线性代数之行列式的性质及其计算

.\第二节 行列式的性质与计算§2.1 行列式的性质aaLa1112L1naaa考虑DL21L22LL2n将它的行依次变为相应的列，得aaLan1n2nnaaLa1121Ln1DTaaa1222Ln2LLLaaLa1n2nnnDT为D的转置行列式.性质1 行列式与它的转置行列式相等.（DTD）谢谢阅读bbLb1112L1n事实上，若记DTbbb则ba(i,j1,2,L,n)21222nLLLLLLLijjibbLbn1n2nnDT(1)(p1p2Lpn)bbLb(1)(p1p2Lpn)aaLaD.1p12p2npnp11p22pnn说明:行列式中行与列具有同等的地位,因此行列式的性质凡是对行成立的结论,对列也同样成立.感谢阅读性质2互换行列式的两行(rr)或两列(cc)，行列式变号.ijij123123例如086351.351086推论 若行列式D有两行（列）完全相同，则D0.谢谢阅读证明:互换相同的两行,则有DD,所以D0.感谢阅读性质3行列式某一行（列）的所有元素都乘以数k，等于数k乘以此行列式，即谢谢阅读aaLaaaLaL11L12LL1nL11L12LL1nkakaLkakaaLaLi1Li2LLinLi1Li2LLinaaLaaaLan1n2nnn1n2nn推论：(1)D中某一行(列)所有元素的公因子可提到行列式符号的外面；精品文档放心下载.\(2)D中某一行(列)所有元素为零，则D0；感谢阅读性质4:行列式中如果有两行(列)元素对应成比例,则此行列式等于零．精品文档放心下载性质5:若行列式某一行(列)的所有元素都是两个数的和，则此行列式等于两个行列式的和.这两个行列式的这一行(列)的元素分别为对应的两个加数之一，其余各行(列)的元素与原行列式相同.即谢谢阅读aaLaaaLaaaLaL11L12LL1nL11L12LL1nL11L12LL1nababLabaaLabbLb.i1i1i2i2Linini1i2ini1i2inLLLLLLLLLLLaaLaaaLaaaLan1n2nnn1n2nnn1n2nn证:由行列式定义D(1)(p1p2Lpn)aaL(ab)La1p12p2ipiipinpn(1)(p1p2Lpn)aaLaLa(1)(p1p2Lpn)aaLbLa.1p12p2ipinpn1p12p2ipinpn性质6行列式D的某一行（列）的各元素都乘以同一数k加到另一行（列）的相应元rkrD)，即素上，行列式的值不变(DijaaLaaaLaL11L12LL1nL11L12LL1naaLrkrakaakaLakaaijLi1Li2LLini1Lj1i2Lj2LinLjnaaLaaaLan1n2nnn1n2nn计算行列式常用方法:利用性质2,3,6,特别是性质6把行列式化为上(下)三角形行列式,从而,较容易的计算行列式的值．例1:计算行列式感谢阅读23243111(1)1232(2)1311D323411310425111312321232解:(1)rr2324r2r0188D1232321086240425r33r10425.\1232123201880188r430143r8r58r3005862[1(1)58]286.32058r44r106214329003037000294666611111111rr13111311rr0200(2)1i66i2i1i2,3,4111311130002此方法称为归边法.例2:计算n阶行列式1a1L1xaLa111aL1axLa(1)D2(2)DLLLLLLLLnn11L1anaaLx(a0,i1,2,L,n)i解:(1)1a11L111L111L11a0L0rir1a0aL01aL012LDa0a02a2(箭形行列式)n1M3MMLLLL1LLLLi2,3,LnMM00La10Laa00Lann1n1L111nc11ciaai2iaL0iaaLaa0i2,3,L,n23n1LL2LL00LanaaLaaaLa(11)aaLa(11)nn23n12na12nai2ii1i(2)注意到行列式各行元素之和等于x(n1)a,有x(n1)aaLa1aLaDccx(n1)axLa[x(n1)a]1xLa1iLLLLLLLLni2,3,L,nx(n1)a a L x 1 a L x.\1aLarir1[x(n1)a]0xaL0LLLL[x(n1)a](xa)n1.i2,3,L,n00LxaaLa111kMM0aLabLbaLa111k111n例3:设Dk1LkkbLb,DMM,DMM,cc1La2Lb111k111nabMMMMk1kkn1nncLcbLbn1nkn1nn证明:DDD.12证:对D作行运算rkr,把D化为下三角形行列式:1ij1p011p;DMOpL1pLp11kkk1kk对D作列运算ckc,把D化为下三角形行列式:2ij2q011qLq.DMO2q L p

nnn1 nk先对D的前kk行作行运算rkr,然后对D的后n列作列运算ckc,把D化为ijij下三角形行列式:p11MO0DpLpq,k1kk111k11OMMMcLcqLqn1nkn1nn故,DpLpqLqDD..11kk11nn12思考练习1.计算行列式.\2512a1a2Lan3714111(2)Da1a2Lan(n2)(1)D92722M25nMMM4612a1a2Lannnnabbccaabc2.证明abbcca2abc111111111abbccaabc2222222223.证明abacaea2(a1)2(a2)2(a3)2(2)b2(b1)2(b2)2(b3)2(1)bdcdde4abcdef0bfcfefc2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2aa4.计算行列式Daa答案151.(1)Dcc171394216

b c dab abc abcd谢谢阅读2ab 3a2bc 4a3b2cd感谢阅读3ab 6a3bc 10a6b3cd感谢阅读221522152234rr,r2r0216rr01132131231657r4r1011302420120012015221522r2r0113rr011311(3)39320304303000r4r200330003a11Ln11(2)Dcca11Ln1aa,n2i12MMMM12n2ni2,3,L,n0,a11Ln1nabbccaabcaca2.左边=abbccaccabcaca21111111111111abbccaabcaca222222222222abca2cabcacccabca2c2abcac321111111111abca2cabcac2 2 2 2 2 2 2 2 2 2.\abacbaccc2abaccc2baccc2312121111111abacbac22222223.证

abc2abc.111abc222111111111(1)左边abcdef111rrabcdef002rrabcdef0204abcdef.2123111r3r1020002a22a14a46a9a22a126(2)左边ccb22b14b46b9c2cb22b126i1320右边i2,3,4c22c14c46c9c3cc22c12642d22d14d46d9d22d1264.解:从第4行开始，后行减前行得，abcdrrabcdabcdD0aababc430aababcrr0aababc430a3ab6a3bcrr00a3ab000a32a4§2.2 行列式按行（列）展开对于三阶行列式，容易验证：aaaaaaaaa111213aaaaaa22232123212321222311aa12aa13aaaaa323331333133313233可见一个三阶行列式可以转化成三个二阶行列式的计算.问题：一个n阶行列式是否可以转化为若干个n－1阶行列式来计算？谢谢阅读一、余子式与代数余子式aaLa1112L1n定义：在n阶行列式Daaa所在的第i行和第j列，余下2122L2n中，划去元素aLLLijaaLan1n2nn的元素按原来的顺序构成的 n1阶行列式，称为元素aij的余子式，记作Mij；而谢谢阅读.\(1)ijM称为元素a的代数余子式.谢谢阅读ij ij ijaaaaa111213例如三阶行列式aaa中元素a的余子式为M1112212223ij23aaaaa3132313232元素a的代数余子式为A(1)23MM2323232310111110251中元素x的代数余子式为A四阶行列式(1)3205151x23320010301二、行列式按行（列）展开aaLa1112L1n定理n阶行列式Daaa等于它的任意一行（列）的各元素与其对应的2122L2nLLLaaLan1n2nn代数余子式的乘积之和，即DaAaALaA(i1,2,L,n)i1i1i2i2inin或DaAaALaA(j1,2,L,n)1j1j2j2jnjnj证（1）元素a一一11a0L011L此时Daaa(1)(j1j2Ljn)aaLa(1)(j1j2Ljn)aaLa21222nLLLLj111j12j2njn1j12j2njnaaLaj11n1n2nna(1)(j2Ljn)aLaaM112j2njn1111(j2j3Ljn)而A(1)11MM,故DaA;1111111111aLaLa11M1jM1nMMM（2）D0LaL0ijMMMMMaLaLan1 nj nn将D中第i行依次与前i1行对调，调换i1次后位于第一行;谢谢阅读.\将D中第j列依次与前j1列对调，调换j1次后位于第一列;精品文档放心下载经(i1)(j1)ij2次对调后，a就位于第一一ijD(1)ij2aM(1)ijaMaA.ijijijijijij(3)一般地aaLaL11L12LL1nDa0L0 0a L0 L 0L 0a精品文档放心下载i1Li2LLLaaLa

inn1 n2 nnaaLaaaLaaaLaL11L12LL1nL11L12LL1nL11L12LL1na0L00aL0L00LaLi1LLLLLi2LLLLLLinaaLaaaLaaaLan1n2nnn1n2nnn1n2nnaAaALaAi1i1i2i2inin同理有DaAaALaA.1j1j2j2jnjnjaaLa1112L1n推论n阶行列式Daaa2122L2n的任意一行（列）的各元素与另一行（列）对LLLaaLan1n2nn应的代数余子式的乘积之和为零，即aAaALaA0(is)i1s1i2s2insn或aAaALaA0(jt)1j1t2j2tnjnt证考虑辅助行列式aLaLaLa11L1jL1jL1nDaaaa21M2jM2jM2n1MMMMaLaLaLan1njnj2ni列j列按第t列展aAaALaA（jt).该行列式中有两列对应元素相等.而D0，所以1j1t2j2tnjnt1.\aAa A LaA（jt)0.精品文档放心下载1j 1t 2j 2t nj nt关于代数余子式的重要性质nD,ij,nD,ij,1,ij，aAD0,ij;aAD0,ij;其中kikjijikjkijij0,ij.k1k1在计算数字行列式时，直接应用行列式展开公式并不一定简化计算，因为把一个n感谢阅读阶行列式换成n个（n－1）阶行列式的计算并不减少计算量，只是在行列式中某一行或精品文档放心下载某一列含有较多的零时，应用展开定理才有意义.但展开定理在理论上是重要的.感谢阅读三、行列式的计算利用行列式按行按列展开定理，并结合行列式性质，可简化行列式计算：计算行列式时，可先用行列式的性质将某一行（列）化为仅含1个非零元素，再按此行（列）展开,变为低一阶的行列式，如此继续下去，直到化为三阶或二阶行列式.谢谢阅读计算行列式常用方法：化零，展开.1234例4:计算四阶行列式D1012.311012051222222cc1000解:D按第2行展1114613213146c42c12171217

r21r2(1)

1 1 11211 4 62 1 7111100135352146c2c12135按第1行展21113939217c3c1239例5已知4阶行列式3040D2222,求MMMM的值.其中M为a的余子式.070041424344ijij5322解:(方法1)直接计算A(i1,2,3,4)的值,然后相加（略）.感谢阅读i(方法2)利用行列式的按列展开定理，简化计算..\M M M M 1A A A A 1A 1A (1)A 1A谢谢阅读41424344142434444142434430403403402222347222141112828.0700111002111111例6:计算n阶行列式xy0L00010L00xyL00002L0(1)DMMMMMM(2)DMMMMMnLxnLn1000y000y00L0xn00L0解：(1)D按第1列展aAaALaAn11112121n1n1Dn

xy0L00y00L000xyL00xy0L00x(1)11MMMMMMy(1)n1MMMMMM000Lxy000Ly0000L0x000Lxyxn(1)n1yn.按第1列展aAaALaA11112121n1n110L0002L00(1)n1nMMMMM(1)n1n!.00Ln2000L0n1ab00ab例7:计算四阶行列式D40abab0.0abab0ab00ab解:按第1行展开，有abab00ababD(ab)(1)11abab0(ab)(1)140abab,40 0 ab ab 0 0.\对等式右端的两个3阶行列式都按第3行展开，得谢谢阅读D[(ab)2(ab)2]abab24a2b2.abab例8:证明范得蒙行列式（Vandermonde)11L1DxxLx(n2),12Ln(xx)nLLL1jinijLxn1xn1xn112n其中(xx)表示所有可能的(x（i)的乘积.ijij1jin证:(用数学归纳法)n2时,D11xx,结论正确;2xx2112假设对n-1n1范得蒙行列式结论成立，以下考虑n阶情形.感谢阅读111L