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1、同济大学线性代数第六版答案(全)第一章队列式1 利用对角线法例计算以下三阶队列式201(1)141183201解1411832(4)30(1)(1)1180132(1)81(4)(1)2481644abc(2) bcacababc解bcacabacbbaccbabbbaaaccc3abca3b3?c3111(3) abca2b2c2111解abca2b2c2bc2ca2ab2?ac2ba2cb2(ab)(bc)(ca)xyxy(4)yxyxxyxyxyxy解yxyxxyxyx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2yx3y3x32(x3y3)2按自然数从小到
2、大为标准序次求以下各摆列的逆序数(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)(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)6264(2个)(2n)2(2n)4(2n)6(2n)(2n2)(n1个)3 写
3、出四阶队列式中含有因子a11a23的项解含因子a11a23的项的一般形式为(1)ta11a23a3ra4s此中rs是2和4组成的摆列这类摆列共有两个即24和42所以含因子a11a23的项分别是(1)ta11a23a32a44(1)1a11a23a32a44?a11a23a32a44(1)ta11a23a34a42(1)2a11a23a34a42?a11a23a34a424 计算以下各队列式4 124(1) 12021052001174124cc412104110解1202232021122(1)4310520c41032141031401177c300104110c2c399100122c11
4、2c3002103141717142141(2) 3121123250622141cc2140rr2140解3121423122423122123212301230506250622140rr2140413122012300000(3)abacaebdcddebfcfef解abacaebcebdcddeadfbcebfcfefbce1114abcdefadfbce111111a100(4)1b1001c1001da100rar01aba0解1b10121b1001c101c1001d001d(1)(1aba0c3dc21abaad1)211c11c1cd01d010(1)(1)321abada
5、bcdabcdad111cd5证明:a2abb2(1) 2aab2b(ab)3;111证明a2abb2c2c1a2aba2b2a22aab2b2aba2b2a111c3c1100(1)31aba2b2a2(ba)(ba)aba(ab)3ba2b2a12axbyaybzazbxxyz(2)aybzazbxaxby(a3b3)yzx;azbxaxbyaybzzxy证明axbyaybzazbxaybzazbxaxbyazbxaxbyaybzxaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbxa2yazbxxb2zxa
6、xbyzaxbyyxyaybzxyzyzxa3yzxb3zxyzxyxyzxyzxyza3yzxb3yzxzxyzxyxyz(a3b3)yzxzxya2(a1)2(a2)2(a3)2(3)b2(b1)2(b2)2(b3)20;c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2证明a2(a1)2(a2)2(a3)2b2(b1)2(b2)2(b3)2(c4?c3?c3?c2?c2?c1得)c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2a22a12a32a5b22b12b32b5(c4?c3?c3?c2得)c22c12c32c5d22d12d32d5a22a1
7、22b22b1220c22c122d22d1221111abcd(4)a2b2c2d2a4b4c4d4(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明1111abcda2b2c2d2a4b4c4d411a1a1a0bcd0b(ba)c(ca)d(da)0b2(b2a2)c2(c2a2)d2(d2a2)(ba)(ca)(da)111bcda)b2(ba)c2(ca)d2(d(ba)(ca)(d1c1d1a)0ba)d(dbba)0c(cb)(cbb)(d(ba)(ca)(da)(cb)(db)c(c11a)ba)d(db=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)x10000x100xna1xn1an1xan(5)00x10anan1an2a2xa1证明用数学概括法证明当n2时D2x1x2a1xa2命题建立a2xa1假定对于(n1)阶队列式命题建立即
