川大版高数第三册答案(1)教学文案

1、川大版高数第三册答案(1)第一章行列式1.()231541 1 0 1 0 3该数列为奇排列n(n 1)()631254 =5 2 0 0 1 0=8该排列为偶排列(3)n(n 1) 321 (n 1) (n 2) (n 3)当n 4m或n 4m 1时,n(n 1) 321为偶数，排列为偶排列当n 4m 2或n 4m 训，n(n 1) 321为奇数，排列为奇排列(其中 m 0,1,2 )(4)135 (2n 1)246 (2n)0 12 3当 n 4m或n 4m 1时， 135 (2n 1)246(n 1) n2(2n)为偶数，排列为偶排列当n 4m 2或n 4m 训，135 (2n 1)24

2、6 (2n)为奇数，排列为奇排列(其中 m 0,1,2 )2.解：已知排列i1i2 in的逆序数为k,这n个数按从大到小排列时逆序数为(n 1) (n 2) (n 3)n(n 1)个.2设第x数ix之后有r个数比ix小，则倒排后ix的位置变为in x 1,其后n x r个数比in x 1小，两者相加为n x故inin 1 i1n(n 1)2ii2 in3证明：.因为：对换改变排列的奇偶性，即一次变换后，奇排列改变为偶排列，偶排列改变为奇排列当n 2时，将所有偶排列变为奇排列，将所有奇排列变为偶排列因为两个数列依然相等，即所有的情况不变。偶排列与奇排列各占一半。4(1) &3a24a33

3、a41不是行列式的项为4a23a31a42是行列式的项 因为它的列排排列逆序列=(4321)=3+2+0+0=5为奇数，应带负号(2) a51a42a33a24a51 不是仃列式的项a13a52a41a35a24 = a13a24a35a41a52 因为匕的列排排列逆序列 (34512)=2+2+2+0+0=6为偶数应带正号。a11 a23 a32 a445 解：a,2 a23 a34 a41利用 为正负数来做，一共六项，为正，则带正号，为ai4 a23 a31 a，2负则带负号来做。6 解：(1)因为它是左下三角形雨00 .0a21a220 .0a31 .a32a33.0.an1an2an3

4、.annai1a22a33ann =a11a22a33123 na11a21a31a41.an10a22a32a42.an200a33a43.an3000a44 .an4 .000.0.anna12(2)a2ia51a21'a31a41a51(3)ai2a22a42%2a23123421a3a23a24ai4a24a25a15a25=a11=a11a22a22a23a24a25a32000a42000a5200011+a12 a210=01212 121133451二3217(4)0 y 0y 02 3 125510 x y=x yx yy 0 xaiia12al naiia21a2a

5、227.证明:将行列式转化为兀多于n2 nan1an2an1an2个时，行列式可变为8.(1)a21anlan2.0.0故可知行列式为.00.233206133411232312062341331101431236231133010145123923143300014315 21 21 013 70630X2x b代入（x，yi）yijxiXiX2b b yiyiV2 XiX X2程X2yiXI X2第一章 高数3册9.(i). y mx b.经过(%, yi)(x2, y2).斜率m y_2Xiy y_lXi X2又丫2“yi又由XiX2XiX2XiX2y yi y2左边=yi V2 xy

6、XiX2Xiy2X2yi0右边Xi问题特征:X2xyXi“yiX2i0. i利用性质分成六个行列式相加其余结合为零故原式=2sin sin sinb b b2性质22 cos2 cos2 coscos2cos2cos2i2 cos2 coscos22coscos2cos2cos2i2 cosi2 cos2 coscos2-(2)列+(i)列 _2cos2i2cosi2 cos2 coscos22cos2i2 coscos2cos2cos22 cos2 cos2 cos0性质5cos2cos2cos2xyxyz xyzyz xz11.1xy2a3a1列-1加到-3-1-1-2-1-2-3降阶1-

7、11+12列yz3列xz4列xz3a6a2a3a0xyzxyz1x02 xzyz xzxyy2yz0z2y z2 x z2 x2 z2 z2 x2b3b3a6a2b3b4a10a-2 + 2 列-3 + 3 列-n + n 列2n2n2n4a10a2a-1-1-13b6b3b6b2c3c2c3cn!xa12a13La1n1a12a13LAnxx2a23La2 n1Xza23La2 nxx3La3n1x2X3La3nLLLLLLLLLLxx2x3Lxn1X2X3Lxn000Ln3-x + 2 列2n2n-xn + n 列1a12x2al3x3La1nxn10a23x3La2nxn100La3 n

8、xnLLLLL100L0降阶x1xyz2xy2x y-11+1a123ax2a13a232a6ax33ba3nan 1nxnxnxn习题一13 (1)xy0L000xyL00MMMLMMD根据“定义法'DnI(2.3.4.5n) n nn 1 nx (1) y x ( 1) y123L110L(2) 022LLLLL000L0000DL Ln 1 1 n根据“降阶法'D223Ln(n+1)34L2LLLLn(n+1)12L将第2n列加到 第(1)列上得n(n+1)2n-1 nn1LLn-2 n-1123L_ n(n+1)134L-2 L L L L112Ln-1nn1L Ln

9、2 n 11201将前一彳r乘以-1加 n(n+1)到后一行得2 01L L3 L1 L1 LL Ln-1 n11 n1 n 1L L11变为(口-1)阶=11L11-n11L 1-n1LLLLL11-n1-n1L L1111将(2)(n)歹U力口到(1)列上得-11Ln(n+1) -11L LLL-11L11-n1-n1L L11n(n+1)211L1111L 1-n1LLLLL11-nL111-n1L11-1 (1)列加到(2)(n)列1 1 n(n 1) L 211 L 0 n1 L n 0L L L Ln LL000 L011 n 1(n 1)(n 2)1)(1)n(n 1)n2 3n

10、 2(1)2n 2-2_ _n2 n1 n 1"Vn(n 1)(1)(3)2a(a 1)(a 2)M(an 1)2范达蒙行列式(a(a(an (n 1)(-1尸 1!2!L (nn 1 a1)2)M转置n 1n 1)1)!注：根据范达蒙行列式原式=(1)g( 2)L(1)g( 2)L ( n2)=(n(n 1)1)1!2!L (n 1)!(4)na1n a2Lnann n na a2 L an 1n n n=a1 a2 L an1)n 1.abn 1.a2 b2n&na2nanL1bn 1a1 1bb2a2Lbn 1bia1b2a2Lbn 1an 1Ln 2, 2an 1bn

11、 12,2a bib222a2Lb212an 1b122 a1bfa2Lbn 12an 1(a 1)2Ln 1(a 1)1)1 2 3 L (n(a22)2L(a 2)n1)1!2!L(n-11)!(a(a11)21)n 1a1bna2bnLn-1an 1。11 n. nab1n 1b2n-1a2Ln 1 bn 1nanbn bn L bn1b；n a1bnn a2Lbn1nan 1第n行提出a：得n 1b1n 1a1n 1b2_n-1a2Ln 1 bn 1n 1 an 1b1nn a1b；a；n n n n=a1 a2a3 L an 1(“aibjaj(ajbiaibj)L14 (1)证明:

12、= cos2+cos 2cos2cos2cossin2sin 2sincos2cos2+ cos2sin 2sin2sin2一.十 sin2cos2cos2cos2+ cos2=cos(sincos cos222cos(sin 22+cos一 2cossin21 / -sin(21 ./ 一 sin(2(2)证明:(3)x1 0Mx2Mcos2sin2+ sin2cos 2+ cos2sin) cos(sincos22222cos2sin)2+,+cossin 2221 . / - sin (2sin(x12K4 为x22x2乂32乂3X1x2 Mxnacossin21)二 sin(2x42x

13、4sin(X2X3xn ax1x2Lxnacossin 22x4 1最后一行乘以(-1)加到(1)(n)行得ax1x2x3L xn(4)ao a“递推法” Man 2an 11 0 L x 1 LM M00 L00 L0 00 0M Mx 10 xa。降阶(-1)n+nx a1Man 21 0 Lx 1 LM M00 Lan 1M M MxDn1an由此类推DmxDn 2anD2xD1a1n 1a°xn&xan 115.(1)(1)n=(ab+1)(cd+1)-a(-d)=(ab+1)(cd+1)+ad(2)=(4-6)(-1-15)=32=-a(c-d)a(d-b)-a(d

14、-c)=abd= abd(c-b)(d-b)(c-d)=(16.范达行列式V()=(x3 x1)L(xn x1) (x3 x2)L(xn x2 )L (xn xn 1)11 1L 1转量行列式n1a1a2 La12M LMMLn1n1a1 a2(a3-a 2) L (an 1 a2)L (an 1-an 2)1x2 xLn1 x1a12 a1Ln1 a11a2LLn1 a2n1an 1MM1 an 1 L Lan 1MMn1an 1(a1x)(a2x)LL(an1x)(a2-a1)L L( an1a1)(1)因为a1a2LL an 1为常数。所以p(x)是n-1次的多项式 令 p(x)=0.得

15、 x= a1 .x= a2 an 1 即 p(x)的根为 QaLL an 1第二章 矩阵代数4.计算下列矩阵乘积3201242110013*2 0*( 2)0*21*012*24*01*2 0*03*1(0*12*11*12)( 1)1( 1)4( 1)0( 1)3( 1) ( 2)20*( 1) 1*22*( 1) 4*2( 1)( 1) 0*265701242621131*2 2*1( 1)2 1*3 2( 1) ( 1)42312*21*1 0*22*3 1( 1) 0*437(2)4 _ 1*2 0*13*21*3 0( 1) 3*4_ 81503(3). (1,-11=(1*2+ (

16、-1) *1+2*4,1*1+(-1 ) *1+2*2, 1*0+ (-1) *3+2*1 =(9, 4, 1)a11碗bix(a12a21)a21a22bzy(4) (x,y,1)b1b2c 1=(x,y,1)x(a11xa11xa21xb1xa12 y b1 a22 y b2 b2y ca12 y bi)y(a21xa22 yb2)b1xb2y c(aux222a12xy a22 y2blx 2b2 y c)11011 0111(5)22102 111 22010 0111011011 0111 1二225. 设 A=B=A2B26322A2B23528A2,B2, A2B2与(AB)2A

17、B = 26624(AB)220346.cossin(1)A=sincosn=1 时A=n=2 时A2cossinn=3 时cossincossinsincoscossinsincosA3 = A2 A=假设Ansincossincossincossin 3cos 3cossincos 2sin 2sincossincoscossinsincos=1时，A1cossinsincoscos2 假设当n 2 时（ n 为自然数）成立，令n=k，Ak= sinsincos 成立；当n=k+1时sin kcoscos ksink sincos kk sinsin k(k 1)(k 1)cos nsin

18、sin ncoscos kAk 1 = Ak a= sin kcos k sin ksinsink coscoscos(k 1)sinsin(k 1)cos上当n微自然数时Ansincossin sin k cossin cos k cos1 1 0当 n=1 时，A10 1 1110 110当n=2时，A20 110 110 0 10 0 112 1110当n=3时，A30 110 110 0 10 0 11 n n(n- 2假设An= 0 1 n0 011 1 01当 n=1 时 A = 0 1 10 0 1假设n=k+1时1 1 00 1 1Ak 1AKAk(k 1)2k0010 0 1

19、k(k 1)1 1 k k -2=01 k 100111k 42=011 k 成立001. n(n 1)1 n 2综上当n为自然数时，An0 1n0 01a 1 0 00 a 1 0A0 0 a 10 0 0 a当A=2时A2a2 2a 100 a2 2a 100a2 2a000a2a3 3a2 3a 1n=3 时 A30 a3 3a2 3a00a3 3a2000a3n=4 时 A4a44a36 a24a0a44a36a200a44a3000a4a55a410a310a2n=5 时 A50a55a410a35400a55a4000a5假设 n 3 时成立Ann a000n nan a00Cn2

20、an 2n nan a0Cn3an 3Cn2an 2n1 nanan=3 时 A33 a0003a23 a003a3a23 a013a3a23 a假设n=k 时成立Akk a000kakn a00Ck2ak nak a0k2Ck3ak 3Ck2ak 2kakk an=k+1 时 akk a000kakk a00Ck2ak 2kak 1ka0k3k2k1 kakaa0000010a00a整理得k a000k1 kak1a00kakCk2a kak k1 a0k1Ck2akkak 1k1 a(kk1 a1)akk1a00Ck21a(k1)(k 1)akk1a0所以Ann nan a0Cn2an n

21、an an2Cn3aCn2aCk3ak 2Ck2a kak k1 aCk3 1a(kCk21a(kk11)1)(k 1)akk1an1 na2成立(n 3)000anaAn =000000a(n2 a2a10n an1 naCn2an 2Cn3an 302 a2a10n an1 naCn2an 22(n2)nn100a2a00ana0002 a000n a1)(n 3)7、已知B=证明BnE, 当 n 为偶数；B,n 为奇数证明：.1B200B2k(B2)kEkB2k 12kB2kB EB B- Bn= E,当n为偶数;B,当n为奇数8、证明两个n 阶上三角形矩阵的乘积仍为一个上三角形矩阵。证

22、明：设两个n 阶上三角形矩阵为 A,B,a11a12a1n-0a22'''a2n卜''''''''''''00'''annb11b12'''b1n0b22'''b2n''''''''''''00'''bnn、B=且根据矩阵乘法，有a11b11a11b12a12b22、a11b1na12b

23、2 n、annbnn0 AB=a22b22、a22b2nannbnnannbnn则可知 AB 为上三角形矩阵BA 也为上三角形矩阵。C 为同阶矩阵，且9、若AB=BA,AC=CA, 证明： A 、 B 、A(B+C)=(B+C)A,A(BC)=BCA.证：设A= (aij )m n ， B= ( Bij )n t ， C= (Cij )n s由题知AB、BA有意义，则可知必有 m=s,又由于AB=BA,且AB为mXn阶矩阵，则可知m=n所以A、B均为n阶矩阵。同理可知A C均为n阶矩阵，故可得A、B、C为同阶矩阵AB CAB ACBCABA CA又由于 ABBA, AC CABA CA=AB

24、AC A BA BCAB C B AC B CABCA10、已知n 阶矩阵 A 和 B 满足等式AB=BA1)2A222AB B22)3)Amm12 m2 2mA B CmA B LBm22A2 B 2m为正整数解： 1 A BABAB=A A AB BA B B22A2 AB BA B2222由于AB BA,则原式 AB A 2 AB B2ABABA2 AB BA B2由于 AB BA,则 AB BA 0故 A B A B =A2 B23 数学归纳法当 m 2时，A B 2 A2 2AB B2成立n1AB A设m n时成立,n 1 An 2B Cn2 1An 3B2L B当m n时，A B

25、n A B A Bn1n2n1A BAn 1 n 1 An 2B LBn 1Ann12 n2 23 n3 3n 1 A B Cn 1A B Cn 1A B LAn 1Bn 1 An 2B2 L BnAn nAn 1B Cn2 1n 1 An 2B2Cn3 1 Cn2 1 An 3B3 L BnAnnAn 1BCn2An2B2 LBn综上，A B m AmmAm 1BCm2Am2B2L Bm11、b11b12Lb1n解：由题知B必为n阶矩阵，设Bb21 L bn1b22 L bn2b2n L bnnBAaibiibi2Lbina2b2ib22Lb2nOLLLL则 ABbnna1b11a2b22O

26、anbnb11b12Lb1nb21b22Lb2nLLLL bn1bn2Lbnna2Ob11a1b21 a1Lb12 a2Lb22a2LLLb1nanb2 nanLbn1a1bn2 a2Lbnn an由于AB BA,且加a2,L ,小两两互不相等,则必有除bii, b22,L , %等元之外的元均为零,biibnn即B必为对角矩阵。i Aaij12、证明mn块,B分成一行为一块aiiai2Lainia2ia22La2n2,BMMMMamiam2La mnn即Amn,Bbijn s则ABalla12aina21a22a2namiam2amnaii1a122b1s Aainna2i1a222La2n

27、nMam11am22LamnnAB的第i个行向量为ai11ai22 Lain n,i 1,2,L ,m即A AA,L ,An ,Bb12Lb1sb22Lb2sM Mbn2LbnsT2若将A分成一列为一块，B分成n耿bnbi2Lb1sb21b22Lb2 sMM Mbn1bn 2Lbns11.b21ABA1, A2,L , AnMbn1b11 Alb21A2Lbn1Anb12Ab22A2Lbn2 Anb2sA2bnsAnAB的第j个列向量为bijA1b2jA2 LbnjAn, j1,2,L ,s13、a b c dbad cAc d a bd c b aAATb22 c000从而14、d2022b

28、 c00d2d2000222b c dAATb2d2AATxyiX2WMb2d2 2xy2X2y2MxynX2ynM记为Dncos 1Xn ynXn%Xn，2 时，D2 =x/X2y1xy2X2y2X1X2y1y23时,X1DnX2My140/IMy20 My30 Myn0 MXn故原行列式DnX1X2y10,ny231cos12cos121Dncos13cos23MMcos1ncos2ncos13Lcos1ncos23Lcos2n1Lcos3nMMcos3nL1当n=2时，D21cos12212当n 3时,则Dn1121Mn11222n21 n2nMnncos 1 sin 1 0 L 0co

29、s 1 cos 2 L L cos ncos 2 sin 2 0 L 0sin 1sin 2 L L sin nDnMMMM00 L L0MMMMMMMcos n sin n 0 L 000 L L0一一2 一故Dsin 12 , n2n0,n 301n. na b11n. na b2L1n. na bn1岫1ab1ah1n. na2 b11n. na2 b2L1n. na2 bnDn1a2b11a2b21a2bnMMM1n. nan b11n. nan b2L1n. nan bn1anb11anb21anbn22n 1 n 11 aibjaj bj Lajbj14La1n10当 n=1 时，

30、A1011001110 110121当 n=2 时，A1 20 110 110120 0 10 0 1001 12 1110当 n=3 时，A30 110 1111L11a2La2n1blb?LbnMM M M MMn 1 n 1 n 1n 11 an Lanb1b2Lbnaj aibj bi1 i j n1 i j naj ai bj bi1 i j n15、a b1 A, ad bc 0c d则A1-1-A*A1 d b ad bc c adbad bc bc adcabc ad ad bc 、 * 、 一 、一 . .其中A为A白T伴随矩阵（下同）cossin2 A sincoscoss

31、insincos123*A0121,A001. n(n 1)1 n -2假设An = 0 1n0 011 11当 n=1 时 A = 0 10 0k(k 1)21 1 0k0 1 110 0 1假设n=k+1时1 1 kAk 1 AK A 0100k(k 1)1 1 k k 2=01 k 100111kg2=011 k 成立0011 n n2综上当n为自然数时，An 0 1n0 01a 1 0 00 a 1 0A0 0 a 10 0 0 aa2 2a 10当A=2时A20 a2 2a 100a2 2a2000a2n=3 时 A3a33a23a10a33a23a00a33a2000a3n=4 时

32、 A4a44a36a24a0a44a36a200a44a3000a4n=5 时 A5a55a410a310a20a55a410a300a55a4000a5假设 n 3 时成立 Ann n1 2 n2 3 n3 a naCnaCnann1 2 n20anaCnann100 ana000ann=3 时 A3a33a23a10a33a23a00a33a2000a3假设n=k 时成立Akk k1 2 k2 3 k3 a ka Ck aCkank1 2 k20 a na Ck a 00akkak 1000akn=k+1 时 ak 1k k1 2 k2 3 k3a kaCkaCkaa 1 0kk1 2 k

33、20akaCka0a100akkak 100a000ak000000 akk1a a kak10a00k1 2 k1 kaCk akka kak1aCk2ak 2kak 1Ck3ak 2Ck2ak 1kk a ka000ak 1整理得k1 a(k 1)ak2 (k 1) 2Ck 1a3 (k 1) 3Ck 1a0k1 a(k 1)akk1a2 (k 1) 2Ck 1 ak 成立(k 1)ak00000k1 ak1 an n1 2 n2 anaCnaCn3an 3所以An0annan 100an000Cn2an 2n1 na(n3)x1 x2 3 62 4x3 x4 4 89 18a1002 a

34、2a10n an1 naCn2an 2Cn3an 3n0a1002 a2a10n an1 naCn2an 2A=00a0 (n1)002 a(n 2a2)00n an 1 (n na000a0002 a000n a3)16、( 1)x1 x2解：设2513xx3 x4x1 x24 6x3 x42 12x1 5x3 4 2x2 5x46x1 3x32x2 3x4 1 由得：x12; x223; x3 0; x4 8;223082）设xx1x2x33x1 4x22 6x1 8x2 4 3x3 4x4 96x3 8x4 1811x1 x1;x2(2 3x1); x3 x3；x4(9 3x3)441、

35、x1;(2 3x1)得：x 41 一八、x3;(9 3x3)4x1设x x2x32 31 x1212 0 x211 22 x332 X 3 x2 x32x1 2x2 01x1 2x2 2x33由方程组，得：x1 1;x21;x33x1x2(4)设 xx3x4x5x6312x1x2433x3x4130x5%3 91 117 53x1x32 x53x2 X4 2x6 94 X1 3X3 X514X2 3X4 3X6 11x1 3x37x2 3x4 5XiX4Xi；X2 X2；X3(7 X1)；311-(7 X2)； X5-(8 5x1); X6331-(8 5X2)；3(3)1、1、7日(7 Xi )(7 X2)行：x 3、173、2711-(8 5Xi) -(8 5x2) 33(5)XiX2X3设 XX4X5X6X7X8X9010X|x2x3100X4X5x6001X7X8x9x4x5x61