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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−ccc=3abc−a3−b3−c3.111(3)abc;a2b2c2111解abca2b2c2=bc2+ca2+ab2−ac2−ba2−cb2=(a−b)(b−c)(c−a).xyx+y(4)yx+yx
2、.x+yxyxyx+y解yx+yxx+yxy=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);n(n−1)解逆序数为:232(1个)52,54(2
3、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
4、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)ta111a23a32a44=(−1)a11a23a32a44=−a11a23a32a44,(−1)ta211a23a34a42=(−1)a11a23a34a42=a11a23a34a42.4.计算下列各行列式:41241202(1);1052001174124c2−c34−12−104−1−10解1202======1202=122×(−1)4+3105201032−14c−7c10
5、3−1401174300104−110c2+c39910=12−2======00−2=0.10314c+1c17171412321413−121(2);123250622141c−c2140r−r214042423−1213−1223−122解==========123212301230506250622140r−r2140413−122======0.12300000−abacae(3)bd−cdde;bfcf−ef−abacae−bce解bd−cdde=adfb−cebfcf−efbc−e−111=adfbce1−11=4a
6、bcdef.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)(−1)3+21+abad=abcd+ab+cd+ad+1.−11+cd5.证明:a2abb2(1)2aa+b2b=(a−b)3;111证明a2abb2c−ca2ab−a2b2−a2212aa+b2b=====2ab−a2b−2a111c3−c
7、1100222=(−1)3+1ab−ab−a=(b−a)(b−a)ab+a=(a−b)3.b−a2b−2a12ax+byay+bzaz+bxxyz33(2)ay+bzaz+bxax+by=(a+b)yzx;az+bxax+byay+bzzxy证明ax+byay+bzaz+bxay+bzaz+bxax+byaz+bxax+byay+bzxay+bzaz+bxyay+bzaz+bx=ayaz+bxax+by+bzaz+bxax+byzax+byay+bzxax+byay+bzxay+bzzyzaz+bx22=ayaz+bxx+bzxa
8、x+byzax+byyxyay+bzxyzyzx33=ayzx+bzxyzxyxyzxyzxyz33=ayzx+byzxzxyzxyxyz33=(a+b)yzx.zxy2222a(a+1)(a+2)(a+3)2222b(b+1)(b+2)(b+3)
