线性代数习题解答-第三版-郑宝东-哈工大习题1.pdf

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1、习习题题一一1按自然数从小到大的自然次序，求解各题.(1)求1至6的全排列241356的逆序数.解解：t(241356)002100 3.(2)求1至2n的全排列135解：解：t(13(2n1)246(2n)的逆序数.210n(n1).2(2n1)242n)000(n1)(n2)(3)选择i与j，使由1至9的排列，91274i56 j成偶排列.解：解：由91274i56 j是从 1 至 9 的排列，所以i,j只能取3或8.当i 8,j 3时，t(912748563)0 111 2 1 3 3 6 18，是偶排列.当i 3,j 8时t(912743568)0 111 2 3 2 2 1 13，是

2、奇排列，不合题意舍去.(4)选择i与j，使由1至9的排列71i25j489成奇排列.解解：由71i25j489是从1至9的排列，所以i,j只能取3或6.当i 3,j 6时，t(713256489)0 11 2 11 3 0 0 9，是奇排列.当i 6,j 3时，t(716253489)0 11 2 2 3 3 0 0 12，是偶排列，不合题意舍去.2计算下列行列式(1)9a1018b81226b13a；(2)32153320537528475184abaccdcf；2aede.ef(3)153；(4)bdbf203212解解：(1)(2)9a18b26b13a3215332053913a2b2

3、ba3205310032053117(a24b2).320533205375184751841003205310075184752847518410812275184100751842223 0.075184003205300 4313100.(3)153 54 3320321248121ab(4)bdaccdcfaeef111111111102200de abcdef 11 abcdef 0bf111 abcdef 0020 4abcdef.02x3已知3yz02 1，利用行列式性质求下列行列式.y3yzx1y 1z 134xy111x(1)3x3x2x3z 2；(2)y2z 2y3yy2z

4、z 2xyz01z2.3解：解：(1)3x33z 2 302 2 302 2.222111x2(2)x1y1z 13401xyz1112 302 3023413413xyz11111102 302 10 1.31114用行列式定义计算：12(1)00；(2)1000020000.3450n12n10解解：(1)345(1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5t(54321)a15a24a33a42a51(1)1012345 120.(1)201(2)020nn10(1)t(p1p2pn)a1p1a2p2anpn(1)(1)5用行列式的定义证明：t(2 3n1)n1

5、a12a23a(n1)nan1n (1)n1n!123a11a12a21a22(1)000000a13a23000a14a24a34a44a54a15a25a35 0；a45a55a11a1200a33a4400a11a12a33a34a21a22a43a45a13a23000a14a24a34a44a54a21a22(2)a31a32a41a42a34.a44a11a12a21a22证：(1)D 000000a15a25a35(1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5a45a55假设有a1P1a2P2a3P3a4P4a5P5 0，由已知p3,p4,p5必等于4或

6、5，从而p3,p4,p5中至少有两个相等，这与p1,p2,p3,p4,p5是1,2,3,4,5的一个全排列矛盾，故所有项a1P1a2P2a3P3a4P4a5P5 0，因此D 0.a11a1200a33a4300(1)t(p1p2p3p4)a1p1a2p2a3p3a4p4，由已知，只有当p1,p2a34a44a21a22(2)a31a32a41a42取1或2时，a1p1a2 p2a3p3a4 p4 0，而p1,p2,p3,p4是1,2,3,4的一个全排列，故p3,p4取3或4，于是3D (1)t(1234)a11a22a33a44(1)t(1243)a11a22a34a43(1)t(2134)a

7、12a21a33a44(1)t(2143)a12a21a34a43 a11a22a33a44a11a22a34a43a12a21a33a44a12a21a34a43从而a11a21a12a33a22a43a34(a11a22a12a21)(a33a44a34a43)a44 a11a22a33a44a11a22a34a43a12a21a33a44a12a21a34a43 D6计算a30(1)523；(2)1022114；0b012c111214000dxa(3)Dn21110aax123n00；1111.1a30bax110；(4)Dn 101100111aaa000a0(5)Dn 00a1a0

8、解：(1)101(2)101110；(6)Dn 111a52 按第43行展开d(1)4403b201000c0141a30d 0b012c1612020311按第3列展开11166(1)33dc abcd.221210361160111021110331221421110003412 312112116 312011212 31201121271 9.01020037003003xa0036000ar1r2(n1)a x(n1)a x(n1)a x(3)Dr1r3axnaxaaaxr1rnaa111axa(n 1)a x aaaaa11110 xa00(n1)a x 00 xa0000 xa(

9、n1)a x(xa)n1.123n123n1100c1c20(4)Dn 1010c1c301001c1cn0123n n(n1)2.a0010a00(5)Dn 00a0100a23100100axn0015a000a(1)1n0a0000100a00按第1行展开(1)11a0a00a0 a(1)a ann2n1n(1)n11an2.11111111002111(6)Dn 1111(2)7证明10201 00210001n11(1)n12n1.a2(1)2aab1b21ab2b (ab)21a2abb2a2ababb2b2c1(1)c22aabab2b2b证证：2aab2bc2(1)c31110

10、01(1)33a(ab)b(ab)abab11ab(ab)3(a2)2(b2)2(c2)(d 2)22(ab)2(2)a2b2cd22(a1)2(b1)2(c1)(d 1)2a2b2c2d22(a3)2(b3)2(c3)(d 3)22 0证证：等式左端a22a1a24a4b22b1b24b4a26a9b26b9c22c1c24c4c26c9d22d 1d24d 4d26d 962a2a1c2(1)c1b22b1c3(1)c12c2c1c4(1)c12d2d 14a46a9a24b46b9c3(2)c2b222a1262b1264c46c9c4(3)c3c2c1264d 46d 9d22d 1

11、2601x1a11x2a1(3)1x3a11x4a1x12b1x1b22x2b1x2b22x3b1x3b22x4b1x4b2x13c1x12c2x1c332x2c1x2c2x2c332x3c1x3c2x3c31x332x4c1x4c2x4c31x41x11x2x122x2xx2324x133x2xx3334c2(a1)c11x11x2证证：等式左端c(b)c3211x3c4(c3)c11x4c3(b1)c21x2c4(c2)c21x31x4端.8解关于未知数x的方程x12b1x12x2b1x22x3b1x32x4b1x4x13c1x12c2x132x2c1x2c2x232x3c1x3c2x33

12、2x4c1x4c2x41x1x122x22x32x4x13c1x121x1x122x22x32x4x133x2等式右3x33x432c(c)cx2c1x2x2413132x3c1x31x332x4c1x41x4x(1)31x2026 0 x10 x解解：31x2026(x1)x10 x13x22(x1)x(x2)3(x1)x 2x3(x1)(x3)(x1)0所以x11,x2 3,x3 1.a(2)maxxbmm 0(m 0)b7a解解：maxxbaabx00 xa1bmm m 111 m 11xbbxb m(xa)因m 0，所以x1 a,x2 b.11bx m(xa)(xb)0a119设a12

13、a22an2a1na2nannanna2na1na1p2a2p2anp2a1pna2pnanpn，其中“”是对1,2,；(2)a21an1 a，求下列行列式：an1an2(1)a1na2nanna12a22an2a11a21an1；a21a11a22a12a1p1a2p1pn(3)p1p2,n的所有全排列p1p2pnanp1取和，n 2.解：解：（1）经行的交换得a11an1原式(1)n1a12an2a32a22a1nanna31a21a3na2na11(1)(n1)(n2)21a12a22a1na2nanna21an1an2(1)n(n1)2a.(2)与(1)类似，经列的交换得8n(n1)原

14、式(1)2a.(3)经列的交换，得a1p1a1p2a1pna11a12a1na2p1a2p2a2pn(1)(p1p2pn)a21a22a2n(1)(p1p2pn)aanp1anp2anpnan1an2ann1 11故原式1)(p1p2pn)a a1 11 0.p1p(2pn1 1110计算行列式aa000100b1a11aa00(1)0a12b200b；(2)011aa0；3a30b011aa400a0400011a611111000161110100(3)11611；(4)0010.11161000111116k000a100b1a100b1a1b100解：解：(1)0a2b2000a4b4

15、a40b3a30 b40b3a300000ab33b400a40a2b2000b2a2a1b1a3b3ba(a1a4b1b4)(a2a3b2b3).44b2a2(2)将前 4 行依次加到第 5 行，再按第 5 行展开得1aa00011aa001aa00原式011aa0 a511aa00011aa011aaa00010011a91aa001aa0 a511aa0011aa a5a4 11aaa001011a1aa0 a5a4 11aa a5a4a31aaa0111a1aa2a3a4a56111110101010101611116111(3)11611 1161111161111611111611

16、116111111111116111050001011611 10 00500111610005011116000051054 6250.(4)按最后一行展开得1000010010001000010 k1000100001010001k000001000 k 511计算行列式1111x1111x11x1mx1x1(1)11x111；(2)x2x2mx21x1111x3x3x3mx11111x4x4x4解：解：(1)依次将第2,3,4,5列加到第 1 列得10 x1x2x3x4mx1111x1x111x11原式 x11x111x1x1111x111111111x1111x11(x1)11x111

17、1x1111111111000 x100 x0(x1)10 x001x000100004(41)(1)2(x1)x4 x4(x1)(2)依次将第2,3,4行加到第 1 行得4444xim1ximi1ximi1ximii1原式x2x2mx2x2x3x3x3mx3x4x4x4x4m11114(x mx2x2mx2x2i)i1x3x3x3mx3x4x4x4x4m1111(4x0m00im)i100m0000m4(mxi)m3i112计算行列式11a1b1a1b2a1b3a2b3a3b3a4b31a1b3a1b4a2b4；a3b4a4b41a1b4a2b1a2b2(1)a3b1a3b2a4b1a4b2

18、1a1b11a1b21a2b11a2b21a2b31a2b4(2)1a3b11a3b21a3b31a3b41a4b11a4b21a4b31a4b4(3)11000100a1a211；(4)1391xx2121411a301a4a1b11827x3解：（1）依次将第3,2,1行乘1加到第4,3,2行得a1b2a2a1a3a2a4a3a1b3a2a1a3a2a4a3a1b4a2a10a3a2a4a3a2a1原式a3a2a4a3(2)依次将第3,2,1行乘1加到第4,3,2行得1a1b11a1b21 a1b31 a1b4b1(a2a1)b2(a2a1)b3(a2a1)b4(a2a1)原式b1(a3a

19、2)b2(a3a2)b3(a3a2)b4(a3a2)b1(a4a3)b2(a4a3)b3(a4a3)b4(a4a3)1a1b11a1b21a1b31a1b4(a2a1)(a3a2)(a4a3)b1b1b1b2b2b21000b3b3b301b4b4b4101100(3)按最后一列展开得1原式a4101010000110a3110a2011a1001101a1a2a3a4(4)由 Vandermonde 行列式的计算公式得12原式(x3)(x2)(x1)(32)(31)(21)2(x1)(x2)(x3)13证明a11000a2x100(1)Da30 x00n2n a11xna2xan1xanan

20、100 x1an000 xa110000a2x1000ra0 x000证：等式左端n(x)rn13an200 x10an1000 x1anan1x000 x20a110000a2x1000rn(x2)rn2a30 x000rx3n()rn3rn1n(x)rnan200 x10an1000 x1f(x)000001x1(1)n1f(x)f(x)1x1(n1)阶其中f(x)an11xan1xan.13210121(2)D 0000002112 n1012000000证证：1n 1时，D1 2 112假 设 当n k时 结 论 成 立，当n k 1时，若k 1 2，D22112 413 21结论成立

21、.若k 13，将Dk1按第一行展开得21121Dk1 2DkDk1 2(k 1)(k 11)(k 1)1112由数学归纳法，对一切自然数n结论成立.1a1(3)D 11111an1111an1a10010a20111a2111ai(1i1i1nn1),ai 0,i 1,2,ai,n.证证：(用加边法)111101a111等式左端 011a2101111a1 1010100an10a20100an111a1a200011an11411(1a1a21)a1a2ananai(1i 1i 1nn1)等式右端.aixy1(4)Dnxyxy1000 xyxy00000 xy1000 xyxyxn 1yn

22、1，其中xy.xy000 x2y2证证：当n1时，D1xy，等式成立.xy假 设nk时 等 式 成 立，当nk 1时，若k 12，则x3y3Dk 1D2xxyy，等式成立.若k 13，将Dk 1按一列展开，得xyxyxy00022Dk 1(xy)(1)1 1100 xy10 xyxy000001xyk阶(1)2 1xy1000 xy100 xyxy00000001xyk阶xk 1yk 1xkykx(k 1)1y(k 1)1(xy)DkxyDk 1(xy)xyxyxyxy由归纳法原理，等式对一切自然数n都成立.14 设f(x)是一个次数不大于n 1的一元多项式，证明如果存在n个互不相同的数a1,

23、a2,an使f(ai)0,i 1,2,n.则f(x)0.k1xk0，依题意有n 1n 2证：证：设f(x)kn 1xkn 2xk0a1k1ka kn10因a1,a2,a1n 1kn 10（1）n 1ankn 10,an互不相同，故(1)的系数行列式151a1D 1a21an所以关于k0,k1,a122a2a1n1n1a21i jn2ann1an(ajai)0，,kn1的线性方程组(1)只有零解，所以k0k1kn10,f(x)0.15用 Cramer 法则解方程组5x14x211(1)6x 5x 2012解解：D 2524 1 0，方程组有唯一解.65114511D15580 25，D210066 34，由克莱姆法则，205620DDx11 25，x2234DD5x16x21(2)x15x26x3 0 x 5x 0235605630530解解：D 156 1519 119015010 5(19)(30)1 65 0，方程组有唯一解.545105616D 056 256 19D 106 5，1，2150501500515D 150 1.301010所以由克莱姆法则得，x1160561D119D11，x22，x3.D1365D6516