《线性代数》(同济_第四版)课后习题答案

://《线性代数》（同济第四版）课后习题答案（完整版）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(−nn32(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+−×−−−=://−−=01417172001099323211=−++======cccc(2);2605232112131412−解2605232112131412−260503212213041224−−=====cc041203212213041224−−=====rr.0000003212213041214=−−=====rr(3);efcfbfdecdbdaeacab−−−解efcfbfdecdbdaeacab−−−ecbecbecbadf−−−=.abcdefadfbce4111111111=−−−=(4).dcba100110011001−−−解dcba100110011001−−−dcbaabarr10011001101021−−−++=====dcaab101101)1)(1(12−−+−−=+01011123−+−++=====cdcadaabdcc=abcd+ab+cd+ad+1.cdadab+−+−−=+111)1)(1(23://证明:(1)=(a−b)3;1112222bbaababa+证明1112222bbaababa+00122222221213ababaabaabacccc−−−−−−======(a−b)3.abababaab22)1(22213−−−−−=+21))((abaabab+−−=(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(c4−c3,c3−c2,c2−c1得)2222222222222222)3()2()1()3()2()1()3()2()1()3()2()1(++++++++++++ddddccccbbbbaaaa(c4−c3,c3−c2得)5232125232125232125232122222++++++++++++=ddddccccbbbbaaaa.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)=xn+a1xn−1+⋅⋅⋅+an−1x+an.1221100000100001axaaaaxxxnnn+⋅⋅⋅−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−−−⋯证明用数学归纳法证明.当n=2时,,命题成立.2121221axaxaxaxD++=+−=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://⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11112nnnnaaaaD⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=11113aaaaDnnnn⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=证明,D3=D.DDDnn2)1(21)1(−−==证明因为D=det(aij),所以nnnnnnnnnnaaaaaaaaaaD2211111111111)1(⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=−⋅⋅⋅=⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−−=−−)1()1(331122111121nnnnnnnnaaaaaaaa.DDnnnn2)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),其中对角线上元素都是a,未写出的元素aaDn11⋅⋅⋅=都是0;解://(按第n行展开)aaaaaDn00010000000000001000⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅=)1()1(100000000000010000)1(−×−+⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−=nnnaaa)1()1(2)1(−×−⋅⋅⋅⋅−+nnnaaa=an−an−2=an−2(a2−1).nnnnnaaa+⋅⋅⋅−⋅−=−−+)2)(2(1)1()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=[x+(n−1)a](x−a)n−1.axaxaxaaaanxDn−⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅−⋅⋅⋅−⋅⋅⋅−+=0000000000)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⋅⋅⋅⋅⋅⋅⋅