四川大学高数第三册

第1章练习题（3）1.计算下列排列的反序数，从而判断奇偶性。2)1(12)2()1(321)1(nnnnnn（4）2)1()1(210)2(246)12(135nnnnn解：对于排列中的数字，设排列中有个小于它的数字，设这些小于它的数字中，位于其右边的有个，则位于其左的有个。2.已知排列的反序数，求的反序数。niii2111iiinn)(jirji)(jil则：njjnjjnjjjnniriliriliii11111)()()()()(niii21)()(jjiril)(2)1()(2111nnniiinniii对于任意n个不相等的自然数，其中最大的数字有n-1个小于它的，次大的数字有n-2个小于它的，……因此，2)1(01)2()1()(1nnnnilnjj解：四阶行列式中的项为5.写出四阶行列式中含因子且带负号的项。23a43214321jjjjaaaa含因子时，令23a32j是数字1、2、3、4的组合。4321jjjj则可能的组合有：1324，1342，2314，2341，4312，4321其中奇排列为：1324，2341，4312则含因子且带负号的项为：4321jjjj423123144134231244322311,,aaaaaaaaaaaa23a分析，无论如何组合，在中都至少有一个数字≥3，使得中出现，使得因此该行列式的值为0.（2）6.利用行列式的定义计算51543215154321)(52514241323125242322211514131211)1(000000000jjjjjjjjjaaaaaaaaaaaaaaaaaaaaa5432154321jjjjjaaaaa54321jjjjj543jjj5432154321jjjjjaaaaa)3,3(jiaij05432154321jjjjjaaaaa（4）6.利用行列式的定义计算51543215154321)()1(000000000000000jjjjjjjjjaaaaaxyyxyxyxyx其中非0项为：555145342312)23451(5544332211)12345()1()1(yxaaaaaaaaaa（3）8.利用行列式的性质计算00000111212111211222111324214bacacbcbabaaccbbacacbcbabaaccbbacacbcbarrrr（1）9.不展开行列式，证明下列等式成立。'''''''''2''''''''''''''''''cbacbacbabaaccbbaaccbbaaccb证明：右边左边'''''''''''''''2'''''''''''''''2'''''''''''''''''''''213123212)(cbcbacbcbacbcbacbcbacbcbacbcbabaaccbabaaccbabaaccbaccccccc（2）02coscossin2coscossin2coscossin222222证明：右边左边0coscossincoscossincoscossinsin-coscossinsin-coscossinsin-coscossin22222222222222222222213cc（3）)0(,01010111100000222222xyzxyxzyzxyzxzyyzxzyx证明：000111010101011110)(2222222432432432xyxyzxzxyzyzxyzxyzyzxxzyyzxxyzxzyxyzxyzxyzrxyzrxyzrzryrxrzcycxc左边右边01010111102222221xyxzyzxyzc（1）10.计算行列式。dcbacbabaadcbacbabaadcbacbabaadcba3610363234232解：cbabaacbabaacbabaadcbaiirri36302320012,3,4cbabaacbabaacbabaaa363232按第一列展开baabaacbabaaaiirri302012,3baabaaa322按第一列展开原式4a（2）解：0303322021111nnn------!2203630022000113,2nnnnnccnii原式（3）nnnnnnnnnnnxxxxxaxxxxaaxxxaaaxxaaaax1321)1(1321313321212232111113121解：nnnnnnnnnnnnnnnnnnnnrrnniaxaaaxaaaaaxaaaaaaaxaaaaxii)1()2()1()1)(2(12312132331211121323122111131212,1,00000000001)())(()1(2331221nnnaxaxaxx（4）nnnnnnbababababababababa212221212111解：202)()(00000000012121131213121113,2111113131313121212121312111,3,211nnbbaaaaaaaabbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaababababannccninnnnnrrniii原式（5）nnnnnnyxyxyxyxyxyxyxyxyx111111111212221212111解：原式BAyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxnnnnnnnnnnn1111111111111112122212121112222121202111212222221213,21nnyxyxyxyxyxyxyxyxAnnnnnccnii∴原式302)(1111111111112112113,222222121111nnxxyxxxyyxyxxyxyxxyxyxxyBncycninnnnnnii302))((2121nnyyxxBA（6）mxxxxmxxxxmxnnn212121解：mxmmmmxmxxxmxxxmxniinniixccninnnniiccciin1113,22221)(0101001111121原式11.利用行列式的性质求方程：1,01111112111112111111111111nxnxnxx解：左边0)2)(3()1)((20000030000010000001111113,2xnxnxxxnxnxxrrnii则方程的根为22,1,0nx12.计算下列n阶行列式。xyyxxyxyx0000000000000000解：（1）yxyyxyyxyxxyxxn0000000000)1(00000000001按第一列展开原式nnnyx1)1(nnnnn110000200000220000111321解：（2）nnnnnnnnccc110000200000220000101322)1(21原式)!1(2)1()!1()1(2)1(11000200002-200012)1(111nnnnnnnnnnn列按第展开121212121111222111111nnnnnananaaaaaaaaaa（3）111122221211211211111nnnnnaaaanaaaanaaaa转置原式)2()1()1()1()1()2()1()2()1(nanaanaaaanaaaaa范德蒙行列式!1)!2()!1()1(!1)1()!2()1()!1()1(2)1(121nnnnnnnn)0(,121111122111212122222121111212111121innnnnnnnnnnnnnnnnnnnnnnabbbbababababababababaaaa（4）nnnnnnnnnniniarniabababababababababanii1121111222222211211111112,1111原式nnnnnnnnniniababababababababababababa112211223311111133112211范德蒙行列式)()()()())((1112213223111131132112nnnnnnnnabababababababababababab13.证明下列等式)sin()sin()sin(212cos2sin2cos2cos2sin2cos2cos2sin2cos（1）2cos2sin2cos2cos2sin2cos2cos2sin2cos2cos2cos2sin2cos2cos2sin2cos2cos2sin左边)sin()sin()sin(412cos2sin2cos22sin212cos2sin22sin212cos2cos2sin其中其中)sin()sin()sin(412cos2sin2cos22sin212cos2sin22sin212cos2sin2cos)sin()sin()sin(21)sin()sin()sin(41)sin()sin()sin(41)sin()si