《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3811411022(4)30(1)(1)1180132(1)81(4)(1)2481644 (2)bacacbcba解bacacbcbaacbbaccbabbbaaaccc3abca3b3?c3(3)222111cbacba解222111cbacbabc2ca2ab2?ac2ba2cb2(ab)(bc)(ca)(4)yxyxxyxyyxyx解yxyxxyxyyxyxx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2 yx3y3x32(x3y3) 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 解逆序数为 0(2)4 1 3 2 解逆序数为 4 41 43 42 32(3)3 4 2 1 解逆序数为 5 3 2 3 1 4 2 4 1, 2 1(4)2 4 1 3 解逆序数为 3 2 1 4 1 4 3(5)1 3 (2n1) 2 4 (2n) 解逆序数为2)1(nn3 2 (1 个)5 2 5 4(2 个)7 2 7 4 7 6(3 个)(2n1)2 (2n1)4 (2n1)6 (2n1)(2n2) (n1 个 )(6)1 3 (2n1) (2n) (2n2) 2 解逆序数为n(n1) 3 2(1 个 )5 2 5 4 (2 个)(2n1)2 (2n1)4 (2n1)6 (2n1)(2n2) (n1 个 )4 2(1 个 )6 2 6 4(2 个)(2n)2 (2n)4 (2n)6 (2n)(2n2) (n1 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?a11a23a34a424 计算下列各行列式(1)71100251020214214解71100251020214214010014231020211021473234cccc34)1(14310221101414310221101401417172001099323211cccc(2)2605232112131412解2605232112131412260503212213041224cc041203212213041224rr0000003212213041214rr(3)efcfbfdecdbdaeacab解efcfbfdecdbdaeacabecbecbecbadfabcdefadfbce4111111111(4)dcba100110011001解dcba100110011001dcbaabarr10011001101021dcaab101101)1)(1(1201011123cdcadaabdcccdadab111)1)(1(23abcdabcdad1 5 证明 :(1)1112222bbaababa(ab)3;证明1112222bbaababa00122222221213ababaabaabaccccabababaab22)1(2221321))((abaabab(ab)3(2)yxzxzyzyxbabzaybyaxbxazbyaxbxazbzaybxazbzaybyax)(33;证明bzaybyaxbxazbyaxbxazbzaybxazbzaybyaxbzaybyaxxbyaxbxazzbxazbzayybbzaybyaxzbyaxbxazybxazbzayxabzayyxbyaxxzbxazzybybyaxzxbxazyzbzayxa22zyxyxzxzybyxzxzyzyxa33yxzxzyzyxbyxzxzyzyxa33yxzxzyzyxba)(33(3)0...
