线性代数(复旦大学出版社周勇)课后习题答案(一)

第一章课后 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 一、1.；5)1(12221122.；1)1)(1(111232222xxxxxxxxxx3.baabbaba22224.536158273255984131115.比例）第一行与第三行对应成(,000000dcba6.。186662781132213321二．求逆序数1.55124300122即2.521340023即3.2)1(12)2()1(12)1(01)2()1(nnnnnnnn即4.2)1(*2]12)2()1[()]1(21[24)22()2()12(310121110nnnnnnnnnnn三．四阶行列式中含有的项为2311aa4234231144322311aaaaaaaa四．计算行列式值1.071108517002021459001577110202150202142701047110025102021421443412321rrrrrrrr2.31000010000101111301111011110111113011310131103111301111011110111104321cccc3.abcdefadfbceefcfbfdecdbdaeacab41111111114.dcdcbadcba1010011101101100110011001按第一行展开adcdabdcdadcab)1)(1(11011115.baccbcabaacbaccbcabaabbacccbcabbaaabacccbcabbaacba202022202022222222222222其中)3)(()(3522)(22)(12221222122)(2202022202022222220222200222202222222222222acabacabaababcbaccaacabbaabaabcbaccaacabccbbaaaccbbaacccbbbaaabaccbcbaabaccbabaabacccbbbaaabacccbcabbaaa其余同法可求。方法2:cccabbabaccbbabaccbcabcbabacccbcabbaacba2222222222)(2222226.27011203840553004222321502321353140422141312rrrrrr7.)!2(22000010022220001)2(2222322222222212nnirrni8.221111)1()1(00010000000000)1(000000000000001000000100nnnnnnnnnaaaaaaaaaaaaaaa按第一行展开五．证明下列等式1.322322232122)(111)(20)(0211122)(011122baababababaarrbbaaababararbbaababa2.222222222222122222222222222222)3()2(12)3()2(12)3()2(12)3()2(12)3()2()1()3()2()1()3()2()1()3()2()1(ddddccccbbbbaaaaccddddccccbbbbaaaa03212321232123212229644129644129644129644122222242322221413ddccbbaaccccddddccccbbbbaaaacccc3.nnnnnnnnnnnnnnnnaxDaxDaxaaaaxxxaxaaaaxxxxxaxaaaaxxxxD121111232123211221)1()1(1000000000100001010000000001000011000000000100001按第一行展开同理，返回代入得11kkkaxDDnnnnnnnnnnaxaxaxaaxDxaxDD111121)(六、计算下列各式1.设为方程的三个根，321,,xxx03qpxx则由三次方程根的性质023dcxbxax,/321abxxx，adxxx/321acxxxxxx/313221得的三个根满足：03qpxx,0321xxxqxxx321所以。033)(3321321321333231132213321xxxqxxxpxxxxxxxxxxxxxxx2.中的系数为-1.xxxxxxf21123232101)(3x3.求的的值，cdbaacbdadbcdcbaD44342414AAAA0111144342414dbacbddbccbaAAAA4.求的的值，xaaaxaaaxDnnnnAAA21121)(1111000000000)(1111nninnnnaxaxaxaxniarraxaaaaxaaaaxAAA七.计算行列式值1.1121121212121212121))((0000),,2(nniinniiinniinniinniinjjnnnnmmxmmxxmxnirrmxxmxxmxmxxxmxccmxxxxmxxxxmxD2.1100000110000111321!11100000220000111321nnnnnnnDn）（各行提出公因子2)!1()1(!12)1()1()1(!110000001000001012!1)1,,1(11113211nnnnknnnkkknnjccnnnknnknknkjj）（）（）（3.121212200)1(00nnnncdcdcbababddcdcbabaadcdcbabaD按第一行展开,同理，…，，2222nnbcDadD4222)(nnDbcadDbcaddcbaD2所以nnbcadD)(24.nnninnaaaaaaaaaaaanirraaaD0110001110111),,2(111111111121121211121八．克莱姆法则解方程1.321202zyxzyxzyx解：8211112121D42131111201D42311121012D123111120213D即；2/3,2/1,2/1zyx2.33713344324324214324321xxxxxxxxxxxxx解：，161370103111104321D12813731031111343241D4813301011113043412D9613701131131044213D，033701031311043214D0,6,3,84321xxxx九．参数取值使有非零解，,0200321321321xxxxxxxxx当即或时，方程组有非零解。0)1(1211111D01