第一章作业讲解(1).pdfVIP免费

第一章作业选讲（A）1（5）、2222222222111a()()()100100=a-a()()a11100()()a10()()()bcbacacbabcbcabacaabacaabacabaVacabandcemonacabebabdc行列式1（6）、22222(y)y2(y)2(y)12(y)1112(y)002(y)()2(y)2(y)xyxyxyxyxyxxxyxxyxyxxyyxyxxyxxyyxyxxyxyxxxyxyxxxyyx6、31122344312314426556312412342+0=2,3214651342564+2=6,aaaaaaaaaa前面带正号前面带正号10（1）、121212212121221221..........................................1...1...............1...1...0...0...nnnnknknknknknknnnkknnnkkabaaaabaaaababaaababaabaabaaabaabaabaabab111.........00...(1)nnnkkbbab10（3）、1122111211110....000...00..........000....111....1100....0000...00..........000....0123....11nnnnnkkaaaaaaaaannna10（5）、方法一((1)1123223123122313111...10...001...0...............00...1...0...0010...00...00101...01..........................................000...100...0011111nnnnnnnnnnnnaaaabDbbaaaabbbbababD3...(2))211111nnnnnnnnnnbDaDabb1nnnnDDab12112334411122334411......nnnnnnnnnnnnnnnnnDDabDababDababababaababababab方法二12323123212...10...001...0k-b...............00...1...010...0001...0...............000...1nnknnkknknkkkaaaabDbbaabaaaaab将第列的倍加到第一列11（1）、11123...123...113...013...()121...002.............123...1000...(1)nknnxnxnxkxnxnxxn1,2,3,...,1xn11（2）、210111...11111...11112...11......111...(2)1111...1(1)111...1100...00001...00(1)()......000...(3)0000...0(2)nnkxxnxnxxxxknxnx0,1,2,3,...,2xn13（1）、225500000000000000000000000000000000000000000bbbabababababababaaaabbbabbaababababaaba13（3）、110...