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1、第一章行列式1.利用对角线法则计算下列三阶行列式:201(1)1−4−1;−183201解1−4−1−183=2×(−4)×3+0×(−1)×(−1)+1×1×8−0×1×3−2×(−1)×8−1×(−4)×(−1)=−24+8+16−4=−4.abc(2)bca;cababc解bcacab=acb+bac+cba−bbb−aaa−ccc333=3abc−a−b−c.111(3)abc;222abc111解abca2b2c2222222=bc+ca+ab−ac−ba−cb=(a−b)(b−c)(c−
2、a).xyx+y(4)yx+yx.x+yxyxyx+y解yx+yxx+yxy333=x(x+y)y+yx(x+y)+(x+y)yx−y−(x+y)−x32333=3xy(x+y)−y−3xy−x−y−x33=−2(x+y).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⋅⋅
3、⋅(2n);n(n−)1解逆序数为:232(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
4、)(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t(−1)a11a23a3ra4s,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t1(−1)a11a23a32a44=(−1)a11a23a32a44=−a11a23a32a44,t2(−1)a11a23a34a42=(−1)a11a23a34a42=a11a23a34a42.4.计算下列各行列式:41241202(1);105200117412
5、4c2−c34−12−104−1−10解1202======1202=122×(−)14+3105201032−14c−7c103−1401174300104−110c2+c39910=12−2======00−2=0.10314c+1c17171412321413−121(2);123250622141c−c2140r−r214042423−1213−1223−122解==========123212301230506250622140r−r2140413−122======0.12300000−
6、abacae(3)bd−cdde;bfcf−ef−abacae−bce解bd−cdde=adfb−cebfcf−efbc−e−111=adfbce1−11=4abcdef.11−1a100−1b10(4).0−1c100−1da100r+ar01+aba012−1b10−1b10解=====0−1c10−1c100−1d00−1d1+aba0c3+dc21+abaad2+1=(−1)(−)1−1c1=====−1c1+cd0−1d0−10=(−1)(−)13+21+abad=abcd+ab+cd+a
7、d+1.−11+cd5.证明:22aabb3(1)2aa+b2b=(a−b);111证明a2abb2c−ca2ab−a2b2−a2212aa+b2b=====2ab−a2b−2a111c3−c1100222=(−)13+1ab−ab−a=(b−a)(b−a)ab+a=(a−b)3.b−a2b−2a12ax+byay+bzaz+bxxyz(2)+++=3+3;aybzazbxaxby(ab)yzxaz+bxax+byay+bzzxy证明ax+byay+bzaz+bxay+bzaz+bxax+byaz+
8、bxax+byay+bzxay+bzaz+bxyay+bzaz+bx=ayaz+bxax+by+bzaz+bxax+byzax+byay+bzxax+byay+bzxay+bzzyzaz+bx=a2++2+yazbxxbzxaxbyzax+byyxyay+bzxyzyzx33=ayzx+bzxyzxyxyzxyzxyz=3+3ayzxbyzxzxyzxyxyz=3+3.(ab)yzxzxya2(a+)12(a+)22(a+)32b2(b+)12(b+)22(b+)32
