工程数学线性代数同济大学第五版课后习题答案.doc

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1、第一章行列式1利用对角线法则计算下列三阶行列式(1)201141183201解1411832(4)30(1)(1)1180132(1)81(4)(1)2481644abc(2)bcacababc解bcacabacbbaccbabbbaaaccc3abca3b3c3(3)解111abca2b2c2111abca2b2c2bc2ca2ab2ac2ba2cb2(ab)(bc)(ca)(4)解xyxyyxyxxyxyxyxyyxyxxyxyx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2yx3y3x32(x3y3)2

2、按自然数从小到大为标准次序求下列各排列的逆序数(1)1234解逆序数为0(2)4132解逆序数为441434232(3)3421解逆序数为532314241,21(4)2413解逆序数为3214143(5)13(2n1)24(2n)解逆序数为n(n1)232(1个)5254(2个)727476(3个)解逆序数为n(n1)32(1个)5254(2个)(2n1)2(2n1)4(2n1)642(1个)(2n1)(2n2)(n1个)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)(6)13(2n1)(2n)(2n2)26264(2个)

3、(2n)2(2n)4(2n)6(2n)(2n2)(n1个)3写出四阶行列式中含有因子a11a23的项解含因子a11a23的项的一般形式为(1)ta11a23a3ra4s其中rs是2和4构成的排列这种排列共有两个即24和42所以含因子a11a23的项分别是(1)ta11a23a32a44(1)1a11a23a32a44a11a23a32a44(1)ta11a23a34a42(1)2a11a23a34a42a11a23a34a4241241202052001174计算下列各行列式(1)14124105200117解1202cc4121232000

4、1c47c310321041212014103102(1)4314411012210314c2c3cc11239910002017171421(2)3514112123206231214132解12215062c4c22315140122230062r4r22312140122230140123000003r4r121401220abacae(3)bdcddebfcfefab解bdbfacaecddecfef11bceadfbcebce1adfbce11114abcdef11a10001c1(4)1b10001d1ba1

5、解010000r110c11dar201aba01b1001c1001d(1)(1ab1)2110a0c3c11ddc21ab10aadc1cd10(5证明:1)(1)321abad11cdabcdabcdad1a2abb2(1)2a1证明ab2b11(ab)3;a2abb2c2c1a2aba2b2a22aab112b1c32abac1102b2a031aba2b2a2aba3(1)ba2b2a(ba)(ba)12(ab)ax(2)ayaz证明byaybzazbxaxbzazbxbxaxby

6、byaybz(a3xb3)yzyzzx;xyaxbyayaybzazazbxaxbzazbxbxaxbybyaybzxayayazzaxxaybzazbxaxbyaybzzbxyaybzazbxbybzazbxaxbybzxaxbyaybzyzazbxa2yazzaxbxxbyyb2zxxaxbyyaybzxyza3yzxzxyxyza3yzxzxyyzxb3zxyxyzxyzb3yzxzxy(a3xyzb3)yzxzxy(a1)2(a2)2(a3)2(b1)2(b2)2(b3)2(c1)2(c2)2(c3)2(d1)2(

7、d2)2(d3)2a2(3)b20;c2d2证明(a1)2(a2)2(a3)2(b1)2(b2)2(b3)2(c1)2(c2)2(c3)2(d1)2(d2)2(d3)2a2b2c2d2a22a12a32a5(c4c3c3c2c2c1得)b22b12b32b5(cccc得)c22cd22d2a1222b1222c1222d122a2b212c12d32c5432d5332c20d21111a2b2c2d2(4)abcda4b4c4d4(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明11aba2b2a4b

8、41011cdc2d2c4d4111bacada0b(b0b2(b2a)a2)c(cc2(c2a)a2)1d(dd2(d2a)a2)11(