工程数学线性代数(同济大学第六版)课后习题答案(全)

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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201(1)141183201解1411832(4)30(1)(1)1180132(1)81(4)(1)2481644abc(2)bcacababc解bcacabacbbaccbabbbaaaccc3333abcabc111(3)abc222abc111解abc222abc222222bccaabacbacb(ab)(bc)(ca)xyxy(4)yxyxxyxyxyxy解yxyxxyxy333x(xy)yyx(xy)(xy)yxy(xy)x323333xy(xy)y3xyxyx332(xy)

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)n(n1)解逆序数为232(1个)5254(2个)727476(3个)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)(6)13(2n1)(2n)(2n2)2解逆序数为n(n1)32(1个)5254(2个)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)42(1个)626

3、4(2个)(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t(1)a11a23a3ra4s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t1(1)a11a23a32a44(1)a11a23a32a44a11a23a32a44t2(1)a11a23a34a42(1)a11a23a34a42a11a23a34a424计算下列各行列式41241202(1)1052001174124cc4121023411012021

4、20243解122(1)10520103214c7c1031401174300104110c2c399101220020110314c12c317171421413121(2)123250622141cc2140rr21404242312131223122解123212301230506250622140rr2140413122012300000abacae(3)bdcddebfcfefabacaebce解bdcddeadfbcebfcfefbce111adfbce1114abcdef111a1001b10(4)01c1001da100

5、rar01aba0121b101b10解01c101c1001d001d1aba0c3dc21abaad21(1)(1)1c11c1cd01d010321abad(1)(1)abcdabcdad111cd5证明:22aabb3(1)2aab2b(ab);111证明a2abb22222c2c1aababa2aab2b2aba2b2a111c3c110022231ababa(ba)(ba)aba(ab)3(1)ba2b2a12axbyaybzazbxxyz33(2)aybzazbxaxby(ab)yzx;azbxaxbyaybzzxy证明a

6、xbyaybzazbxaybzazbxaxbyazbxaxbyaybzxaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbx22ayazbxxbzxaxbyzaxbyyxyaybzxyzyzx33ayzxbzxyzxyxyzxyzxyz33ayzxbyzxzxyzxyxyz33(ab)yzxzxy2222a(a1)(a2)(a3)2222b(b1)(b2)(b3)(3)22220;c(c1)(c2)(c3)2222d(d1)(d2)(d3)证明2222a(

7、a1)(a2)(a3)2222b(b1)(b2)(b3)2222(c4c3c3c2c2c1得)c(c1)(c2)(c3)2222d(d1)(d2)(d3)2a2a12a32a52b2b12b32b52(c4c3c3c2得)c2c12c32c52d2d12d32d52a2a1222b2b12220c2c1222d2d1221111abcd(4)2222abcd4444abcd(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明1111abcd2222abcd4444abcd11110bacada0b(ba)c(ca)d(da

8、)2222222220b(ba)c(ca)d(da)111(ba)(ca)(da)bcd222b(ba)c(ca)d(da)111(ba)(ca)(da)0cbdb0c(cb)(cba)d(db)(dba)