山东建筑大学线性代数作业答案

班级姓名学号1第一章行列式1.利用对角线法则计算下列三阶行列式：（1）bacacbcbacccaaabbbcbabacacb3333cbaabc（2）222111cbacba222222cbbaacabcabc))()((accbba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）2413；（2）13…)12(n24…)2(n；（3）13…)12(n)2(n)22(n…2.解（1）逆序数为3.（2）逆序数为2)1(nn.（3）逆序数为)1(nn.3.写出四阶行列式中含有因子2311aa的项.解由定义知，四阶行列式的一般项为班级姓名学号243214321)1(pppptaaaa，其中t为4321pppp的逆序数．由于3,121pp已固定，4321pppp只能形如13□□，即1324或1342.对应的t分别为10100或2200044322311aaaa和42342311aaaa为所求.4.计算下列各行列式：解(1)260523211213141224cc260503212213041224rr041203212213041214rr0000032122130412=0(2)efcfbfdecdbdaeacab=ecbecbecbadf班级姓名学号3=111111111adfbce=abcdef4(3)dcba10011001100121arrdcbaab100110011010=12)1)(1(dcaab10110123dcc010111cdcadaab=23)1)(1(cdadab111=1adcdababcd5、证明：(1)bzaybyaxzbyaxbxazybxazbzayxa分开按第一列左边班级姓名学号4bzaybyaxxbyaxbxazzbxazbzayyb002ybyaxzxbxazyzbzayxa分别再分bzayyxbyaxxzbxazzybzyxyxzxzybyxzxzyzyxa33分别再分右边233)1(yxzxzyzyxbyxzxzyzyxa(2)2222222222222222)3()2()12()3()2()12()3()2()12()3()2()12(dddddcccccbbbbbaaaaa左边9644129644129644129644122222141312ddddccccbbbbaaaacccccc964496449644964422222ddddccccbbbbaaaa分成二项按第二列班级姓名学号5964419644196441964412222dddcccbbbaaa949494949464222224232423ddccbbaacccccccc第二项第一项06416416416412222dddcccbbbaaa(3)444444422222220001adacabaadacabaadacaba左边=)()()(222222222222222addaccabbadacabadacab=)()()(111))()((222addaccabbadacabadacab=))()((adacab)()()()()(00122222abbaddabbaccabbbdbcab班级姓名学号6=))()()()((bdbcadacab)()()()(112222bdabbddbcabbcc=))()()()((dbcbdacaba))((dcbadc(4)用数学归纳法证明.,1,2212122命题成立时当axaxaxaxDn假设对于)1(n阶行列式命题成立，即,122111nnnnnaxaxaxD:1列展开按第则nD1110010001)1(11xxaxDDnnnn右边nnaxD1所以，对于n阶行列式命题成立.6、计算下列各行列式（kD为k阶行列式）：班级姓名学号7(1)aaDn11,其中对角线上元素都是a未写出的元素都是0解aaaaaDn00010000000000001000)1()1(100000000000010000)1(nnnaaa)1()1(2)1(nnnaaannnnnaaa)2)(2(1)1()1(anan2an2(a21)(2)xaaaxaaaxDn;解将第一行乘(1)分别加到其余各行得班级姓名学号8axxaaxxaaxxaaaaxDn0000000再将各列都加到第一列上得axaxaxaaaanxDn0000000000)1([x(n1)a](xa)n1(3)nnaaaD11111111121,,433221ccccccnnnnaaaaaaaaaa10000100010000100010001000011433221展开（由下往上）按最后一列班级姓名学号9))(1(121nnaaaannnaaaaaaaaa00000000000000000000000000022433221nnnaaaaaaaa000000000000000001133221nnnaaaaaaaa000000000000000001143322nnnnnnaaaaaaaaaaaa322321121))(1()11)((121niinaaaa班级姓名学号10(4)nnnnndcdcbabaD00011112nnnnnnddcdcbabaa0000000011111111展开按第一行000000)1(1111111112cdcdcbababnnnnnnn2222nnnnnnDcbDda都按最后一行展开由此得递推公式：222)(nnnnnnDcbdaD即niiiiinDcbdaD222)(班级姓名学号11而111111112cbdadcbaD得niiiiincbdaD12)((5)jiaij0432140123310122210113210)det(nnnnnnnnaDijn,3221rrrr0432111111111111111111111nnnn,,141312cccccc1524232102221002210002100001nnnnn=212)1()1(nnn班级姓名学号127.用克莱姆法则解下列方程组：解11213513241211111D81207350321011111450081300321011111421420005410032101111112105132412211151D1121051329050111511210233130905091512331309050112109151120230046100011210915114200038100112109151142112035122412111512D811507312032701151班级姓名学号13313900112300231011512842840001910023101151426110135232422115113D14202132132212151114D1,3,2,144332211DDxDDxDDxDDx8.齐次线性方程组取何值时问,,0200321321321xxxxxxxxx有非零解？解12111113D，齐次线性方程组有非零解，则03D即0,得班级姓名学号1410或不难验证，当,10时或该齐次线性方程组确有非零解.第二章矩阵及其运算1﹑已知两个线性变换,zzyzzyzzy,yyyxyyyxyyx32331221132133212311323542322求从变量321z,z,z到变量321x,x,x的线性变换。解由已知221321514232102yyyxxx321310102013514232102zzz321161109412316zzz所以有3213321232111610941236zzzxzzzxzzzx2﹑设,B,A150421321111111111求AAB23及BAT.解AAB2315042132111111111131111111112班级姓名学号150926508503111111111222942017222132150421321111111111BAT092650850.3﹑计算；⑴127075321134解：127075321134102775132)2(7111237449635.⑵21312解：2131223)1(321)1(122)1(2632142。4．设101A,求kAAA,,,32.解12011011012A;1301101120123AAA班级姓名学号16利用数学归纳法证明:101kAk当1k时,显然成立,假设k时成立,则1k时1)1(01101101kkAAAkk由数学归纳法原理知:101kAk.5﹑设,A001001求kA.解首先观察0010010010012A