbdbbacadacbcadbda222211()()()()()()()()()()()()()()()()()()()()()()()()()()()()(bacadacbdbcbcadbdabacadacbdbdbdacbcabacadacbdbdadbdabcacbcabbacadacbdbdadbdcacbcba)()()()()()()cadacbdbdcabcd右.1-5．计算行列式xyyxyxyx000000000000.解记000000000000nxyxyDxyyx,当1n时，1Dx;当2n时，按第1列展开得000000000000000000nxyxyxyxyDxxyxyx9100000(1)00000nyxyyyxy1(1)nnnxy.1-6．计算4阶行列式（1）2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(ddddccccbbbbaaaa.解21314122222222222222222222(1)(2)(3)212223(1)(2)(3)212223(1)(2)(3)212223(1)(2)(3)212223ccccccaaaaaaaabbbbbbbbccccccccdddddddd4332222221112111021112111ccccaabbccdd.（2）1122334400000000ababbaba.解112222221114133133334444000000(1)0(1)000000000ababababababbabaabba2222333114143333(1)(1)ababaabbbaba142323142323aaaabbbbaabb14142323()()aabbaabb.1-7．如果行列式10nnnnnnaaaaaaaaa212222111211，试用表示行列式nnnnnnnaaaaaaaaaaaa11211213323122221的值．解112212122211121313232122211121211121(1)(1)nnnnrrnrrnnrrnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaaa.1-8．证明：nnnnaaaaaa212)1(21)1(000000.证1(1)2((1)21)212120000(1)(1)00nnnnnnnaaaaaaaaa.1-9.利用克莱姆法则解线性方程组067452296385243214324214321xxxxxxxxxxxxxx.解方程组的系数行列式112151130627002121476D，由克莱姆法则知，方程组有惟一解.进一步计算，有1815193068152120476D，22851190610805121076D，3218113962702521406D，4215813092702151470D,方程组的解为12343,4,1,1xxxx.1-10.问取何值时，齐次线性方程组可能有非零解？121200xxxx解方程组的系数行列式211(1)(1)1D,当1或1时，0D，方程组可能有非零解．补充题B1-1．计算行列式1111321321121121nnnaaaaxaaaaaxaaaax.12解12112111231231111nnnnxaaaaxaaDaaaxaaaa111221111223123123110100()000100001nnnnnnncaccacnnncaciinxaaaaaaaaaaaxaxaxaxa.B1-2．计算行列式nnnnnaaabaaabaaabaaa22222111111111.解111111222221111nnnnnnaaabaDaabaaabaaa21132111121111000000000nnrarrarrarnbbb(3)((1)21)21212(1)1()()()(1)nnnnnnbbbbbb.B1-3．计算行列式13xaaaaaxaaaaann21020210.解012021012nnnaaaaaxaaDaaax213111012101000()000nrrrrnnrriinaaaaxaaxaxa.