高等代数习题及答案(1)

精编WORD文档下载可编缉打印下载文档，远离加班熬夜高等代数习题及答案(1)篇一：高等代数习题解答(第一章)高等代数习题解答第一章多项式补充题1．当a,b,c取何值时，多项式f(x)?x?5与g(x)?a(x?2)2?b(x?1)?c(x2?x?2)相等？6136提示：比较系数得a??,b??,c?.555补充题2．设f(x),g(x),h(x)??[x]，f2(x)?xg2(x)?x3h2(x)，证明：f(x)?g(x)?h(x)?0．证明假设f(x)?g(x)?h(x)?0不成立．若f(x)?0，则?(f2(x))为偶数，又g2(x),h2(x)等于0或次数为偶数，由于g2(x),h2(x)??[x]，首项系数（如果有的话）为正数，从而xg2(x)?x3h2(x)等于0或次数为奇数，矛盾．若g(x)?0或h(x)?0则?(xg2(x)?x3h2(x))为奇数，而f2(x)?0或?(f2(x))为偶数，矛盾．综上所证，f(x)?g(x)?h(x)?0．1．用g(x)除f(x)，求商q(x)与余式r(x)：1）f(x)=x3-3x2-x-1，g(x)=3x2-2x+1；2）f(x)=x4-2x+5，g(x)=x2-x+2．1）解法一待定系数法．由于f(x)是首项系数为1的3次多项式，而g(x)是首项系数为3的2次多项式，精编WORD文档下载可编缉打印下载文档，远离加班熬夜1所以商q(x)必是首项系数为的1次多项式，而余式的次数小于2．于是可设31q(x)=x+a,r(x)=bx+c3根据f(x)=q(x)g(x)+r(x)，即1x3-3x2-x-1=(x+a)(3x2-2x+1)+bx+c3右边展开，合并同类项，再比较两边同次幂的系数，得21?3?3a?,?1??2a??b,?1?a?c337262解得a??,b??,c??，故得99917262q(x)?x?,r(x)??x?.3999解法二带余除法．3-211-3-1-11???213374?-1337147?399262?9917?39?得17262q(x)?x?,r(x)??x?.39992）q(x)?x2?x?1,r(x)??5x?7.r(x)??2．m,p,q适合什么条件时，有1）x2?mx?1x3?px?q;2）x2?mx?1x4?px2?q.?1除x3?px1）解x2?mx得余式为：?q262x?.99精编WORD文档下载可编缉打印下载文档，远离加班熬夜r(x)?(p?m2?1)x?(q?m)，?p?m2?1?0;令r(x)?0，即??q?m?0.故x2?mx?1x3?px?q的充要条件是?m?q;?2p?m?1?0.??1除x4?px2?q得余式为：2）解x2?mxr(x)??m(p?m2?2)x?(q?p?m2?1)，2???m(p?m?2)?0;令r(x)?0，即?2??q?p?m?1?0.解得x2?mx?1x4?px2?q的充要条件是?m?0;?或p?q?1??q?1;?2p?2?m.?3．求g(x)除f(x)的商q(x)与余式r(x)：1）f(x)?2x5?5x3?8x,g(x)?x?3;2）f(x)?x3?x2?x,g(x)?x?1?2i.1）解法一用带余除法（略）．解法二用综合除法．写出按降幂排列的系数，缺项的系数为0：-320-50-80+-618-39117-3272-613-39109-327所以q(x)?2x4?6x3?13x2?39x?109,r(x)??327.2）解法一用带余除法（略）．解法二用综合除法．写出按降幂排列的系数，缺项的系数为0：精编WORD文档下载可编缉打印下载文档，远离加班熬夜f(x)1-2i1-1-10+1-2i-4-2i-9+8i1-2i-5-2i-9+8i所以q(x)?2i8.x?2ix?(5?2i),r(x?)??94．把f(x)表成x?x0的方幂和，即表成c0?c1(x?x0)?c2(x?x0)2??的形式：1）f(x)?x5,x0?1;2）f(x)?x4?2x2?3,x0??2;3）f(x)?x4?2ix3?(1?i)x2?3x?7?i,x0??i.注设f(x)表成c0?c1(x?x)?c(x?20x)??的形式，则c0就是f(x)被x?x0除02所得的余数，c1就是f(x)被x?x0除所得的商式c1?c2(x?x)?c(x?20x)??再被03x?x0除所得的余数，逐次进行综合除法即可得到c0,c1,?,cn.1）解综合除法进行计算1100000+111111111111精编WORD文档下载可编缉打印下载文档，远离加班熬夜+1234123451+136136101+141410115所以x5?1?5(x?1)?1x0(?21?)x10?3(2）3）略5．求f(x)与g(x)的最大公因式：1）f(x)?x4?x3?3x2?4x?1,g(x)?x3?x2?x?1;2）f(x)?x4?4x3?1,g(x)?x3?3x2?1;3）f(x)?x4?10x2?1,g(x)?x4?3?6x2??1.1）解用辗转相除法g(x)f(x)11q2(x)?11-1-111-3-4-1q1(x)10244?5(?x1)5?(?1x)1).13111-1-12284??-1r1(x)-2-3-1q3(x)2233131???-2-224433r2(x)??-1-144-1-1精编WORD文档下载可编缉打印下载文档，远离加班熬夜r3(x)0所以(f(x),g(x))?x?1.2）(f(x),g(x))?1.3）(f(x),g(x))?x2??1.6．求u(x),v(x)使u(x)f(x)?v(x)g(x)?(f(x),g(x)):1）f(x)?x4?2x3?x2?4x?2,g(x)?x4?x3?x2?2x?2；2）f(x)?4x4?2x3?16x2?5x?9,g(x)?2x3?x2?5x?4；3）f(x)?x4?x3?4x2?4x?1,g(x)?x2?x?1．1）解用辗转相除法g(x)f(x)q2(x)1111-1-2-212-1-4-2q1(x)110-2011-1-211-2-2r1(x)10-2q3(x)110-2010-2r2(x)10-2r3(x)0由以上计算得f(x)?q1(x)g(x)?r1(x),g(x)?q2(x)r1(x)?r2(x),r1(x)?q3(x)r2(x),因此(f(x),g(x))?r2(x)?x2?2，且精编WORD文档下载可编缉打印下载文档，远离加班熬夜(f(x),g(x))?r2(x)?g(x)?q2(x)r1(x)?g(x)?q2(x)[f(x)?q1(x)g(x)]??q2(x)f(x)?[1?1q(x2)q(x)g]x()所以u(x)??q2(x)??x?1,v(x)?1?q1(x)q2(x)?x?2．0篇二：高等代数期末卷1及答案沈阳农业大学理学院第一学期期末考试《高等代数》试卷（1）一、填空（共35分，每题5分）421．设f(x)?x?x?4x?9,则f(?3)?2．当t?时,f(x)?x3?3x?t有重因式。23.令f(x),g(x)是两个多项式,且f(x3)?xg(x3)被x?x?1整除,则f(1)?g(1)?3?14.行列式11000262?。111103精编WORD文档下载可编缉打印下载文档，远离加班熬夜0??3??9?2?1????。1??9911??4??3?201?4??103?1???15.矩阵的积???2102??2??1?1?5?500????6.031??0???021??0?????1?0??1?1??23???0?x1?2x2?2x3?x4?0?7.?2x1?x2?2x3?2x4?0的一般解为?x?x?4x?3x?034?125?x?2x?x43??13,x3,x4?4?x??2x?x234?3?精编WORD文档下载可编缉打印下载文档，远离加班熬夜二、（10分）令f(x),g(x)是两个多项式。求证(f(x),g(x))?1当且仅当(f(x)?g(x),f(x)g(x))?1。证：必要性.设(f(x)?g(x),f(x)g(x))?1。（1%）令p(x)为f(x)?g(x),f(x)g(x)的不可约公因式，（1%）则由p(x)|f(x)g(x)知p(x)|f(x)或p(x)|g(x)。(1%)不妨设p(x)|f(x)，再由p(x)|(f(x)?g(x))得p(x)|g(x)。故p(x)|1矛盾。(2%)充分性.由(f(x)?g(x),f(x)g(x))?1知存在多项式u(x),v(x)使u(x)(f(x)?g(x))?v(x)f(x)g(x)?1,(2%)从而u(x)f(x)?g(x)(u(x)?v(x)f(x))?1,(2%)故(f(x),g(x))?1。(1%)三、(16分)a,b取何值时，线性方程组ax1?bx2?2x3?1???ax1?(2b?1)x2?3x3?1?ax?bx?(b?3)x?2b?123?1有唯一解、没有解、有无穷解？在有解情况下求其解。解：b21??ab21??a????a2b?131?0b?110?????a??bb?32b?1??00b?12b?2???(5%)01??a2?b精编WORD文档下载可编缉打印下载文档，远离加班熬夜????0b?110??00b?12b?2???当a(b?1)?0时，有唯一解：x1?25?b?22b?2(4%)，x2?,x3?；a(b?1)b+1b?1当b?1时，有无穷解：x3?0,x2?1?ax1,x1任意取值；当a?0,b?5时，有无穷解：x1?k,x2??,x3?,k任意取值；(3%)当b??1或a?0且b??1且b?5时，无解。(4%)四、(10分)设a1,a2,...,an都是非零实数，证明?a111..111?a21..111..1...1...1.....111..?a1a2...an(1??i?1n1?a3...11)ai...11?an1a1精编WORD文档下载可编缉打印下载文档，远离加班熬夜证:对n用数学归纳法。当n=1时,D1?1?a1?a1(1?),结论成立(2%);假设n-1时成立。则n时?a1111?a21..111...11...11?a111..111?a21..111..1...1...1.....000..Dn=1..11?a3...11?.........1...111?a3...1...1an=a1a2...an?1?anDn?1(4%)?n?11?精编WORD文档下载可编缉打印下载文档，远离加班熬夜现由归纳假设Dn?1?a1a2...an?1?1???有?i?1ai??n?11?Dn=a1a2...an?1?anDn?1=a1a2...an?1?a1a2...an?1an?1????i?1ai?n?1?=a1a2...an?1an?1???,(3%)?i?1ai?故由归纳原理结论成立。(1%)五、(10分)证明f(x)?x4?1在有理数域上不可约。证:令x?y?1得(1%)g(y)?f(x)?y4?4y3?6y2?4y?2。(3%)取素数p=2满足2|2,2|4,2|6,2|4,且2不整除1,4不整除2.(2%)再据艾茵斯坦茵判别法知g(y)?y?4y?6y?4y?2在有理数域上不可约，（2%）432从而f(x)?x4?1在有理数域上不可约（2%）六、（9分）令A为数域F上秩为r的m?n矩阵，r?0。求证：精编WORD文档下载可编缉打印下载文档，远离加班熬夜存在秩为r的m?r矩阵F和秩为r的r?n矩阵G,使得A?FG。证:A为数域F上秩为r的m?n矩阵，r?0,则存在m?m可逆阵使P和n?n可逆阵Q?IA?P?r?0进而令0??Q.（3%）0??I?F?P?r?,G??Ir?0?就得A?FG（2%）.0?Q（4%）七、（10分）设A,B是n?n矩阵,且A?B,A?B可逆。求证2n?2n矩阵P??证:|P|??AB??1?可逆,且求P。?BA?ABA?BBA?B精编WORD文档下载可编缉打印下载文档，远离加班熬夜??BAB?AA0B?|A?B