线性代数课后习题答案全习题详解.docx

线性代数课后习题答案全习题详解 ( 总 9 2 页)-本页仅作为文档封面，使用时请直接删除即可-内页可以根据需求调整合适字体及大小-第一章行列式1. 利用对角线法则计算下列三阶行列式：201abc111xyx + y（1） 1- 4-1 ；（2） bca ;（3）abc；（4）yx + yx.-183caba 2b2c2x + yxy解 （1）2011- 4-1 = 2 ´ (-4) ´ 3 + 0 ´ (-1) ´ (-1) + 1´1´ 8 - 0 ´1´ 3 - 2 ´ (-1) ´ 8 - 1´ (-4) ´ (-1)-183= - 24 + 8 + 16 - 4 = - 4abc（2） bca = acb + bac + cba - bbb - aaa - ccc = 3abc - a3 - b3 - c3cab111（3） abc = bc2 + ca2 + ab2 - ac2 - ba2 - cb2 = (a - b)(b - c)(c - a)a2 b2c2xyx + y13（4）yx + yx= x(x + y) y + yx(x + y) + (x + y) yx - y3 - (x + y)3 - x3x + yxy= 3xy(x + y) - y3 - 3x2 y - 3 y 2 x - x3 - y3 - x3= -2(x3 + y3 )2. 按自然数从小到大为标准次序，求下列各排列的逆序数：（1）1234；（2）4132；（3）3421；（4）2413；（5）13(2n - 1)24 (2n) ；（6）13(2n - 1)(2n)(2n - 2)2.解（1）逆序数为 0（2）逆序数为 4：41，43，42，32（3）逆序数为 5：32，31，42，41,21（4）逆序数为 3：21，41，43（5）逆序数为 n(n - 1) ：2321 个52，542 个72，74，763 个(2n - 1) 2， (2n - 1)（6）逆序数为n(n - 1)4， (2n - 1)6， (2n - 1)(2n - 2)(n -1) 个3252，541 个2 个(2n - 1) 2， (2n - 1) 4， (2n - 1) 6， (2n - 1)(2n - 2)(n -1) 个421 个62，642 个(2n) 2， (2n) 4， (2n) 6， (2n) (2n - 2)(n -1) 个3. 写出四阶行列式中含有因子 a a11 23的项.解由定义知，四阶行列式的一般项为(-1)t aa1 p 2 p12aa3 p4 p34，其中t 为 p p p p1 2 3 4的逆序数由于 p1= 1, p2= 3 已固定， p p p p1 2 3 4只能形如13 ，即 1324 或 1342.对应的t 分别为0 + 0 + 1 + 0 = 1或0 + 0 + 0 + 2 = 2 - a a a a11 23 32 44和 a a a a11 23 34 42为所求.4. 计算下列各行列式：ê 41 2 4úê214 1úê- abacae úê 12 0 2úê3 -1 2（1）； （2）1ú ； （3） ê bd- cdde ú ； （4）êúêúê10 5 2 0úê123 2úê bfcf- ef úêë 01 1 7úûêë506 2úûëûê aê-10êêêë 0解10b1-1c0-1410ú0ú1úúd úû1 22 042c - c4-1122 -104- 1 - 104-1 1002(1) 10 5 2 0231032 -14 = 122 ´ (-1)4+3 = 12- 201 1c - 7c7430010103- 1410314c + c23c + 1 c991000- 2 =017 17 1412 3(2)214 1c - c3 -1 2 1214 03 -1 2 2214 0r - r3 -1 2 2214r - r3 -1 20123422123402123401123050625062214000002 =0- abacae- bce- 111(3) bd- cdde = adfb- ce = adfbce1- 11 = 4abcdefbfcf- efbc- e11- 1a10001 + aba0-1b10r + ar-1b100-1c1120-1c100-1d00-1d1 + aba0(4)= (-1)(-1)2+1-1c1c + dc1 + abaad1 + abad0-1 d32- 1c1 + cd = (-1)(-1)3+20- 10- 11 + cd = abcd + ab + cd + ad + 1a2abb25.证明:(1) 2a1a + b12b = (a - b)3 ;1ax + by ay + bz az + bxxyz(2) ay + bz az + bx ax + by = (a3 + b3 ) yzx ;az + bx ax + by ay + bzzxya2(3) b2c2d 2(a + 1)2(b + 1)2(c + 1)2(d + 1)2(a + 2)2(b + 2)2(c + 2)2(d + 2)2(a + 3)2(b + 3)2 = 0 ;(c + 3)2(d + 3)2(4) (4)1111abcd= (a - b)(a - c)(a - d )(b - c)(b - d ) × (c - d )(a + b + c + d ) ;a4b4c4d 4x-10L000x-1L00(5) LLLLLL= xn + a xn-1 + L + ax + a .a2 b2c2d 2000Lx-1n-1n1aann-1aL an-22x + a1证明c - cc - c2131a2ab - a2b2 - a2ab - a2b2 - a2a b + a(1) 左边 =2ab - a2b - 2a = (-1)3+1100= (a - b)3 = 右边b - a2b - 2a= (b - a)(b - a) 12(2) 左边按第一列分开x ay + bz az + bxy ay + bz az + bxa y az + bxax + by+ b zaz + bxax + byz ax + byay + bzxax + byay + bz分别再分x ay + bzzyzaz + bx分别再分xyzyzxa2 y az + bxx + 0 + 0 + b zx ax + bya3 yzx + b3 zxy zax + byyxy ay + bzzxyxyzxyzxyz= a3 yzx + b3 yzx (-1)2 = 右边zxyzxyc - cc - c21c - c3141a2a2 + (2a + 1)(a + 2)2(a + 3)2b2b2 + (2b + 1)(b + 2)2(b + 3)2c2c2 + (2c + 1)(c + 2)2(c + 3)2d 2d 2 + (2d + 1)(d + 2)2(d + 3)2a22a + 14a + 46a + 9b22b + 14b + 46b + 9c22c + 14c + 46c + 9d 22d + 14d + 46d + 9(3) 左边 =a2a4a + 46a + 9a21 4a + 46a + 9按第二列2 b2 分成二项 c2b 4b + 4c 4c + 4d 2第一项c- 4cc - 6c3242第二项c- 4cc - 9c324210左边 =ab -a2a4b2 -b4 -d 4d + 46b + 9 + b26c + 9c26d + 9d 21 4b + 41 4c + 41 4d + 46b + 96c + 96d + 9a2ab2bc2cd 2d4 9a24 9 + b24 9c24 9d 21 4a 6a1 4b6b = 0 1 4c6c1 4d 6d00ac - ad - ab - ac - ad - a(4) (4)a2c2 - a2d 2 - a2 =b2 - a2c2 - a2d 2 - a2a4c4 - a4d 4 - a4b2 (b2 - a2 ) c2 (c2 - a2 ) d 2 (d 2 - a2 )111= (b - a)(c - a)(d - a)b + ac + ad + ab2 (b + a) c2 (c + a) d 2 (d + a)100= (b - a)(c - a)(d - a) ´b + ac - bd - bb2 (b + a) c2 (c + a) - b2 (b + a) d 2 (d + a) - b2 (b + a)= (b - a)(c - a)(d - a)(c - b)(d - b) ´11(c2 + bc + b2 ) + a(c + b) (d 2 + bd + b2 ) + a(d + b)= (a - b)(a - c)(a - d )(b - c)(b - d ) (c - d )(a + b + c + d )(5) 用数学归纳法证明x- 1当n = 2时, D = x2 + a x + a , 命题成立.2ax + a1221假设对于(n -1) 阶行列式命题成立，即Dn-1= xn-1 + a xn-2 + L + a1n-2x + a,n-1则D 按第1列展开 :n- 10L00x- 1 L00D = xDnn-1+ a (-1)n+1nL1L L L1LxL = xD- 1+ an-1n= 右边所以，对于n 阶行列式命题成立.6. 设n 阶行列式 D = det(aij) ,把 D 上下翻转、或逆时针旋转90 o 、或依副对角线翻转，依次得aD = Mn11a11L a2Mnn ， DL a1na= M1na11L aMnnL an1a， D =Mnn3an1L aM1n ，L a11证明 D = D12= (-1)n(n-1)2D, D3= D .证明Q D = det(a )ijLaL a111aL aaa n111nLaan1nnaL a212n D = MM= (-1)n-1 n1nn = (-1)n-1 (-1)n-2 aa= L1aL aMMn1nnMM111naL a21aL a2naL a313n111nn( n-1)= (-1)n-1 (-1)n-2 L(-1) MM= (-1)1+2+L+( n-2)+( n-1) D = (-1) 2DaL an1nnn( n-1) 2aL a11n1n( n-1)n( n-1)同理可证 D2= (-1)MM= (-1) 2DT = (-1) 2DaL a1nnnD = (-1)3n(n-1)2D2= (-1)n(n-1)2(-1)n(n-1)2D = (-1)n(n-1) D = D7. 计算下列各行列式（ D 为k阶行列式 ）：k(1) Dna1=O,其中对角线上元素都是a ，未写出的元素都是 0；1x aa xaLLa aLLLLaaLx(2) D =;nan(a -1)nan-1 (a -1)n-1(3) D= LLn+1aa -111L(a - n)nL (a - n)n-1LL;提示：利用范德蒙德行列式的结果La - nL1abn O0Nnab(4) D= 0110 ;2ncdN1 0 1 Ocdnn(5) Dn= det(aij), 其中aij= i - j ;1 + a11L1(6) D =n解11 + a2LL11L1LLL 1 + an, 其中a a La1 2n¹ 0 .=a00L010a0L0000aL00LLLLLL000La0100L0a(1) Dn按最后一行展开( 再按第一行展开)0a(-1)n+1 0L0a00 L00 La0 LL L L00 L010000L La0(n-1)´(n-1)a+ (-1)2n × aOa (n-1)(n-1)= (-1)n+1 × (-1)nOa(n-2)(n-2)+ an = an - an-2 = an-2 (a 2 - 1)(2) 将第一行乘(-1) 分别加到其余各行，得xaaLa a - x x - a0L0D = a - x0x - anLLLa - x00L0LL0x - a再将各列都加到第一列上，得x + (n -1)aaaLa0x - a0L0D =00x - anLLL000L0= x + (n - 1)a(x - a)n-1LL0x - a(3) 从第n +1行开始，第n + 1行经过n 次相邻对换，换到第 1 行，第n 行经(n -1) 次对换换到第 2 行，经n + (n - 1) + L + 1 = n(n + 1) 次行交换，得2Dn+1n(n+1)1aL1a - 1LLLL1a - nLan-1(a - 1)n-1L(a - n)n-1an(a - 1)nL(a - n)n= (-1) 2此行列式为范德蒙德行列式Dn+1n(n+1)= (-1) 2Õ (a - i + 1) - (a - j + 1)n+1³i> j³1n( n+1)= (-1)2Õ-(i - j) = (-1) n( n+1) · (-1)n+( n-1)+L+1 ·2n+1³ i > j³1Õ(i - j)n+1³ i > j³1=Õ (i - j)n+1³i> j³1a0bn ONnab(4) D= 0112ncdN11 Oc0dnn按第一行aa0n-1 ON0ab11cdb0n-10 M+ (-1)2n+1 b00an-1abON011cdn-1b0展开nNcn-111 O0d0n-1nNcn-111 O0dn-10L0dc0nn都按最后一行展开a d D- b c Dn n2n-2n n2n-2由此得递推公式：D= (a d - b c )D2 nn nn n2 n-2即D= Õn (a d2ni ii=2ab- b c )Di i2而D =11 = a d- b c2cd111 11 1D= Õn (a d - b c )得(5) aij= i - j2ni ii i i=1230123Ln - 1- 1111L11012Ln - 2- 1- 111L12101Ln - 3r - r- 1- 1- 11L13210Ln - 412r - r ,L- 1- 1- 1- 1L1D = det(a ) =nijLLLLLLn - 1n - 2n - 3n - 4L0LLLLL Lc + c , c + c2131c + c ,L41n - 1 n - 2 n - 3 n - 4 L0- 1000L0- 1- 200L0- 1- 2- 20L0- 1- 2- 2- 2L0LLLLLLn - 12n - 32n - 42n - 5Ln - 11 + a1L11 1 1 + aL1ac - c , c - c1223c - c ,L-00L001a0L0011a-L001L001220aa332(6) D =nLL11LLL 1 + a00- a34LL4LLLLLn000L -100L0n-1- n-1n1 + an00L000a0L000aa0aa-1aL000L00022按最后一列(1 + a展开（由下往上）n)(a a1 2L an-10- aa-a) -00 33LL4LLLLL000000LL- an-20an-20- an000L00- aa0L00a0L0020- 2aaL00a-1a223333+ 0- aaLLLL00LLL+ L +00- a4LLLL00LLL000L000L- an-10a-n-1000L - aLn-1an-1a000n0- an= (1 + an)(a a1 2L an-1) + a a1 2L an-3aan-2 n+ L + a a L a2 3n= (a a La1 2n)(1+ ån 1 )ai=1i8. 用克莱姆法则解下列方程组：ì5x+ 6x= 1,ìx + x + x + xï 1234= 5,ï12ï x + 5x + 6x= 0,ïx + 2x - x + 4x = -2,ï123(1)í1234(2)íx + 5x+ 6x= 0,234ï2x- 3x- x - 5x= -2,ï2341îï3ïx + 5x + 6x = 0,345îx + x12+ 2x3+ 11x4= 0;ïx4+ 5x5= 1.111111111 1111 111解(1) D = 12- 14 = 01- 23= 0 1- 23 = 0 1 - 23= -1422 - 331- 1 - 50 - 5 - 3 - 70 0 - 132110 - 2 - 180 0- 580 0 - 1 - 54140 00142511151111- 5-1- 91- 5-1- 9D = - 22- 14 = 0509 = 0509= 012111- 2 - 301- 1 - 5- 2 - 3 21101- 1 - 50 -13 - 3 - 230509211012110 -13 - 3 - 231- 5- 1- 9012111 - 5 - 1 - 900-10- 4600- 1380023120000142= 01211= -1421511151115111511D = 1 - 2 - 14 = 0- 7- 23 = 0 -132 = 0 - 132= -28422 - 2 - 1 - 50 - 12 - 3 - 700302110 - 15- 180023 110039 3100- 1- 190- 284115112- 242- 3- 2- 531011D = -426;D3411= 122 - 33115- 1 - 2 = 142- 1 - 220DDDDx =1 = 1 ,x =2 = 2 ,x =3 = 3 ,x =4 = -11D2D3D4D按最后一行展开5 6 0 01 5 6 05600015600015600015600015(2)D =5D¢ - 0 1 5 0 = 5D¢ - 6D ¢0 0 1 6= 5(5D ¢ - 6D ¢) - 6D ¢ = 19D ¢ - 30D ¢ = 65D ¢ -114D ¢¢ = 65 ´19 - 114 ´ 5 = 6651116000056000156000156100155100010600005600015601015116000560015600156( D¢为行列式D中a 的余子式, D ¢为D¢中a¢ 的余子式, D ¢, D ¢ ¢类推)D =按第一列1展开D¢ += D¢ + 64 = 19D ¢ - 30 ¢ ¢ + 64 = 15071 6 0D =按第二列 - 0 5 605 00 - 1 60 05 6 00 0 = 1 5 6 - 5 ´ 632展开0 1 50 0 160 550 16 00 1 55 6按第三列展开5610015000010600005600115= -65 - 1080 = -1145D =31 5 00 1 60 0 50 0 105 6 0 016= 05010 + 1 5 0 060 1 6 050 0 5 605 6 06 + 6 1 5 050 1 6= 19 + 6 ´114 = 7035601015600D =0150040010600015按第四列展开1 5 6- 0 1 50 0 10 0 005 6 0 0560= -5 - 6156= -3950150 - 1 5 6 060 1 5 050 0 1 65 6 01 5 6D = 0 1 5001500001150 110 06 0 按最后一列0展开005 6 01 5 60 1 50 0 1+ D¢ = 1 + 211 = 212x = 1507 ;1665x = - 1145 ;2665x = 703 ;3665x = - 395 ;4665x = 212 4665ìlx + x + x = 0xí9. 问l, m取何值时, 齐次线性方程组 ï1 m 2 + 3 = 0 有非零解+ xxï 123îx + 2mx + x = 0l11123解D = 1m1 = m - ml ，齐次线性方程组有非零解，则 D = 031 2m 13即m - ml = 0得m = 0或l = 1不难验证，当m = 0或l = 1时, 该齐次线性方程组确有非零解.ì(1 - l)x- 2x+ 4x = 010. 问l 取何值时, 齐次线性方程组ï2x + ( 1 l) 2 + x 3í3 -x=0有非零解ï123îx + x + (1 - l)x = 0123解D =1 - l2- 2 3 - l1141 - l - 3 + l41=21 - l11 - l101 - l= (1 - l)3 + (l - 3) - 4(1 - l) - 2(1 - l)(-3 - l) = (1 - l)3 + 2(1 - l)2 + l - 3齐次线性方程组有非零解，则 D = 0得l = 0, l = 2或l = 3不难验证，当l = 0, l = 2或l = 3 时，该齐次线性方程组确有非零解.第二章矩阵及其运算1 已知线性变换ìïx = 2y + 2y + yíx1 =3y1 + y 2+5y3ïîx2 =3y1 + 22y +33y求从变量xx x 到变量y3123y y 的线性变换123123解 由已知æ x öæ2 2 1öæ y öç x1 ÷=ç3 1 5÷ç y1 ÷ç x2 ÷ç3 2 3÷ç y2 ÷32èøèøèøæ y öæ2 2 1ö-1æ x öæ-7 - 49 öæ y öç y1 ÷=ç3 1 5÷ ç x1 ÷ =ç 63-7÷ç y1 ÷故ç y2 ÷ç3 2 3÷ ç x2 ÷ç 32- 4÷ç y2 ÷2øèøèø è 3 øèøè 3ìïy =-7x - 4x +9xíy1 =6x +1 3x -2 7x 3ïîy2 =3x1+ 2x2 - 4x33123252 已知两个线性变换ìïx = 2y + yìïy =-3z + zíx1