高等代數(shù)教案北大版行列式計算方法

高等代數(shù)教案北大版行列式計算方法行列式計算方法1.利用行列式的定義直接計算：適用于行列式中零比較多的情形．2.化行列式為三角形行列式——初等變換法1）保留某行（列）不動，將其它的行（列）分別乘上常數(shù)加到這一行（列）上。2）將某行（列）的倍數(shù)分別加到其它各行（列）3）逐行（列）相加4）加邊法——在原行列式的邊上增加一行一列，使行列式級數(shù)增加1，但值不變。計算行列式例1amaaa12aamnD12nnaaamn123.利用行列式展開定理。適用于某行（列）有較多零的行列式．4.其他方法（一）析因子法——利用多項式的性質(zhì)112312x223例：計算D23152319x2解：由行列式定義知D為x的4次多項式．又，當x1時，1，2行相同，有D0，x1為D的根．當x2時，3，4行相同，有D0,x2為D的根．故D有4個一次因式，x1,x1,x2,x2設Da(x1)(x1)(x2)(x2),1/11高等代數(shù)教案北大版行列式計算方法11231223令x0,則12，D23152319即，a1(1)2(2)12.a3.D3(x1)(x1)(x2)(x2)（二）箭形行列式abbbnca0001211Dc0a0,a0,i1,2,3n.n122ic00anncia解：把所有的第列的倍加到第1列，得：i1(i1,2n)ibcnDaaa(a)n0iian112i1i可轉(zhuǎn)為箭形行列式的行列式：1a1111a11axx1xax1)2)221x11anxan（第2至第n行分別減去第1行，轉(zhuǎn)為箭形行列式）（三）所有行（列）對應元素相加后相等的行列式abba(n1)bbb1bb1abbaba(n1)baba(n1)b1)bbaa(n1)bba1ba2/11高等代數(shù)教案北大版行列式計算方法1brr(i1,2n)0abb0()abn1a(n1)bi100123n1n234n1ab123n1n134n(n1)n12)n1n1n3n2ccc21n1n3n2112n2n112nn12n2n2rrnrn1123011n1n11nrn1n21111nrr1n(n1)n(n1)211n1122011n01n111111n1111111nrr(i2,3n1)100nnn(n1)i2nn00111n(n1)00n2cccn2n11nn000nnnn(n1)(1)1(n1)(1)n(n1)(1)n1(1)()nn2(n2)(n1)32122nn0n2(1)2n1(n1)nn12(n1)nn1．(n2)(#)(1)2(1)2n(n1)2（四）加邊法（適用于除主對角線上元素外，各行對應的元素分別相同，可轉(zhuǎn)為箭形行列式的行列式——加邊法是計算復雜行列式的方法，應多加體會）3/11高等代數(shù)教案北大版行列式計算方法abaaa112naab21）Dn,bbb012n12naaab12nn0aa12aa02）D2n,aaa01n12naaaa0n1n2解：1aa2aaa1aaa12n1nnn0aba1D0aabrr(i2,3n1)1b0001211)i110b2n1220aaab100bn12nnn1ain1aa1nbi1anbi0b10bbb(1i).12ncci1(i1,2n1)bi1i1i00bn2)1aaanaa1aa21aaana1a2121200aa1D0aa0rr(i2,3n1)1aa2n11i11aan12n220aaaa01aaann1n1n2nnn1100001aa0101101aa1anaa1a1aa21a12a02n001a1aa111a11cc(i3,4n2)a102a222222i1a1aanann2a100n2ann2nn4/11高等代數(shù)教案北大版行列式計算方法1n1n122a111i1i12a1naana2cc(i3,4n2)i12ni101i02a01002a00c1c(j1,2n)002a2j2j0002ann1n1122aani(2)naa(2)n2aaa[(n2)2i]i1a1n1n12nna1i,j1j22ii1（五）三角型行列式——遞推公式法950049501）D0490n009500495005495按c展開解：D9D4n19D20D,n1n21n49n1即有,orD5D4(D5D)D4D5(D4D)nn1n1n2nn1n1n2于是有同理有D5D42(D5D)4n2(D5D)4n2(6145)4n,nn1n2n321D4D52(D4D)5n2(D4D)5n2(6136)5nnn1n2n321D5D4D5n14D4D5nn即n1nn1nnn1（先將行列式表示兩個低階同型的行列式的線性關系式，再用遞推關5/11高等代數(shù)教案北大版行列式計算方法系及某些低階（2階，1階）行列式的值求出D的值）abab01abab01ab0000002）D.n000000abab1ab解：按c展開1D(ab)DabDDaDb(DaD)bn2(DaD)n2nn1n2nn1n121同理而DbDa(DbD)an2(DbD).nn1n1n221Da2abb2,Dab21DaDbn2(a2abb2a2ab)bn;nn1DbDan2(a2abb2a2ab)an.nn1an1bn1ab由以上兩式解得Dabn(n1)anab（六）拆項法（主對角線上，下元素相同）axa1aaxaa1)D2naaaxn解：x010x0a0aaxa1aaxaaxaa0a01aaax2DnxDnn12200xan1aaxxxaxDnn1aaaaxn000a12n1DxxxaxDxxxaxD.n112n2n1n212n3n2n36/11高等代數(shù)教案北大版行列式計算方法繼續(xù)下去，可得DxxaxxxaxxxxaxxxaxxxxxD.n1n112n2n12n3n1124nnn132（Daxaxxx12）212xxxa(xxxxxxxxxxxxx)12n12n112n2n13n23n1n當xxx0時,Dxxx(1a)12xi12nnni11）也可以用加邊法做：1aa1aaa1x00ax，D11n0aax10xnn1an1aaxii1當x0時,i1,2n,D0x0in10xnabbbcabb2）Dccabnccca解：cbbbacbbb1bbbcabb0abb1abbDccab0cabc1cab(ac)Dnn1ccca0cca1ccan7/11高等代數(shù)教案北大版行列式計算方法1bb0ab0b0①c0cbab0(ac)Dc(ab)n1(ac)Dn1n10cbcbabnbbbbab000cabbcabb又Dccabccabncccaccca1111cabbbccab(ab)Dn1ccca②b(ac)n1(ab)Dn1①（ab）-②(ac)，得．(cb)Dc(ab)nb(ac)nn當cb時，D[c(ab)nb(ac)n]/cbn當cb時，D[a(n1)b](ab)n1n（七）數(shù)學歸納法（第一數(shù)學歸納法，第二數(shù)學歸納法）1）（用數(shù)學歸納法）證明：1a1111a11Dnaaa(11)2a12ni111an1證：當時，，結(jié)論成立．n1D1aa(1)a11111k假設nk時結(jié)論成立，即后一列拆開，得，對，將按最k1Daaa(1)nk1Dak12ni1i8/11高等代數(shù)教案北大版行列式計算方法1a1111a111a11010011111a22Dk11111a11111a0k111111111akk1a11011a2100aDaaaaDk1kk1k12k111a0k1111111kakaaaaaaa(1)aaa(1)a12kk112k12k1i1i1ii所以nk1時結(jié)論成立，故原命題得證．cos10012cos2）證明：Dn2coscosn112cos證：n1時，Dcos.，結(jié)論成立．1假設nk時，結(jié)論成立．當nk1時，D按第k1行展開得k1cos100112cos2cos2cosDDkk1D2cosD(1)k1kk1k12cos由歸納假設D2coscoskcos(k1)2coscoskcoskk12coscoskcoskcossinksincoskcossinksincos(k1)于是nk1時結(jié)論亦成立，原命題得證．（八）范德蒙行列式9/11高等代數(shù)教案北大版行列式計算方法11xx1x1xx2nx2221）D12nnxn2xn2xn2n12xxxnnn12n解：考察n1階范德蒙行列式11xx11xxnx2x21xx2(xx)(xx)(xx)(xx)12nij22f(x)12n1jinxn1xn1xn1xn1nxnxnn12xxnn12顯然D就是行列式f(x)中元素xn1