利用對(duì)角線法則計(jì)算下列三階行列式

1、第一章行列式1 .利用對(duì)角線法則計(jì)算下列三階行列式2 01(1) 1-4-1；- 183201解1-4-1- 183=2(-4)30(-1)(-1)118- 013-2(-1)8-1(-4)(-1)- 24816-4-4abc(2) bca；cababc解bcacab=acbbaccba-bbb-aaa-ccc333=3abc-a-b-c1(3)a2a11bc；.22bc111解abc2.22abc-bc2ca2ab2-ac2-ba2-cb2二(a-b)(b-c)(c-a)xyx+y解yx+yxx+yxy二x(xy)yyx(xy)(xy)yx-y3-(xy)3-x3=3xy(xy)-y3-3x

2、2y-x3-y3-x3-2(x3y3).2 .按自然數(shù)從小到大為標(biāo)準(zhǔn)次序.求下列各排列的逆序數(shù)(1)1234解逆序數(shù)為0(2)4132解逆序數(shù)為4：41.43.42.32.(3)3421解逆序數(shù)為5：32-31-42-41,21(4)2413解逆序數(shù)為3：21-41-43(5)13-(2n-1)24-(2n)解逆序數(shù)為丑尸：23 2(1個(gè))5 2-54(2個(gè))7 2-74-76(3個(gè))(2n-1)2.(2n-1)4.(2n-1)6.一,(2n-1)(2n-2)(n-1個(gè))(6)13(2n-1)(2n)(2n-2)2逆序數(shù)為n(n-1):32(1個(gè))52.54(2個(gè))(2n-1)2.(2n-1)

3、4.(2n-1)6，(2n-1)(2n-2)(n-1個(gè))42(1個(gè))62.64(2個(gè))(2n)2.(2n)4,(2n)6.-(2n)(2n-2)(n-1個(gè))寫出四階行列式中含有因子ana23的項(xiàng),含因子ana23的項(xiàng)的一般形式為(一1)a11a23a3a4s其中rs是2和4構(gòu)成的排列.這種排列共有兩個(gè).即24和42.所以含因子ana23的項(xiàng)分別是(一1)411a23a32a44=(-1)1ana23a32a44-ana23a32a44(一1)111223a34a42=(-1)2ana23a34a42=ana23a34a424.計(jì)算下列各行列式：41241202105200117(4)001。c

4、lIdO2II0-4321大O9大O9IIO0-24II-21一4324002。a-1001b-1001c-1001d=(-1)(-1)21r1ar20-1001abb-10a1c-1001d1+ab-10a0c1-1dc3dc21+ab-10ac-1ad1cd0=(-1)(-1)321+ab-1ad1cd=abcdabcdad15證明:2a2a1abb22b1=(a-b)3;證明2a2a1abab1b22b1C2一C1C3-Ci2a2a1ab-a2b。a022b-a2b-2a0=(-1)31ab2b-2a二(b-a)(ba)aba12=(a-b)3azbxaxbyaybzaybzazbxax

5、byax+by(2)ay+bzaz+bxx=(a3+b3)yz證明axbyaybzazbxaybzazbxaxbyazbxaxbyaybzaz+bxyay+bzaz+bxax+by+bzaz+bxax+byay+bzxax+byay+bzxaybzyazbxzaxbyx2=ayzaybzazbxaxbyx+b2zzazbxxaxbyyaybzx3=ayzy十b3zxx3=ayzx+b3yzxyz=(a3+b3)yzxzxy2.2.一2,一2a(a+1)(a+2)(a+3)b2(b+1)2(b+2)2(b+3)2c2(c+1)2(c+2)2(c+3)2d2(d+1)2(d+2)2(d+3)2證明

6、a2(a1)2b2(b1)2c2(c1)2d2(d1)2(a2)2(b2)2(c2)2_2(d2)(a3)2(b3)2(c3)2(d3)2(G4-c3-c3-c2c2-6得)a22a1b22b1c22c1d22d12a32b32c32d32a十52b+52c+52d十5(04-63c3-Q得)a22a122b22b122c22c122d22d1221a2a4a1bb2b41c2c4c1dd2d4=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(abcd);證明1a2a4a1bb2b41c2c4c1dd2d410001b-ab(b-a)222、b(b-a)1c-ac(c-a)222

7、、(c-a)1d-ad(d-a)d2(d-a2)二(b-a)(c-a)(d-a)二(b-a)(c一a)(d一a)b2(ba)c2(ca)1c-b1dd2(da)0c(c一b)(c+b+a)d(db)(d+b+=(b-a)(c-a)(d-a)(c-b)(d-b)c(cba)d(dba)=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(abcd)-1x0-1=xn,a1xn”an-1x'an'a)0an0anixa2-1xa1證明用數(shù)學(xué)歸納法證明當(dāng)n=2時(shí).D2x-1a2xa1=x2+ax+a2命題成立,假設(shè)對(duì)于(n-1)階行列式命題成立.即Dn二Xn，31Xn4-3

8、nX3n-1則Dn按第一列展開(kāi).有Dn=XDn3n(-1)n1-1000x-10011X-1xDnanxna1xn_1anX-an.因此.對(duì)于n階行列式命題成立6.設(shè)n階行列式D=det(aij),把D上下翻轉(zhuǎn)、或逆時(shí)針旋轉(zhuǎn)90口、或依副對(duì)角線翻轉(zhuǎn).依次得3n1-3nnD1=311,1'a1n31n一annD2=a11'1'an1ann'"am-D3=an1'''a11n(n)證明D1=D2=(-1)一口.D3=D.證明因?yàn)镈=det(aij所以an1anna11D1a11a1n=(-1尸(-1廣=(-1產(chǎn)a11a21an1an

9、1a21TTTOa1naa2nTTTOann一ann32na31.一a3nn(n7)十42D=.1尸d-同理可證n(n)aiianin（n）n(n)D2=(-1)2二(-1)Dt=(一1)d.aln.-annn(n_1)n(n_J)n(n_1)D3-(-1)2D2-(-1)2(-1)2D=(-1)n(n口D=D，7.計(jì)算下列各行列式（Dk為k階行列式）:(1)Dn,其中對(duì)角線上元素都是未寫出的元素都是0Dn(-1)2n（按第n行展開(kāi)）n1=(一1)(nJ)(n-1)（n）（n）n1=(-1)(-I)"nnn-2a二a-a二an-2,2(a1)(n二)(nN)(2)Dn解將第一行乘（-

10、1）分別加到其余各行.得xaaaa-xx-a0.0a-x0x-a.0*一a-x000x-aDn再將各列都加到第一列上DnDnx(n-1)a00nanaa-11解根據(jù)第n(nF)Dn1=(F2n-1anan.1=x(n-1)a(x-a)(a-n)nn1(a-n)1a-1(a-1)n(a-1)n/一nU(a-n)一(a-n)n此行列式為范德蒙德行列式n(n-1)Dn1=(-1尸(a-i1)-(a-j1)n1J.j.1n(n1)=(-1尸II-(i-j)n1_i>j_1n(n1)=(-1)(一1)11(i-j)n1_i.j_1anbna1b15didnanD2n=a1b1(按第i行展開(kāi))cid

11、1Cndnan1bn0a1匕anC1d1Cn1dn。00dna1b1C1d1bn(-1)2nhCn1Cn再按最后一行展開(kāi)得遞推公式D2n=andnD2n_2一bnCnD2n-2即D2n=(andnbnCn)D2n-2,于是D2n=【(a4-biG)D2.i=2而D2=a1"=a1dl-b1cl.2IIIICidin所以D2n=/(aidi-biQ).i1(5)D=det(a。其中aj=|i-j|;解aj=|i-j|Dn=det(a。)二0123101223120110n-1n-2n-3n-4n1，-n21n3，-n4011110-1111- 1-111- 1-1-11- 1-1-1-

12、1*n7n2n3n4-1000-1-200-1-2-20-1-2-2-2*n-12n-32n-42n-50000n-11+a1(6皿=111a2=(-1)n-1(n-1)2n-211,其中1aa2,an#0,-1+an1+a1Dnc11a2a100一001-a2a20一0010一a3a3,',001*000一_anan1000一0-an1+a1C2-c3n100001an_1a1二a1a2an=a1a2an-101-1a2a3-1-1an.an_1=(aa2an)(1八8用克萊姆法則解下列方程組x1x2x3x4=5-1a2-1a3_1ann1Ja(1)x12x2-x34x42x1-3x

13、2-x33x1x22x311x4=0解因?yàn)?12312-311-1-1214-511二-142.D1D3所以(2)5-2_2012-311-1-1214-51111235-2_201-1_1214-511-284112312-315-2-2014-511-426D4112312-311-1-125-2-20=142x2D2=2DX3D3=35x16x2x15x26x3x25x36x4x35x46x解因?yàn)?100065100065100065100065=665D11000165100065100065100065-15075100010001065100065100065=7145所以D3D515076655100065100100010065100065二703D45100065100065101000100065-395.51000651000651000651100011145665703X3二665-395X4:665X4212665X1x2x3=0問(wèn)九.N取何值時(shí).齊次線性方程組，x1+Nx2+x3=0有非X12x2x3=0零解?系數(shù)行列式為-1112令D=0.得N=