大學數學-線性代數習題答案第一章行列式1利用對角線法則計算下列三階行列式201(1)141183201141183解2(4)30(1)(1)1180132(1)81(4)(1)2481644abc(2)bcacababcbcacab解acbacbabbbaaaccc3abcabc333111(3)abcabc222111abcabc222解bccaabacbacb222222(ab)(bc)(ca)xyxy(4)yxyxxyxyxyxy解yxyxxyxyx(xy)yyx(xy)(xy)yxy(xy)x333yx3xy(xy)y3x2yx33332(xy33)2按自然數從小到大為標準次序求下列各排列的逆序數(1)1234解逆序數為0(2)4132解逆序數為441434232(3)3421解逆序數為532314241,21(4)2413解逆序數為3214143(5)13(2n1)24(2n)n(n1)解逆序數為232(1個)5254(2個)727476(3個)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1個)(6)13(2n1)(2n)(2n2)2解逆序數為n(n1)32(1個)5254(2個)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1個)42(1個)6264(2個)(2n)2(2n)4(2n)6(2n)(2n2)(n1個)3寫出四階行列式中含有因子aa的項1123a的項的一般形式為解含因子a1123(1)taaaa11233r4s其中rs是2和4構成的排列這種排列共有兩個即24和42a的項分別是所以含因子a1123(1)taaaa(1)aaaaaaaa1112332441123324411233244(1)taaaa(1)aaaaaaaa21123344211233442112334424計算下列各行列式41241202(1)1052001174124120210520011741210cc41101202122(1)1032141031423解43c7c400103cc411099101222300210314cc1717140112321413121(2)123250622141cc214021403122rr3121312212324212304212302140解5062506221403122rr41123000000abacae(3)bdcddebfcfefabacaebcebdcddeadbfce解bfcfefbce111adf1bc1e14abcdef111a1001b1001c1(4)001da10001aba0rar1b1001c11b1001c112解001d001dcdc1aba01abaad01d(1)(1)1c13ccd211210101abadabcdabcdad1(1)(1)3211cd5證明:aabb22(ab)3;(1)2aab2b111證明aabb2cca2ababa222aab2b2212aba2b2acc100111231aba2baaba(ab)(ba)b(a)ba2b2a(1)3122312axbyaybzazbxxyz(2)aybzazbxaxby(ab)yzx;33azbxaxbyaybzzxy證明axbyaybzazbxaybzazbxaxbyazbxaxbyaybzxaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbxayazbxxb2zxaxby2zaxbyyxyaybzxyzyzxayzxbzxy3zxyxyz3xyzxyzayzxbyzx3zxyzxy3xyz(ab)yzx33zxya2(a1)(a2)(a3)222b(b1)(b2)(b3)20;22(3)22c(c1)(c2)(c3)222d2(d1)(d2)(d3)222證明a2(a1)(a2)(a3)222b2(b1)(b2)(b3)2(cccccc得)22c(c1)(c2)(c3)4332212222d2(d1)(d2)(d3)222a22a12a32a5b22b12b32b5c22c12c32c5(cccc得)4332d22d12d32d5a22a122b22b1220c22c122d22d1221111abcdabcd(4)2222a4b4c4d4(ab)(ac)(ad)(bc)(bd)(cd)(abcd);證明1111abcda2b2c2d2abcd444411110bacada0b(ba)c(ca)d(da)0b(ba)c(ca)d(da)222222222111(ba)c(a)d(a)bcdb2(ba)c2(ca)d2(da)111(ba)c(a)d(a)0cbdb0c(cb)c(ba)d(db)d(ba)11(ba)c(a)d(a)c(b)d(b)c(cba)d(dba)=a(b)(ac)(ad)(bc)(bd)(cd)(abcd)x10000x100x(5)xxaaan1nnn11000x1aaaaxann1n221證明用數學歸納法證明x1axaDx2axa命題成立當n2時21221假設對于(n1)階行列式命題成立即Dxxxaaan2n1n1n2n11按第一列展開有則Dn1000DxDa(1)n1x10011x1nn1naxxxaxDaann1n1n1n1n因此對于n階行列式命題成立6設n階行列式Ddet(a),把D上下翻轉、或逆時針旋轉ij90、或依副對角線翻轉依次得aaaaaann1nn1nn1nnnDDDaaaaaan111123111n11n1n(n1)(1)D3證明DDDD122)所以證明因為Ddet(aijaa111naaaan1nnD(1)1n1n1nnaa111naa212naa111naa(1)(1)212naan1nnn1n2aa313nn(n1)2(1)D(1)D12(n2)(n1)同理可證aan(n1)2n(n1)(1)D(1)DT22n(n1)11n1D(1)2aa1nnnn(n1)2n(n1)n(n1)D(1)D(1)(1)D(1)DDn(n1)22327計算下列各行列式(D為k階行列式)ka1,其中對角線(1)Dn上元素都是a未寫出的元素1a都是0解a00010a00000a00(按第n行展開)Dn000a01000a00001aa0000(1)0a000(1)aan12n000a0(n1)(n1)(n1)(n1)a(1)(1)aa(aaa1)nn2n22an1nn(n2)n(2)xaaaxa(2)D;aaxn解將第一行乘(1)分別加到其余各行得xaaaaxxa00Dax0xa0nax000xa再將各列都加到第一列上得x(n1)aaaa0xa00Dn00xa0[x(n1)a](xa)n10000xaa(a1)(an)nnna(a1)(an)1(3)D;nn1n1n1aaan1111解根據第6題結果有111aa1ann(n1)D(1)2a(a1)(an)n1n1n1n1a(a1)(an)nnn此行列式為范德蒙德行列式D(1)[a(i1)(aj1)]n1(1)nn2n1ij1(1)[(ij)]nn(1)2n1ij1(ij)(1)nn(1)nn(1)1(1)22n1ij1()ijn1ij1abnnabcd(4)D2ncd;1111nn解abnnabcdD2ncd(按第1行展開)1111nnan1b0n1abcd11an11cdn100dn10n0abn1n1ab11(1)b2n1ncd11cdn1n10cn再按最后一行展開得遞推公式DadDbcD即D(adbc)D2nnn2n2nn2n22nnnnn2n2()nDadbcD于是2niiii2i2ab而Dadbc11cd1121111()nDadbc所以2niiiii1(5)Ddet(a)其中a|ij|;ijij|ij|解aij0123n11012n22101n3Ddeat)(3210n4nijn1n2n3n401111111111rr111111211111rr23n1n2n3n401000012000cc211220012220ccn12n32n42n5n131(n1)2(1)n1n21a11111a1,其中aaa0(6)Dn2111an12n解1a11111a12Dn111ana000011aa0001cc12220aa00133cc23000aa1n1n1a1a0000nn10000a111000a1101100a2aaa1312n00011a1000011na11n10000a101000a11200100a13aaa12n00001a1n1n000001a1ii11an(aaa)1()12ni1i8用克萊姆法則解下列方程組xxxx5x122x3x44x24(1)2x21x3x23x5x3x122x1x10341234解因為111112142315142D31211511122141511121428422152315142DD12302111115012111151122412121422312426DD232534310113120DDDD3x1所以1