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−cb=(a−b)(b−c)(c−a).2考试点www.kaoshidian.com(4)yxyxxyxyyxyx+++.解yxyxxyxyyxyx+++=x(x+y)y+yx(x+y)+(x+y)yx−y3−(x+y)3−x=3xy(x+y)−y33−3x2y−x3−y3−x=−2(x33+y32.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)考试点www.kaoshidian.com⋅⋅⋅⋅⋅⋅(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解含因子a的项.11a23(−1)的项的一般形式为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(−1)的项分别是ta11a23a32a44=(−1)1a11a23a32a44=−a11a23a32a44(−1),ta11a23a34a42=(−1)2a11a23a34a42=a11a23a34a424.计算下列各行列式:.考试点www.kaoshidian.com(1)71100251020214214;解71100251020214214010014231020211021473234−−−−−======cccc34)1(143102211014+−×−−−=143102211014−−=01417172001099323211=−++======cccc.(2)2605232112131412−;解2605232112131412−260503212213041224−−=====cc041203212213041224−−=====rr0000003212213041214=−−=====rr.(3)efcfbfdecdbdaeacab−−−;解efcfbfdecdbdaeacab−−−ecbecbecbadf−−−=abcdefadfbce4111111111=−−−=.考试点www.kaoshidian.com(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+00...
