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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abc222abc222222=bc+ca+ab-ac-ba-cb=(a-b)(b-c)(c-a).xyx+y(4)yx+yx.x+yxyxyx+y解yx+yxx+yxy3
2、33=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×××(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,××
3、×,(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
4、是t1(-1)a11a23a32a44=(-1)a11a23a32a44=-a11a23a32a44,t2(-1)a11a23a34a42=(-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-7c103-1401174300104-110c2+c39910=12-2======00-2=0.10314c+1c17171412321413-121(2);123250622141c-c
5、2140r-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=4abcdef.11-1a100-1b10(4).0-1c100-1da100r+ar01+aba012-1b10-1b10解=====0-1c10-1c100-1d00-1d1+aba0c3+dc21+abaad=(-1)(-1)2
6、+1-1c1=====-1c1+cd0-1d0-10=(-1)(-1)3+21+abad=abcd+ab+cd+ad+1.-11+cd5.证明:a2abb23(1)2aa+b2b=(a-b);111证明a2abb2c-ca2ab-a2b2-a2212aa+b2b=====2ab-a2b-2a111c3-c1100222=(-1)3+1ab-ab-a=(b-a)(b-a)ab+a=(a-b)3.b-a2b-2a12ax+byay+bzaz+bxxyz(2)ay+bzaz+bxax+by=(a3+b3)yzx;az+bxax+byay+bzzxy证明ax+byay+bzaz+bx
7、ay+bzaz+bxax+byaz+bxax+byay+bzxay+bzaz+bxyay+bzaz+bx=ayaz+bxax+by+bzaz+bxax+byzax+byay+bzxax+byay+bzxay+bzzyzaz+bx=a2yaz+bxx+b2zxax+byzax+byyxyay+bzxyzyzx=a3yzx+b3zxyzxyxyzxyzxyz=a3yzx+b3yzxzxyzxyxyz=(a3+b3)yzx.zxya2(a+1)2(a+2)2(a+3)22222b(b+1)(b+2)(b+3)
