線性代數(shù)同步練習(xí)冊(cè)第一章答案

第一章行列式第一章行列式00010020020(1)50302(1)403(1)24.40030040001、利用對(duì)角線法則計(jì)算行列式.400mabma2nb2.nba(1)12220112210011332991333001（2）221982211002020(2)(3)(4)1034.351100335110320412210012212011333001331000.22110022121414.1833511003513112311231aaa2a2a36a26.aa312310114(3)31120471123110157xyxy(5)yxyxxyx1123112301140114153.y00327003270063xy(xy)(xy)3x3y32(x3y)3.300051２、利用行列式的性質(zhì)計(jì)算行列式．1第一章行列式123410234a0b00x0ya0x0y0b02341rrrr10341（4）3412104120cx0y1234d(6)c0d00u0vu0vu0v41231012312341341adxycbxy(adcb)(xuyv).uvuv=10141211231x111xx0011x11rr,rr11x111234（7）1234011302220111111y111y1100xy11x110011001100yyrr,rr,rr111111y213141101100rr,rr1xy0x112141234r2r,rr100113160.21111y0011y324.004411000004rr0x11x2y2.432300012300xy0011000y230D01200123234.(5)行列式121b0001b00000023000120121111bb01rrb21221（8）011bb011bb32320011b30011b32第一章行列式1b0001000111...1...b1011rrrr01baxaaab1.322432xnaaax[(1)]...001b3001b3cac,cac,...,cac0011b...............00012131n13aaa...x3、計(jì)算n階行列式123...n111...10...00xa120...0[x(n1)a]0xa...0(xa)[x(n1)a]n1.........0...xa0rr,rr,...,rr(1）103...0...12131n......00100...na10...00(n1)00...01aax1...000x...001120...02303...0...(1n)n!.(3)Dn..................an1a00...x100...0x100...nnxaa...aaxa...a...解:將行列式按第n行展開,有rrr....rDxD(1)(1)n1xDan1nn1n1a123n（2）aaxann...............遞推得aaa...xDxDaDxDa...,Daxa,n1n2n1,212n1n2n1,所以Daxn1axn2...axa.2n1nn13第一章行列式a22a126b22b1260.c22c1264、證明：(1)設(shè)a,b,c為互異實(shí)數(shù),證明行列式:abcDac2為零的充要條件是abc0.bccaabb22d22d126axbyaybzazbx(3）aybzazbxaxby(a3y3)yzx.azbxaxbyaybzzxyxyz證明:abcabcrrDabccaabbcabc22223122證明:：將左式按第一列拆開得abcabcabcaxbyaybzazaxbyaybzbx左式aybzazbxaxaybzazbxbyabc(abc)a2b2c2(abc)(ca)(cb)(ba).azbxaxbyayazbxaxbybz111（按第二列拆開，因有兩列成比例得一項(xiàng)為零）為零的充要條件是abc0.所以,行列式axbyayazbyaybzbxaybzazaxbzazbxbyazbxaxaybxaxbybza2(a1)2(a2)2(a3)2b2(b1)2(b2)2(b3)2c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2(2)0.axayazbybzbxxyzayazaxbzbxby=(a3b3)yzx.azaxaybxbybzzxya22a14a46a9b2b14b46b92證明:左式c22c14c46c9d22d14d46d94第一章行列式2a100...0002a10...0001000...000a22a10...000a20a22a1...0005、設(shè)行列式D........................0000...a22a102a1...000a2a2(1)21........................0000...a2a120000...0a22a0000...02aa2k1證明D(n1)an.n2aDk1a2Dk22a(kak1)a2((k1)ak2)(k1)ak.證用歸納法.當(dāng)n1時(shí),D2a,命題正確;所以D(n1)an.n11111當(dāng)n2時(shí),D3a2,命題正確.26、設(shè)D1012，求A2A3A4A14，設(shè)nk時(shí),D(n1)an,命題正確.n31101112131205當(dāng)nk時(shí),按第一列展開,則有2a100...000其中A為行列式中元素a的代數(shù)余子式.a22a10...0000a22a1...000ijij12341012D2a........................n解:A2A3A4A311024.0000...a22a111121314120540220000...0a22ak130227、求行列式D0700的第四行各元素的余子式之和.53225第一章行列式當(dāng)1且2時(shí),齊次線性方程組只有零解.30402222MMMM0解:x10xx700414243442310、問,取何值時(shí)，齊次線性方程組x0xx1111有非零解.123x2xx0123340解:方程的系數(shù)行列式744228110011(1),D130xxkx3121124xkx0有非零解8、如果齊次方程組,k應(yīng)取什么值?12當(dāng)0或1時(shí),齊次線性方程組有非零解.xkx013解:方程的系數(shù)行列式31k12111x2311、解方程00x20.D4k02k(k2),10k002x當(dāng)k0或k2時(shí),齊次線性方程組有非零解.12111x23(x2)2(x2)0,解:00x2002xxx0120只有零解2339、為何值時(shí)，齊次線性方程組xxx2xxx0.1x123解:方程的系數(shù)行列式xx2,x4.解之得1231112、利用范德蒙行列式計(jì)算行列式1(1)(2),D12116第一章行列式11111111213（1）xy2513x7y2.12131（1）1419(1)222123218127(1)3231333251，D1511D273，D3321，解:371=[2(1)][1(1)][3(1)](12)(32)(31)48.xD13,yDD21.D11...1222...2nxy2z3332...3n..............nn2...n2(2)Dn5x2y7z22.(2)2x5y4z4解:從各行中提出公因子,231211111...11222...2n1解:D52763,D222763,125445432n1n!n!(aa)D133...3ni1jinj13211...............D5227126,D5222189,2442541nn...nn12n!(21)(31)...(n1)(32)(42)...(n2)...(n(n1))n![12(n1)][12(n2)]2!1!1!2!3!(n2)!(n1)!n!.xD11,yD22,zD33.DDDf(x)14、求三次多項(xiàng)式，13、用克萊姆法則解下列線性方程組：使得f(1)0,f(104,f(2)3,f(3)16.7第一章行列式(C)k1且k3k21(D)k1或k3解:設(shè)f(x)ax3bx2cxdf(1)abcd02、行列式k111200的充要條件是（D）.f(1)abcd4因?yàn)閒(2)8a4b2cd3(A)k2(C)k2且k3（B）k3(D)k21或k3f(3)27a9b3cd16解方程組得a2,b5,c0,d7,四階行列式A0，則A中（C）.3、設(shè)(A)必有一行元素全為零；所以f(x)2x35x27.15、已知m階行列式Aa,n階行列式Bb,OA(B)必有兩行元素對(duì)應(yīng)成比例；(C)必有一行元素可以表示為其余各行對(duì)應(yīng)元素的線性關(guān)系.(D)對(duì)角線上元素全為零.0105求DB*的值.02264、行列式3307的值為(A).次與其前面的n逐列交換,共交換mn次,得解:將A所在的列一0408(A)72;(B)24;(C)36;(D)12.OAAO(1)mnab.*B(1)mn二、填空題B*ab1、設(shè)行列式ac1acabc2，則1111，3.1ab11第一章行列式自測(cè)題abc2222222一、選擇題：2A2，則A1三階行列式2、設(shè)4.k122k10的充分必要條件是（C）.1、行列式(A)k1（B）k38第一章行列式a00bab00aaa3122ba2ba2ba6，00000(acbd)2.3、若三階行列式abcdcd1c11223c33（1）（2）000ab00cdc2c00daaa312111x1bbb3.則行列式12ccc3311x111x111x111112000100a0114、設(shè)，則a2.x11x102001x11x100accccxx11x11111234aa12a1211Da5、若行列式aa1，2131223223111x1100xaaa33x11x1110x0x4a2a3aa121x111cccccc1x00,,11則行列式111222322131414a2a3aa–12.111110002121234a2a3aa31313300xx0x0x4.x001234A2341,則A2A3A4A0.6、設(shè)3412212223244123三、計(jì)算四階行列式9第一章行列式0111...11234...n01x11...11123...n11x12...n2001x1...1......001x............00...1xx...11xx1...n3.四、計(jì)算n階行列式..................1xxx...21xxx...1111...1111...111x(1)n101x1...111234...n1123...n11x12...n21xx1...n3..................1xxx...21xxx...1...0...............00...1x1解：00...000...00...00x1xxrr,rr,...,rr...01n11223nxx..................00...0......1x11234...n1123...n11x12...n2n2.(1)n1xx2x1x2x3rr,rr,rr,...,rr1...n3122334n1n1xx2x22x12x22x3五、設(shè)f(x)3x33x24x53x5，4x4x35x74x3..................1xxx...21xxx...1求方程f(x)0根的個(gè)數(shù)？10第一章行列式(x1)(x2)(x2)(21)(22)(21)解:求方程f(x)0有幾個(gè)根f(x)是x幾次多項(xiàng)式.,也就是問(x1)(x2)(x2).將第1列的-1倍分別加至第2、3、4烈,得x21011f(x)0的根方程為1、2、–2.f(x)2x2103x31x224x3x73xxx20xkxx0有非零解如果齊次線性方程組1123七、,k應(yīng)取什么值?x21000023kxxx01232x21cc3x31x21.21