行列式练习题1.pdf

该【行列式练习题1 】是由【小屁孩】上传分享，文档一共【8】页，该文档可以免费在线阅读，需要了解更多关于【行列式练习题1 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..第二章行列式练****题(1)一、判断题:(在括号里打“√”或“×”,每小题2分,共20分),其逆序数必增加1或减少1.(×)()12n?1nnn?121a?ba??12?12(×)a?ba?,则行列式的值一定是整数.(√),则A?B.(×),同时把第一行的3倍加到第二行上去,所得的行列式与abca?2ab?2bc?2c111121212原行列式相等即:()abc?a?3ab?3bc?,k是任意常数,则kA?kA或kA??kA;(×),则abcd与badc的奇偶性相同;(√),则该线性方程组无解;(×)=21222n,D=2k2k2k,其中是1、2、3、??、n的一个排列,112n12naaaaaan1n2nnnknknk12n????kkk?则D??112nD()1二、填空题(每小题2分,共20分)n(n?1)(n?1)321的逆序数为,当n是时为奇排列;,则54321的逆序数是。?(2n-1)246?(2n)的逆序数为,排列(2k)1(2k-1)2?(k+1)k的逆序数为;、、变为排列25341,这些对换并不唯一,但所作的对换的次数与逆?序数(12435)具有相同的奇偶性。?1a?2a?31??2300006.①03000②0bf0③b?1b?2b?3?;④??21=;?_______;00300?______;0gc0c?1c?2c?3?21?00073h00d111941:..?a0bbb1123?a??_________;2112?a?b?c0d321?a?b?c?=0113利用拉普拉斯定理按前两行展开D=;求A?A?A?A?________;111213144?10?2?38?;g(x)?123x41234x?ax?x?x?0??ax?x?0有非零解,那么=;?123?x?x?ax?0?12323121?(1)与方程(2)27?x212的全部根分别为12?x34?0?0123?x453861234?x53815?x2和(重根按重数计算);111123000111112.(1)2345(2)5862=;(3)?14000?_______;491625222218079120?________;,x4,x3的系数项和常数项分别为()p(x)?13?x114x3x(A)-6,2,-6;(B)-6,-2,6;(C)-6,2,6;(D)-6,-2,-,其值不为零,A经若干次初等变换后,其行列式值()(A)保持不变;(B)保持不为零;(C)可变为任何值;(D)保持相同符号。,那么()(A)列式与它的转置行列式相等;(B)D中两行互换,则行列式不变符号;(C)若D?0,则D中必有一行全是零;(D)若D?0,则D中必有两行成比例。()?bbbb;?bbbb;C.(aa?bb)(aa?bb);D.(aa?bb)(aa?bb)123412341234123412123434141423232:..?kx?x?x?0??kx?x?0仅有零解则()?123??2x?x?x??4或k??1;??4或k??1;?4且k??1;?4且k?1?x?2x?4x?1??x?0的解为()?23???x?23(A).(x,x,x)?(1,0,?2)(B).(x,x,x)?(?7,2,?2)123123(C).(x,x,x)?(?11,2,?2)(D).(x,x,x)?(?11,?2,?2)()00?1??????2????,?,?均为方程x3?1?0的根,则行列式???的值为()???(A)1;(B)-1;(C)3;(D)0四、计算行列式1、用定义计算0001xy000aa01000001213(1)0020;(2);3)0xy00(4)aaaaa00202122232425aaaaa0n?100000xy3132333435000n?1aa0424300n000n000y000xaa052530011?1(5)由11?1说明:奇偶排列各半?0????11?12、用行列式的性质a2?a?1?2?a?2?2?a?3?2?aa?babc(1)bca1(2)b2?b?1?2?b?2?2?b?3?2(3)证明:?aa?b?2abccab12??2??2??2111111111cc?1c?2c??aa?babc?aa?b??2??2??22222222221d2d?1d?2d?32223:..a?ba?b?a?b11121n(4)a?ba?b?a?b21222n????a?ba?b?a?bn1n2nn3、利用性质化上三角或按行(列)展开(降级)1111122?2xy0?001?00?001211?3222?20xy?00?11???0?00(1)(2).(3).4)12223?200x?000?11????00122523??????????????????4321000???0000?1???222?nn?1ny00?0x0000??11??na?xaax?mx?x12n123?n?1n12naa?xa1?1000xx?m?x12n?逐步下加(4)12n(5)(6).02?2?00??????????xx?x?maaa?x12n12n000?2?n0000?n?11?n123n?1n234n1(7)(循环行列式,后列减前列)D?34512nn12n?2n?14、各行(列)元素之和相等abb1111?abab111?a1(1)(2)(n级)3、(4)(5)D?n11?a11bba1?a1115、将一行各元素拆成两项的和a?xxx11?aa00012xa?xx?11?aa00223(1)(2)0?11?a003xxa?xn0001?aan?1n000?11?an6、爪型行列式a11?101a0?01?n1?(1)证明10a?0?aa?a?a???212n?0??a??????i?1i100?an4:..7、加边法122?2abb1111?a222?2bab111?a1(1)(2)D?(3)223?2n?????11?a11bba222?n1?a1111?a11?11a?xaa112n11?a1?11aa?xa2(4)(5)证明12n111?a?11?n1?3?aa?a?1???12n?????????a?i?1iaaa?x12n111?1?a1n?1111?11?an8、递推法、数学归纳法(三对角)1?x2xx?xxx00?0a01121n?1x0?0axx1?x2?xx1(1)2122n(2)0?1x0a?????2?xn?axn?1??ax?a??????n?110xxxx?1?x2n1n2n000?xan?2000??1x?an?1????00053000?????0025300(3)??n?1??n?1(4)0????00?(???)???D?02500??n??n???(n?1)(?)000????00025000?????????0008150001?????00??n?1??n?1181500(5)01???00(???)(6)?D?01800?????n?(n?1)?n(???)?000?????000180001???cos?100012cos?100(7)012cos?00?cosn?0002cos?100012cos?9、利用范德蒙行列式1111xxxx123nx2x2x2x2??(x?x)123nij1?j?i?nxn?1xn?1xn?1xn?1123n还可用因式法证明范德蒙行列式5:..1827641233343111114916**********(1)解原式???(2)D=增加一行一列,设f(x)=22222,可以看出D正好是f(x)中x3的余子式abcdxa2b2c2d2a3b3c3d3x3a4b4c4d4a4b4c4d4x4M?(?1)4?5A,f(x)按最后一列展开,A的值就是f(x)的x3的系数的相反数454545an(a?1)n?(a?n)n1111an?1(a?1)n?1?(a?n)n?1xxxx(3)123n(4)计算D?????x2x2x2x2n?1D?123naa?1?a?nn11?1xn?2xn?2xn?2xn?2123nnnnnxxxx123n????x?x??x??x?x,?n?2?12nij1?j?i?n解(4):+1行,第n行,?,第2行分别向上与相邻行交换n次,n-1次,?,1次,n(n?1)共交换了次;将列也作同样的变换。这样一共交换了n(n?1)次,即偶数次,得211?11a?na?n?1?a?1a(a?n)2(a?n?1)2?(a?1)2a2D?n?1?????(a?n)n?1(a?n?1)n?1?(a?1)n?1an?1(a?n)n(a?n?1)n?(a?1)nan由范德蒙行列式的计算公式得D?1?2???n?1?2???(n?1)???2?1?2n?1?3n?2???(n?1)2?nn?110、利用拉普拉斯定理A*AOOB?A?B?A?Bn?(?1)mnA?BOB*BAOmnmanbnan?1bn?1??D??(ad?bc)abD?112niiii2ncd111?i?n??cdn?1n?1cdnn6:..?,并求A?A???a1?=11?a?1,a≠0,i=1,2,...,????11?1?anaaa?a815000123n?a0a?。?a?a0?aD?0180012nn??????a?a?a?(x),使f(1)?0,f(2)?3,f(?3)?:设多项式为f(x)?ax2?bx?c,则有f(1)?a?b?c?0,D??20?0,D??40,f(2)?4a?2b?c?3,1D?60,D??(?3)?9a?3b?c?28,23DDDa?1?2,b?2??3,c?3?,所求的多项式为f(x)?2x2?3x??x??a?ax?ax2??axnn?1x,x,,x证设有个互异的零点,则有012n12n?1f?x??a?ax?ax2??axn?01≤i≤n?1,.i01i2ini即?a?xa?x2a??xna?0,?011121n?a?xa?x2a??xna?0,?022222n??2n?a?xa?xa??xa??1nn?12n?1n这个关于a,a,,a的齐次线性方程组的系数行列式01n1xx2xn1111xx2xn???222?x?x?0,ij1≤j?i≤n?11xx2xnn?1n?1n?17:..a?a?a??a?0f?x?,a,,,b,,b,存在惟一的次数小于n的多项式12n12nL?x?:nx?aL?x???b?j,ia?ai?1j?iijL?a??b1≤i≤n使得,.iiL?x?L?a??b证从定义容易看出的次数小于n,且,?x??c?cx?cx2??cxn?1设满足012n?1f?a??b1≤i≤n,,ii?c?ac?a2c??an?1c?b,?011121n?11?c?ac?a2c??an?1c?b,即021222n?12???2n?1?c?ac?ac??ac??1n这个关于c,c,c,,c的线性方程组的系数行列式012n?11aa2an?11111aa2an?1222???,?a?a?0ij1≤j?i≤n1aa2an?1nnnc,c,c,,cf?x??L?x?故是唯一的,?