《线性代数》同济大学版 课后习题答案详解.

..《线性代数》同济大学版课后习题答案详解第一章行列式1利用对角线法则计算下列三阶行列式(1)381141102解3811411022(4)30(1)(1)1180132(1)81(4)(1)2481644(2)bacacbcba解bacacbcbaacbbaccbabbbaaaccc3abca3b3c3(3)222111cbacba解222111cbacbabc2ca2ab2ac2ba2cb2(ab)(bc)(ca)(4)yxyxxyxyyxyx解yxyxxyxyyxyxx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2y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)1234解逆序数为0(2)4132解逆序数为441434232(3)3421解逆序数为532314241,21(4)2413解逆序数为3214143(5)13(2n1)24(2n)解逆序数为2)1(nn..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)13(2n1)(2n)(2n2)2解逆序数为n(n1)32(1个)5254(2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解含因子a11a23的项的一般形式为(1)ta11a23a3ra4s其中rs是2和4构成的排列这种排列共有两个即24和42所以含因子a11a23的项分别是(1)ta11a23a32a44(1)1a11a23a32a44a11a23a32a44(1)ta11a23a34a42(1)2a11a23a34a42a11a23a34a424计算下列各行列式(1)71100251020214214解71100251020214214010014231020211021473234cccc34)1(14310221101414310221101401417172001099323211cccc(2)2605232112131412解2605232112131412260503212213041224cc041203212213041224rr0000003212213041214rr(3)efcfbfdecdbdaeacab..解efcfbfdecdbdaeacabecbecbecbadfabcdefadfbce4111111111(4)dcba100110011001解dcba100110011001dcbaabarr10011001101021dcaab101101)1)(1(1201011123cdcadaabdcccdadab111)1)(1(23abcdabcdad15证明:(1)1112222bbaababa(ab)3;证明1112222bbaababa00122222221213ababaabaabaccccabababaab22)1(2221321))((abaabab(ab)3(2)yxzxzyzyxbabzaybyaxbxazbyaxbxazbzaybxazbzaybyax)(33;证明bzaybyaxbxazbyaxbxazbzaybxazbzaybyaxbzaybyaxxbyaxbxazzbxazbzayybbzaybyaxzbyaxbxazybxazbzayxabzayyxbyaxxzbxazzybybyaxzxbxazyzbzayxa22zyxyxzxzybyxzxzyzyxa33..yxzxzyzyxbyxzxzyzyxa33yxzxzyzyxba)(33(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222ddddccccbbbbaaaa;证明2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(ddddccccbbbbaaaa(c4c3c3c2c2c1得)5232125232125232125232122222ddddccccbbbbaaaa(c4c3c3c2得)022122212221222122222ddccbbaa(4)444422221111dcbadcbadcba(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明444422221111dcbadcbadcba)()()(0)()()(001111222222222addaccabbaddaccabbadacab)()()(111))()((222addaccabbdcbadacab))(())((00111))()((abdbddabcbccbdbcadacab)()(11))()()()((abddabccbdbcadacab=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)..(5)1221100000100001axaaaaxxxnnnxna1xn1an1xan证明用数学归纳法证明当n2时2121221axaxaxaxD命题成立假设对于(n1)阶行列式命题成立即Dn1xn1a1xn2an2xan1则Dn按第一列展开有1110010001)1(11xxaxDDnnnnxDn1anxna1xn1an1xan因此对于n阶行列式命题成立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nnnnaaaaD1111111112nnnnaaaaD11113aaaaDnnnn证明DDDnn2)1(21)1(D3D证明因为Ddet(aij)所以nnnnnnnnnnaaaaaaaaaaD2211111111111)1()1()1(331122111121nnnnnnnnaaaaaaaaDDnnnn2)1()1()2(21)1()1(同理可证nnnnnnaaaaD)1(11112)1(2DDnnTnn2)1(2)1()1()1(DDDDDnnnnnnnn)1(2)1(2)1(22)1(3)1()1()1()1(7计算下列各行列式(Dk为k阶行列式)(1)aaDn11,其中对角线上元素都是a未写出的元素都是0解..aaaaaDn00010000000000001000(按第n行展开))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axxaaxxaaxxaaaaxDn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)1111)()1()()1(1111naaanaaanaaaDnnnnnnn;解根据第6题结果有nnnnnnnnnnaaanaaanaaaD)()1()()1(1111)1(1112)1(1此行列式为范德蒙德行列式112)1(1)]1()1[()1(jinnnnjaiaD..112)1()]([)1(jinnnji1121)1(2)1()()1()1(jinnnnnji11)(jinji(4)nnnnndcdcbabaD11112;解nnnnndcdcbabaD11112(按第1行展开)nnnnnnddcdcbabaa00001111111100)1(1111111112cdcdcbababnnnnnnn再按最后一行展开得递推公式D2nandnD2n2bncnD2n2即D2n(andnbncn)D2n2于是niiiiinDcbdaD222)(而111111112cbdadcbaD所以niiiiincbdaD12)((5)Ddet(aij)其中aij|ij|;解aij|ij|..0432140123310122210113210)det(nnnnnnnnaDijn04321111111111111111111112132nnnnrrrr15242321022210022100021000011213nnnnncccc(1)n1(n1)2n2(6)nnaaaD11111111121,其中a1a2an0解nnaaaD11111111121nnnnaaaaaaaaacccc