习题解答习题一(A)1.用消元法解下列线性方程组:(1).5432,9753,432321321321xxxxxxxxx解由原方程组得同解方程组12323234,23,xxxxx得方程组的解为13232,23.xxxx令3xc,得方程组的通解为cxcxcx321,32,2,其中c为任意常数.(2).552,12,12432143214321xxxxxxxxxxxx解由原方程组得同解方程组1234421,44,02,xxxxx所以方程组无解.(3).05,3523,22,1231321321321xxxxxxxxxxx解由原方程组得同解方程组12323321,0,41,xxxxxx得方程组的解为41,41,45321xxx.(4).23,0243,6332,322432143214321421xxxxxxxxxxxxxxx解由原方程组得同解方程组123423434432,310,39,3,xxxxxxxxxx得方程组的解为3,4,1,24321xxxx.2.用初等行变换将下列矩阵化成行阶梯形矩阵和行最简形矩阵:(1)122212221.解122122100212012010221001001rr,得行阶梯形:100210221(不唯一);行最简形:100010001.(2)324423211123.解1102232111232551232041050124442300000000rr,得行阶梯形:0000510402321(不唯一);行最简形:000045251021201.(3)211132.解231110110101120000rr,得行阶梯形:001011(不唯一);行最简形:001001.(4)34624216311230211111.解11001111111111122032102501101002136120070000100426430000000000rr,得行阶梯形:00000007001052011111(不唯一);行最简形:0000000100210010211001.3.用初等行变换解下列线性方程组:(1).3,1142,53332321321xxxxxxxx解2100313357214110109011320019rB,得方程组的解为920,97,32321xxx.(2).2222,2562,134432143214321xxxxxxxxxxxx解114311143121652032101222200001rB,得方程组无解.(3).1837,4,133,44324324324214321xxxxxxxxxxxxx解47100021234415130310101201114230012073118200000rB,得方程组的解为1243447,215,2232.2xxxxx令4xc,得方程组的通解为cxcxcxx4321,2232,215,247,其中c为任意常数.(4).1224,9138436,354236,232254321543215432154321xxxxxxxxxxxxxxxxxxxx解1112112321002226324530010436348139000100421121000000rB,得方程组的解为125354111,22243,0.xxxxxx令2152,xcxc,得方程组的通解为2542312211,0,34,,2122cxxcxcxccx,其中21,cc为任意常数.(B)1.当为何值时,线性方程组2321321321,,1xxxxxxxxx有无穷多解,并求解.解2211111111110101110011rB.当1时,111100000000rB,方程组有无穷多解,且解为1231xxx.令2132,xcxc,得方程组的通解为2312211,,1cxcxccx,其中21,cc为任意常数.3.(联合收入问题)已知三家公司A、B、C具有如下图所示的股份关系,即A公司掌握C公司50%的股份,C公司掌握A公司30%的股份,而A公司70%的股份不受另外两家公司控制等等.3现设A、B和C公司各自的营业净收入分别是12万元、10万元、8万元,每家公司的联合收入是其净收入加上其它公司的股份按比例的提成收入.试确定各公司的联合收入及实际收入.解A公司的联合收入为元,实际收入为元;B公司的联合收入为元,实际收入为元;BACC公司的联合收入为元,实际收入为元.习题二(A)1.利用对角线法则计算下列行列式:(1)cossinsincos.解原式1.(2)22xyxy.解原式()xyyx.(3)123312231.解原式18.(4)000abcaba.解原式3a.(5)000aababc.解原式3a.2.按定义计算下列行列式:(1)0000000000abfcde.解原式1311000(1)0(1)0bcafcababcddede.(2)010000200001000nnLLLLLLLLL.解原式1100020(1)001nnn!)1(1nn.3.利用行列式的性质,计算下列行列式:(1)abacaebdcddebfcfef.解原式111111111022111002abcdefabcdefabcdef4.(2)1111222233334444.解原式1111044419200660008.(3)axaaaaaxaaaaaxaaaaax.解原式1111100000(4)(4)0000aaxaaaxaxaxaaaxaaxaaaaxax3(4)axx.(4)23100120103518510154.解原式1201120112012310001102011020351803518003512510154001510015112351221501151.(5)121111100100100naaaLLLMMMLML,其中0,1,2,,iainL.解原式11121211000100100100iiniirrainnaaaaniniiiaa11)11(.4.利用行列式展开定理,计算下列行列式:(1)1214012110130131.解原式020120100101213212132171013131311310131.(2)5487235472856393.解原式00303230141532231415314431443181434663430663443.(3)123100010000000000001000nnaaaaaLLLMMMMMLL.解原式122131100010000000(1)00000000nnnnaaaaaaa2311(1)1210000(1)(1)00nnnnaaaaaa23112nnaaaaaa2311(1)nnaaaaa.(4)2100012100012000002100012nDLLLMMMMMLL.解将行列式按第一行展开,得122nnnDDD,则11221212112nnnnDDDDDD,所以12112(1)1nnnDDDDnn.5.利用行列式展开定理证明:当时,有1100010001000000001nnnDLLLMMMOMMLL.证将行列式按第一行展开,得12()nnnDDD,则211223()()nnnnnnDDDDDD22221()[()()]nnnDD,所以1nnnDD.(1)由nD关于与对称,得1nnnDD.(2)由(1)与(2)解得11nnnD(类比于高中学过的由数列an与an-1的关系推导通项公式)6.利用范德蒙德行列式计算行列式222abcabcbcacab.解原式222222111()()111abcabcabcabcabcabc()()()()abcbacacb.7.设2142112531335111D,试求14243444A+A+A+A和11121314M+MMM.解14243444A+A+A+A0;111213141112131411111125+31335111MMMMAAAA01002346502134652422422842142626206206120.8.利用克拉默法则解下列线性方程组:(1)12341234123412345,242,2352,32110.xxxxxxxxxxxxxxxx解经计算,得1234142,142,284,426,142DDDDD,所以方程组的解为1,3,2,14321xxxx.(2)123423412423423411,3,30,735.xxxxxxxxxxxxx解经计算,得123416,16,0,32,16DDDDD,所以方程组的解为1,2,0,14321xxxx.9.试问取何值时,齐次线性方程组123123123230,3470,20xxxxxxxxx有非零解.解方程组有非零解,则0D.又2133475(3)12D,所以3.10.试问、取何值时,齐次线性方程组1231231230,0,20xxxxxxxxx有非零解.解方程组有非零解,则0D.又1111(1)121D,所以1