线性代数 同济第五版 习题答案

线性代数(同济四版)习题参考答案黄正华Email:huangzh@whu.edu.cn武汉大学数学与统计学院,湖北武汉430072WuhanUniversity目录第一章行列式1第二章矩阵及其运算17第三章矩阵的初等变换与线性方程组33第四章向量组的线性相关性48第五章相似矩阵及二次型69第一章行列式课后的习题值得我们仔细研读.本章建议重点看以下习题:5.(2),(5);7;8.(2).(这几个题号建立有超级链接.)若您发现有好的解法,请不吝告知.1.利用对角线法则计算下列三阶行列式:(1)∣∣∣∣∣∣∣∣2011−4−1−183∣∣∣∣∣∣∣∣;(2)∣∣∣∣∣∣∣∣abcbcacab∣∣∣∣∣∣∣∣;(3)∣∣∣∣∣∣∣∣111abca2b2c2∣∣∣∣∣∣∣∣;(4)∣∣∣∣∣∣∣∣xyx+yyx+yxx+yxy∣∣∣∣∣∣∣∣.解:(1)∣∣∣∣∣∣∣∣2011−4−1−183∣∣∣∣∣∣∣∣=2×(−4)×3+0×(−1)×(−1)+1×1×8−0×1×3−2×(−1)×8−1×(−4)×(−1)=−24+8+16−4=−4.(2)∣∣∣∣∣∣∣∣abcbcacab∣∣∣∣∣∣∣∣=acb+bac+cba−bbb−aaa−ccc=3abc−a3−b3−c3.(3)∣∣∣∣∣∣∣∣111abca2b2c2∣∣∣∣∣∣∣∣=bc2+ca2+ab2−ac2−ba2−cb2=(a−b)(b−c)(c−a).(4)∣∣∣∣∣∣∣∣xyx+yyx+yxx+yxy∣∣∣∣∣∣∣∣=x(x+y)y+yx(x+y)+(x+y)yx−y3−(x+y)3−x3=3xy(x+y)−y3−3x2y−3y2x−x3−y3−x3=−2(x3+y3).2.按自然数从小到大为 标准 excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载 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4:41,43,42,32.12第一章行列式(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)2:32...........................................................................1个52,54.....................................................................2个72,74,76..................................................................3个..................................................................................(2n−1)2,(2n−1)4,(2n−1)6,...,(2n−1)(2n−2)..............(n−1)个(6)逆序数为n(n−1):32...........................................................................1个52,54.....................................................................2个72,74,76..................................................................3个..................................................................................(2n−1)2,(2n−1)4,(2n−1)6,...,(2n−1)(2n−2)..............(n−1)个42...........................................................................1个62,64......................................................................2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a11a23的项.解:由定义知,四阶行列式的一般项为(−1)ta1p1a2p2a3p3a4p4,其中t为p1p2p3p4的逆序数．由于p1=1,p2=3已固定,p1p2p3p4只能形如13¤¤,即1324或1342.对应的逆序数t分别为0+0+1+0=1,或0+0+0+2=2.所以,−a11a23a32a44和a11a23a34a42为所求.4.计算下列各行列式:(1)∣∣∣∣∣∣∣∣∣∣41241202105200117∣∣∣∣∣∣∣∣∣∣;(2)∣∣∣∣∣∣∣∣∣∣21413−12112325062∣∣∣∣∣∣∣∣∣∣;(3)∣∣∣∣∣∣∣∣−abacaebd−cddebfcf−ef∣∣∣∣∣∣∣∣;(4)∣∣∣∣∣∣∣∣∣∣a100−1b100−1c100−1d∣∣∣∣∣∣∣∣∣∣.解:(1)∣∣∣∣∣∣∣∣∣∣41241202105200117∣∣∣∣∣∣∣∣∣∣r1↔r2======−∣∣∣∣∣∣∣∣∣∣12024124105200117∣∣∣∣∣∣∣∣∣∣r2−4r1=======r3−10r1−∣∣∣∣∣∣∣∣∣∣12020−72−40−152−200117∣∣∣∣∣∣∣∣∣∣线性代数(同济四版)习题参考答案3r2↔r4======∣∣∣∣∣∣∣∣∣∣120201170−152−200−72−4∣∣∣∣∣∣∣∣∣∣r4+7r2=======r3+15r2∣∣∣∣∣∣∣∣∣∣1202011700178500945∣∣∣∣∣∣∣∣∣∣=17×9∣∣∣∣∣∣∣∣∣∣1202011700150015∣∣∣∣∣∣∣∣∣∣=0.(2)∣∣∣∣∣∣∣∣∣∣21413−12112325062∣∣∣∣∣∣∣∣∣∣c4−c2=====∣∣∣∣∣∣∣∣∣∣21403−12212305062∣∣∣∣∣∣∣∣∣∣r4−r2=====∣∣∣∣∣∣∣∣∣∣21403−12212302140∣∣∣∣∣∣∣∣∣∣r4−r1=====∣∣∣∣∣∣∣∣∣∣21403−12212300000∣∣∣∣∣∣∣∣∣∣=0.(3)∣∣∣∣∣∣∣∣−abacaebd−cddebfcf−ef∣∣∣∣∣∣∣∣=adf∣∣∣∣∣∣∣∣−bceb−cebc−e∣∣∣∣∣∣∣∣=adfbce∣∣∣∣∣∣∣∣−1111−1111−1∣∣∣∣∣∣∣∣r2+r1=====r3+r1adfbce∣∣∣∣∣∣∣∣−111002020∣∣∣∣∣∣∣∣=−adfbce∣∣∣∣∣0220∣∣∣∣∣=4abcdef.(4)∣∣∣∣∣∣∣∣∣∣a100−1b100−1c100−1d∣∣∣∣∣∣∣∣∣∣r1+ar2======∣∣∣∣∣∣∣∣∣∣01+aba0−1b100−1c100−1d∣∣∣∣∣∣∣∣∣∣按第1列========展开(−1)(−1)2+1∣∣∣∣∣∣∣∣1+aba0−1c10−1d∣∣∣∣∣∣∣∣c3+dc2======∣∣∣∣∣∣∣∣1+abaad−1c1+cd0−10∣∣∣∣∣∣∣∣按第3行========展开(−1)(−1)3+2∣∣∣∣∣1+abad−11+cd∣∣∣∣∣=abcd+ab+cd+ad+1.5.证明:(1)∣∣∣∣∣∣∣∣a2abb22aa+b2b111∣∣∣∣∣∣∣∣=(a−b)3;证明∣∣∣∣∣∣∣∣a2abb22aa+b2b111∣∣∣∣∣∣∣∣c2−c1=====c3−c1∣∣∣∣∣∣∣∣a2ab−a2b2−a22ab−a2b−2a100∣∣∣∣∣∣∣∣=(−1)3+1∣∣∣∣∣ab−a2b2−a2b−a2b−2a∣∣∣∣∣=(b−a)(b−a)∣∣∣∣∣ab+a12∣∣∣∣∣=(a−b)3.(2)∣∣∣∣∣∣∣∣ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz∣∣∣∣∣∣∣∣=(a3+b3)∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣;4第一章行列式证明:∣∣∣∣∣∣∣∣ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz∣∣∣∣∣∣∣∣按第1列========分裂开a∣∣∣∣∣∣∣∣xay+bzaz+bxyaz+bxax+byzax+byay+bz∣∣∣∣∣∣∣∣+b∣∣∣∣∣∣∣∣yay+bzaz+bxzaz+bxax+byxax+byay+bz∣∣∣∣∣∣∣∣再次====裂开a2∣∣∣∣∣∣∣∣xay+bzzyaz+bxxzax+byy∣∣∣∣∣∣∣∣+0+0+b2∣∣∣∣∣∣∣∣yzaz+bxzxax+byxyay+bz∣∣∣∣∣∣∣∣再次====裂开a3∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣+b3∣∣∣∣∣∣∣∣yzxzxyxyz∣∣∣∣∣∣∣∣=a3∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣+b3(−1)2∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣=(a3+b3)∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣.此题有一个“经典”的解法:∣∣∣∣∣∣∣∣ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bz∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣axayazayazaxazaxay∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣bybzbxbzbxbybxbybz∣∣∣∣∣∣∣∣=a3∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣+b3∣∣∣∣∣∣∣∣yzxzxyxyz∣∣∣∣∣∣∣∣=a3∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣+b3(−1)2∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣=(a3+b3)∣∣∣∣∣∣∣∣xyzyzxzxy∣∣∣∣∣∣∣∣.这个解法“看上去很美”,实则是一个错解!我们强调,行列式不能作这种.形.式.上的加法:∣∣∣∣∣∣∣∣a11...a1n.........an1···ann∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣b11...b1n.........bn1···bnn∣∣∣∣∣∣∣∣=∣∣∣∣∣∣∣∣a11+b11...a1n+b1n.........an1+bn1···ann+bnn∣∣∣∣∣∣∣∣.(3)∣∣∣∣∣∣∣∣∣∣a2(a+1)2(a+2)2(a+3)2b2(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∣∣∣∣∣∣∣∣∣∣=0;证明:∣∣∣∣∣∣∣∣∣∣a2(a+1)2(a+2)2(a+3)2b2(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∣∣∣∣∣∣∣∣∣∣cj−c1======j=2,3,4∣∣∣∣∣∣∣∣∣∣a22a+14a+46a+9b22b+14b+46b+9c22c+14c+46c+9d22d+14d+46d+9∣∣∣∣∣∣∣∣∣∣c3−2c2======c4−3c2∣∣∣∣∣∣∣∣∣∣a22a+126b22b+126c22c+126d22d+126∣∣∣∣∣∣∣∣∣∣两列成比例=========0.线性代数(同济四版)习题参考答案5(4)∣∣∣∣∣∣∣∣∣∣1111abcda2b2c2d2a4b4c4d4∣∣∣∣∣∣∣∣∣∣=(a−b)(a−c)(a−d)(b−c)(b−d)(c−d)(a+b+c+d);证明:∣∣∣∣∣∣∣∣∣∣1111abcda2b2c2d2a4b4c4d4∣∣∣∣∣∣∣∣∣∣cj−c1======j=2,3,4∣∣∣∣∣∣∣∣∣∣1000ab−ac−ad−aa2b2−a2c2−a2d2−a2a4b4−a4c4−a4d4−a4∣∣∣∣∣∣∣∣∣∣展开r1======∣∣∣∣∣∣∣∣b−ac−ad−ab2−a2c2−a2d2−a2b2(b2−a2)c2(c2−a2)d2(d2−a2)∣∣∣∣∣∣∣∣=(b−a)(c−a)(d−a)∣∣∣∣∣∣∣∣111b+ac+ad+ab2(b+a)c2(c+a)d2(d+a)∣∣∣∣∣∣∣∣c2−c1=====c3−c1(b−a)(c−a)(d−a)∣∣∣∣∣∣∣∣100b+ac−bd−bb2(b+a)c2(c+a)−b2(b+a)d2(d+a)−b2(b+a)∣∣∣∣∣∣∣∣展开r1======(b−a)(c−a)(d−a)(c−b)(d−b)∣∣∣∣∣11(c2+bc+b2)+a(c+b)(d2+bd+b2)+a(d+b)∣∣∣∣∣=(a−b)(a−c)(a−d)(b−c)(b−d)(c−d)(a+b+c+d).(5)∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···000x−1···00...............000···x−1anan−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣=xn+a1xn−1+···+an−1x+an.证明:方法一.设法把主对角线上的x变为0,再按第一列展开.Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···0000x−1···000..................000···x−10000···0x−1anan−1an−2···a3a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cn−1+xcn========∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···0000x−1···000..................000···x−10000···00−1anan−1an−2···a3x2+a1x+a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣6第一章行列式cn−2+xcn−1==========∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···0000x−1···000..................000···0−10000···00−1anan−1an−2···x3+a1x3+a2x+a3x2+a1x+a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cj+xcj−1========∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0−1···0000···00............00···−1000···0−1xn+a1xn−1+···+an−1x+anxn−1+a1xn−2+···+an−2x+an−1···x2+a1x+a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=(xn+a1xn−1+···+an−1x+an)(−1)n+1∣∣∣∣∣∣∣∣∣∣∣∣∣−1···000···00.........0···−100···0−1∣∣∣∣∣∣∣∣∣∣∣∣∣=(xn+a1xn−1+···+an−1x+an)(−1)n+1(−1)n−1=xn+a1xn−1+···+an−1x+an.方法二.设法把−1全部变为0,得到一个下三角矩阵.若x=0,则Dn=an.等式成立.若x6=0,则Dnc2+1xc1=======∣∣∣∣∣∣∣∣∣∣∣∣∣x00···000x−1···00...............000···x−1anan−1+anxan−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣c3+1xc2=======∣∣∣∣∣∣∣∣∣∣∣∣∣x00···000x0···00...............000···x−1anan−1+anxan−2+an−1x+anx2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣=···=∣∣∣∣∣∣∣∣∣∣∣∣∣x00···000x0···00...............000···x0anan−1+anxan−2+an−1x+anx2···P2P1∣∣∣∣∣∣∣∣∣∣∣∣∣这里,P2=a2+a3x+a4x2+···++anxn−2,线性代数(同济四版)习题参考答案7P1=x+a1+a2x+a3x2+···++anxn−1.得到下三角阵,所以Dn=xn−1·P1=xn+a1xn−1+···+an−1x+an.方法三.用递归法证明.记Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···000x−1···00...............000···x−1anan−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣,则Dn展开c1======x∣∣∣∣∣∣∣∣∣∣∣x−1···00............00···x−1an−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣+an(−1)n+1∣∣∣∣∣∣∣∣∣∣∣−10···00x−1···00............00···x−1∣∣∣∣∣∣∣∣∣∣∣=xDn−1+an(−1)n+1(−1)n−1=xDn−1+an.所以,Dn=xDn−1+an.由此递归式得Dn=xn+a1xn−1+···+an−1x+an.方法四.按最后一行展开.先看an−i的代数余子式.因为Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−1x−1x......−1x−1x−1x......−1x−1anan−1an−2···an−(i−1)an−ian−(i+1)a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣划掉an−i所在的行和所在的列,左上角是i×i的方块,右下角是(n−i−1)×(n−i−1)的方块,余下全为0.则an−i的::代::数::余::子::式为(注意到an−i处在第n行、i+1列)(−1)n+i+1∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−1x−1x......−1x∣∣∣∣∣∣∣∣∣∣∣∣∣∣i×i∣∣∣∣∣∣∣∣∣∣∣−1x......−1x−1∣∣∣∣∣∣∣∣∣∣∣(n−i−1)×(n−i−1)=xi所以,Dn按最后一行展开,得到Dn=an+an−1x+an−2x2+···+an−ixi+···+a2xn−2+(x+a1)xn−18第一章行列式=xn+a1xn−1+···+an−1x+an.方法五.针对c1作变换.Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣x−10···000x−1···0000x···00...............000···x−1anan−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣c1+xc2======∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0−10···00x2x−1···0000x···00...............000···x−1an+an−1xan−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣c1+x2c3=======∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0−10···000x−1···00x30x···00...............000···x−1an+an−1x+an−2x2an−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=···=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0−10···000x−1···0000x···00...............000···x−1Pan−1an−2···a2x+a1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣,这里,P=an+an−1x+an−2x2+···+a1xn−1+xn.再按第一列展开,得Dn=xn+a1xn−1+···+an−1x+an.6.设n阶行列式D=det(aij),把D上下翻转、或逆时针旋转90◦、或依副对角线翻转,依次得D1=∣∣∣∣∣∣∣∣an1···ann......a11···a1n∣∣∣∣∣∣∣∣,D2=∣∣∣∣∣∣∣∣a1n···ann......a11···an1∣∣∣∣∣∣∣∣,D3=∣∣∣∣∣∣∣∣ann···a1n......an1···a11∣∣∣∣∣∣∣∣,证明D1=D2=(−1)n(n−1)2D,D3=D.证明:D1=∣∣∣∣∣∣∣∣an1···ann......a11···a1n∣∣∣∣∣∣∣∣n−1次行的相邻互换===============使rn换到第一行(−1)n−1∣∣∣∣∣∣∣∣∣∣∣a11···a1nan1···ann......a21···a2n∣∣∣∣∣∣∣∣∣∣∣线性代数(同济四版)习题参考答案9n−2次行的相邻互换===============使rn换到第二行(−1)n−1(−1)n−2∣∣∣∣∣∣∣∣∣∣∣∣∣a11···a1na21···a2nan1···ann......a31···a3n∣∣∣∣∣∣∣∣∣∣∣∣∣=···=(−1)n−1(−1)n−2···(−1)∣∣∣∣∣∣∣∣a11···a1n......an1···ann∣∣∣∣∣∣∣∣=(−1)1+2+···+(n−2)+(n−1)D=(−1)n(n−1)2D.同理可证D2=(−1)n(n−1)2∣∣∣∣∣∣∣∣a11···an1......a1n···ann∣∣∣∣∣∣∣∣=(−1)n(n−1)2DT=(−1)n(n−1)2D.D3=(−1)n(n−1)2D2=(−1)n(n−1)2(−1)n(n−1)2D=(−1)n(n−1)D=D.7.计算下列各行列式(Dk为k阶行列式)：(1)Dn=∣∣∣∣∣∣∣∣a1...1a∣∣∣∣∣∣∣∣,其中对角线上元素都是a,未写出的元素都是0;解:方法一.将cn作n−1次列的相邻对换,移到第二列:Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣a0···010a···00............00···a010···0a∣∣∣∣∣∣∣∣∣∣∣∣∣=(−1)n−1∣∣∣∣∣∣∣∣∣∣∣∣∣a0···0110···0a0a···00............00···a0∣∣∣∣∣∣∣∣∣∣∣∣∣再将rn作n−1次行的相邻对换,移到第二行:Dn=(−1)n−1(−1)n−1∣∣∣∣∣∣∣∣∣∣∣∣∣a10···01a0···000a···0............000···a∣∣∣∣∣∣∣∣∣∣∣∣∣=∣∣∣∣∣a11a∣∣∣∣∣∣∣∣∣∣∣∣∣a···0......0···a∣∣∣∣∣∣∣∣(n−2)×(n−2)=(a2−1)an−2.方法二.Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣a0···010a···00............00···a010···0a∣∣∣∣∣∣∣∣∣∣∣∣∣展开c1======a∣∣∣∣∣∣∣∣a...a∣∣∣∣∣∣∣∣(n−1)×(n−1)+1×(−1)n+1∣∣∣∣∣∣∣∣∣∣∣0···01a···00.........0···a0∣∣∣∣∣∣∣∣∣∣∣(n−1)×(n−1)10第一章行列式展开r1======an+(−1)n+1×1×(−1)(n−1)+1∣∣∣∣∣∣∣∣a...a∣∣∣∣∣∣∣∣(n−2)×(n−2)=an−an−2.(2)Dn=∣∣∣∣∣∣∣∣∣∣∣xa···aax···a.........aa···x∣∣∣∣∣∣∣∣∣∣∣;解:方法一.将第一行乘(−1)分别加到其余各行,得Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣xaa···aa−xx−a0···0a−x0x−a···0............a−x00···x−a∣∣∣∣∣∣∣∣∣∣∣∣∣,再将各列都加到第一列上,得Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣x+(n−1)aaa···a0x−a0···000x−a···0............000···x−a∣∣∣∣∣∣∣∣∣∣∣∣∣=[x+(n−1)a](x−a)n−1.方法二.将各列都加到第一列得Dn=∣∣∣∣∣∣∣∣∣∣∣x+(n−1)aa···ax+(n−1)ax···a.........x+(n−1)aa···x∣∣∣∣∣∣∣∣∣∣∣=[x+(n−1)a]∣∣∣∣∣∣∣∣∣∣∣1a···a1x···a.........1a···x∣∣∣∣∣∣∣∣∣∣∣再将第一行乘以(−1)分别加到其余各行,得Dn=[x+(n−1)a]∣∣∣∣∣∣∣∣∣∣∣∣∣1aa···a0x−a0···000x−a···0............000···x−a∣∣∣∣∣∣∣∣∣∣∣∣∣=[x+(n−1)a](x−a)n−1.方法三.升阶法.Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣1aa···a0xa···a0ax···a............0aa···x∣∣∣∣∣∣∣∣∣∣∣∣∣(n+1)×(n+1)ri−r1=======i=2,3,···∣∣∣∣∣∣∣∣∣∣∣∣∣1aa···a−1x−a0···0−10x−a···0............−100···x−a∣∣∣∣∣∣∣∣∣∣∣∣∣(n+1)×(n+1)线性代数(同济四版)习题参考答案11若x=a,则Dn=0.若x6=a,则将1x−acj加到c1,j=2,3,···,n+1:Dn=∣∣∣∣∣∣∣∣∣∣∣∣∣1+ax−anaa···a0x−a0···000x−a···0............000···x−a∣∣∣∣∣∣∣∣∣∣∣∣∣(n+1)×(n+1)=(1+nax−a)(x−a)n=[x+(n−1)a](x−a)n−1.(3)Dn+1=∣∣∣∣∣∣∣∣∣∣∣∣∣an(a−1)n···(a−n)nan−1(a−1)n−1···(a−n)n−1.........aa−1···a−n11···1∣∣∣∣∣∣∣∣∣∣∣∣∣;(提示:利用范德蒙德行列式的结果.)解: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)+···+1=n(n+1)2次行交换,得(或者直接由题6的结论)Dn+1=(−1)n(n+1)2∣∣∣∣∣∣∣∣∣∣∣∣∣11···1aa−1···a−n.........an−1(a−1)n−1···(a−n)n−1an(a−1)n···(a−n)n∣∣∣∣∣∣∣∣∣∣∣∣∣,此行列式为范德蒙德行列式.对照范德蒙德行列式的写法知,这里的a=x1,a−1=x2,...,a−(n−1)=xn,a−n=xn+1.则xi=a−(i−1),xj=a−(j−1).所以Dn+1=(−1)n(n+1)2∏n+1>i>j>1(xi−xj)=(−1)n(n+1)2∏n+1>i>j>1[(a−i+1)−(a−j+1)]=(−1)n(n+1)2∏n+1>i>j>1[−(i−j)]=(−1)n(n+1)2×(−1)n+(n−1)+···+1×∏n+1>i>j>1(i−j)=∏n+1>i>j>1(i−j).(4)D2n=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣an0bn......0a1b1c1d10......cn0dn∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣;解:方法一.将c2n作2n−1次列的相邻对换,移到第二列;再将r2n作2n−1次行的相邻对换,移到12第一章行列式第二行:D2n=(−1)2n−1(−1)2n−1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣anbncndnan−1bn−1......a1b1c1d1......cn−1dn−1∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣=(andn−bncn)D2(n−1),又n=1时D2=∣∣∣∣∣a1b1c1d1∣∣∣∣∣=a1d1−b1c1,所以D2n=(andn−bncn)···(a1d1−b1c1)=n∏i=1(aidi−bici).这个方法与教材P.15的例11相同.本题的第(1)小题也用到了此方法.方法二.D2n展开r1======an∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣an−10bn−10......0a1b1c1d10.........cn−10dn−100···0dn∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣+(−1)2n+1bn∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0an−10bn−1......0...a1b1c1d10......0cn−1dn−1cn000∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣展开c2n−1=========andnD2n−2−bncnD2n−2.由此得递推公式:D2n=(andn−bncn)D2n−2,即D2n=n∏i=2(aidi−bici)D2.而D2=∣∣∣∣∣a1b1c1d1∣∣∣∣∣=a1d1−b1c1,得D2n=n∏i=1(aidi−bici).(5)Dn=det(aij),其中aij=|i−j|;解:由aij=|i−j|得Dn=det(aij)=∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣0123···n−11012···n−22101···n−3...............n−2n−3n−4n−5···1n−1n−2n−3n−4···0∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣线性代数(同济四版)习题参考答案13ri−ri+1=======i=1,2,···∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣−1111···1−1−111···1−1−1−11···1...............−1−1−1−1···1n−1n−2n−3n−4···0∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣cj+c1=======j=2,3,···∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣−1000···0−1−200··&