线性代数(同济第五版)习题答案.pdf

线性代数(同济四版)习题参考答案黄正华Email: huangzh@武汉大学 数学与统计学院, 湖北 武汉 430072Wuhan University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)flflflflflflflfl2011−4−1−183flflflflflflflfl;(2)flflflflflflflflabcbcacabflflflflflflflfl;(3)flflflflflflflfl111abca2b2c2flflflflflflflfl;(4)flflflflflflflflxyx + yyx + yxx + yxyflflflflflflflfl.解: (1)flflflflflflflfl2011−4−1−183flflflflflflflfl= 2 × (−4) × 3 + 0 × (−1) × (−1) + 1 × 1 × 8 − 0 × 1 × 3 − 2 × (−1) × 8 − 1 × (−4) × (−1)= −24 + 8 + 16 − 4= −4.(2)flflflflflflflflabcbcacabflflflflflflflfl= acb + bac + cba − bbb − aaa − ccc = 3abc − a3− b3− c3.(3)flflflflflflflfl111abca2b2c2flflflflflflflfl= bc2+ ca2+ ab2− ac2− ba2− cb2= (a − b)(b − c)(c − a).(4)flflflflflflflflxyx + yyx + yxx + yxyflflflflflflflfl= 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次序,求下列各排列的逆序数: (1) 1 2 3 4;(2) 4 1 3 2;(3) 3 4 2 1;(4) 2 4 1 3;(5) 1 3···(2n − 1) 2 4···(2n);(6) 1 3···(2n − 1) (2n) (2n − 2)···2. 解(1)逆序数为0.(2)逆序数为4: 4 1, 4 3, 4 2, 3 2.12第一章 行列式(3)逆序数为5: 3 2, 3 1, 4 2, 4 1, 2 1.(4)逆序数为3: 2 1, 4 1, 4 3.(5)逆序数为n(n−1) 2:3 2...........................................................................1个5 2, 5 4 ..................................................................... 2个7 2, 7 4, 7 6..................................................................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):3 2...........................................................................1个5 2, 5 4 ..................................................................... 2个7 2, 7 4, 7 6..................................................................3个..................................................................................(2n − 1) 2, (2n − 1) 4, (2n − 1) 6, ..., (2n − 1) (2n − 2)..............(n − 1)个4 2...........................................................................1个6 2, 6 4 ...................................................................... 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)flflflflflflflflflfl41241202105200117flflflflflflflflflfl;(2)flflflflflflflflflfl21413−12112325062flflflflflflflflflfl;(3)flflflflflflflfl−abacaebd−cddebfcf−efflflflflflflflfl;(4)flflflflflflflflflfla100−1b100−1c100−1dflflflflflflflflflfl.解: (1)flflflflflflflflflfl41241202105200117flflflflflflflflflflr1↔r2= = = = = = −flflflflflflflflflfl12024124105200117flflflflflflflflflflr2−4r1= = = = = = =r3−10r1−flflflflflflflflflfl12020−72−40−152−200117flflflflflflflflflfl线性代数(同济四版)习题参考答案3r2↔r4= = = = = =flflflflflflflflflfl120201170−152−200−72−4flflflflflflflflflflr4+7r2= = = = = = =r3+15r2flflflflflflflflflfl1202011700178500945flflflflflflflflflfl= 17 × 9flflflflflflflflflfl1202011700150015flflflflflflflflflfl= 0.(2)flflflflflflflflflfl21413−12112325062flflflflflflflflflflc4−c2= = = = =flflflflflflflflflfl21403−12212305062flflflflflflflflflflr4−r2= = = = =flflflflflflflflflfl21403−12212302140flflflflflflflflflflr4−r1= = = = =flflflflflflflflflfl21403−12212300000flflflflflflflflflfl= 0.(3)flflflflflflflfl−abacaebd−cddebfcf−efflflflflflflflfl= adfflflflflflflflfl−bceb−cebc−eflflflflflflflfl= adfbceflflflflflflflfl−1111−1111−1flflflflflflflflr2+r1= = = = =r3+r1adfbceflflflflflflflfl−111002020flflflflflflflfl= −adfbceflflflflfl0220flflflflfl= 4abcdef.(4)flflflflflflflflflfla100−1b100−1c100−1dflflflflflflflflflflr1+ar2= = = = = =flflflflflflflflflfl01 + aba0−1b100−1c100−1dflflflflflflflflflfl按第1列= = = = = = = = 展开(−1)(−1)2+1flflflflflflflfl1 + aba0−1c10−1dflflflflflflflflc3+dc2= = = = = =flflflflflflflfl1 + abaad−1c1 + cd0−10flflflflflflflfl按第3行= = = = = = = = 展开(−1)(−1)3+2flflflflfl1 + abad−11 + cdflflflflfl= abcd + ab + cd + ad + 1.5 .证明:(1)flflflflflflflfla2abb22aa + b2b111flflflflflflflfl= (a − b)3;证明flflflflflflflfla2abb22aa + b2b111flflflflflflflflc2−c1= = = = =c3−c1flflflflflflflfla2ab − a2b2− a22ab − a2b − 2a100flflflflflflflfl=(−1)3+1flflflflflab − a2b2− a2b − a2b − 2aflflflflfl= (b − a)(b − a)flflflflflab + a12flflflflfl= (a − b)3.(2)flflflflflflflflax + byay + bzaz + bxay + bzaz + bxax + byaz + bxax + byay + bzflflflflflflflfl= (a3+ b3)flflflflflflflflxyzyzxzxyflflflflflflflfl;4第一章 行列式证明:flflflflflflflflax + byay + bzaz + bxay + bzaz + bxax + byaz + bxax + byay + bzflflflflflflflfl按第1列= = = = = = = = 分裂开aflflflflflflflflxay + bzaz + bxyaz + bxax + byzax + byay + bzflflflflflflflfl+ bflflflflflflflflyay + bzaz + bxzaz + bxax + byxax + 。