线性代数第二章

1、习题习题2 .32 ,512231,402753bababa求已知914984512231402753：ba解5122313402753232ba732813.,2,662479154257,864297510213zbzaba求且已知.16111122223286429751021366247915425721)(21：abz解., 0)(2)2()2( ;3) 1 (，101012121234,432112122121yybyababa求若求设；ba139732828513110101212123412963363663633) 1 ( ：解.22323234034022310310322

2、)2(bay.,2103,1231baabba求.25661231210321031231： baab解计算下列各式:.20413121013143110412)4(;00000010001001) 3( ;127075321134)2( ;123321) 1 (cbak10132231) 1 ( :原式解;00000)3(cbkba原式.6520876)4(原式;49635)2(原式.)(,2412,435214,240031tabccba求设.4226140602412435214240031)( :tabc解.,)(,22baabbababanba证明：若阶方阵均为证明:由题可知:222

3、2()()ab ababaabbab.abba所以已知线性变换,5232232133212311yyyxyyyxyyx323312211323zzzzzyzzy求从变量321,zzz321,xxx到的线性变换.321321321321310102013,511232102:zzzyyyyyyxxx解由此可得32132132116419412316310102013511232102zzzzzzxxx.164941236321332123211zzzxzzzxzzzx即.,10001001,00010001,333231232221131211caacbaab，kckbaaaaaaaaaa求设,

4、.,：333231232221131211333231232221131211kakakaaaaaaabakaaakaaakaaaab解.,333231132312221121131211333232312322222113121211aaakakakaakaaaaacaaakaaaakaaaakaaac.)(4(;)(3( ;)(2();()(1 (1221,0121,1110ttttttababbababcaccbabcacab，cba验证已知).()(;662121451110122101211110)(;662112212201122101211110)(1 ( :bcacabbcac

5、ab所以证.)(;5457214533121221012112211110;545712211231122101211110)(2(bcaccbabcaccba所以.)(;13210211111001211110;1321123101211110)(3(ttttttttttbabababa所以.)(;20211110021111100121;2021220101211110)(4(ttttttttttabababab所以将下列矩阵化为行最简形:12433023221453334311)3(174034301320)2(4311) 1 (10014311) 1 ( :解0000310050101

6、74034301320)2(.0000001000221003201112433023221453334311)3(221( )53,33f xxxa已知( ).f a求2( )53,21212110533333330100.00f xxx解:35)(2aaaf计算下列各式:20012(1);(2) 00.3400nabc212510(1).3415100000(2) 0000 .0000nnnnaabbcc解:.,001001kaa求设由二项式定理可得解,000100010100010001001001a：be nkkkkkkkkkbbkkbkebbekkbekebea2212212) 1(

7、)(2) 1()()()(.0002) 1( 0000001002) 1(000100010100010001222221kkkkkkka：kkkkn所以, 000000010000010001032nb，b，bb其中证明:(1) 两个同阶的上三角矩阵的乘积仍为上三角矩阵;(2) 对于任意矩阵a, aat和ata均为对称矩阵; (3) 设a,b是同阶对称矩阵, 则ab也是对称矩阵的充分必要条件是a与b可交换, 即ab=ba.)(3(.)()()()()2(.) 1 (：baababbaababa，aaaaaaaaa，aaaaaattttttttttttttttt均为对称矩阵所以略证明用逆矩阵定

8、义求下列方阵的逆阵 .;13431) 1 (a.)( ,1012,0121)2(111abbaba求设.1232)(;10201121;10112021)2( 1 ( :21214341112121*12121*1*1abbbbaaaaaa解;12643152) 1 (x.101201325120112)2(x用矩阵乘法 解以下矩阵方程:;8023212643152) 1 ( :1x解.381326101201325120112)2(1x.,323321321323312211的线性变换表示求用已知线性变换zzzyyyzzyzzyzzy.923192,3132,913

9、1913101020133213321232113211321yyyzyyyzyyyzyyyzzz所以,310102013:321321zzzyyy解.,) 3(; 00,)2(;,) 1 ( :1也是对称阵则是对称阵若可逆阵则且为同阶方阵可逆若方阵则其逆阵惟一可逆若方阵证明aabab，baa.,)()(,) 3(; 000,)2(;,) 1 (:1111111也是对称矩阵所以故为对称矩阵存在则可逆所以则的逆矩阵同为设证明aaaaaaabaabaaacecbacbebecaeacbaabacbttt., 0)2(; 0, 0) 1 (:,1*naaaaaaa则若则若证明的伴随矩阵为若阶矩阵.

10、0.0; 00)(0)(, 0,)(, 0, 0; 0, 0, 0) 1 ( :*1*1*1 *aaaaaaaeaaaaaaaaa所以必有的假设矛盾与得由存在则设若从而则若证明.,:,)2(1*nnaaeaaaeaaaaa从而得等式两边同取行列式由定理知设方阵a满足关系式 , 试证a及a+2e均可逆,求出逆阵. 0222eaa;2)2(,2)2(022:12eaaaeeaaeaa且可逆从而得由证明.64)2(,2,6)4)(2( :, 0226)4)(2( :12aeeaeaeeaeaeaaeeaea且可逆从而所以又因为.323432111)4( 602130321)3( 6432)2( 52

11、21) 1 (判别下列矩阵是否可逆，若可逆，求其逆矩阵.;6432, 06432)2(;12255221, 015221) 1 ( :1不可逆故且故矩阵可逆解., 0323432111)4(;431234102147329, 04602130321)3(*1故矩阵不可逆且故矩阵可逆aaa.,2001,1141,111apapp求其中设.68468327322731 242124213120011141 ,2001,:111113133131343111313134311111111111111111apppppppappaapp故其中从而得由解个利用初等变换求下列矩阵的逆矩阵:;1210232

12、112201023) 1 (a.)2(,300341003)2(1ebb求;2210100:)()() 1 ( :56515253535152535111aaeea可得利用初等变换解变换初等行.100001)2()2(2321211eb同上可得;101311022141)2(x解矩阵方程;234311111012112) 1 (x32538122111012112234311) 1 ( :1x解;011110210132141)2(411x.2)3(,21,3,*1*的值求的伴随矩阵为阶方阵为已知aaaaaba.27161)32()32(222)3(,:3131321131*131*11*1a

13、aaaaaaaaaaaaaaa解.23)4( ;)3( ;2)2( ;) 1 (:. 3,3*1*21aaaaaaa求阶方阵为. 92323)4(;93)3(;729822)2(;311, 1) 1 (:*121*3211aaaaaaaaaaaaaaan所以因为解.3, 3, 2,5,1*的值求行列式阶方阵均为设bababa.12961333:451*51*bababa解).0 , 0 , 0 , 1, 1 (),0 , 0 , 1 , 0 , 1 (,4使其中两行为的矩阵求作一个秩为.0000001000001000001100101:所作的矩阵为解求下列矩阵的秩,并求一个最高阶非零子式:0

14、2301085235703273812)2(443112112013) 1 (ba; 01113, 2)(.000056401211564056401211) 1 ( :的最高阶非零子式为：解aara. 0001532712, 3)(00000140000712100230171210024205363002301)2(为的一个最高阶非零子式bbrb; 11110421106312121011 )2( ; 512311124031 ) 1 (.0722134305143212031214013)4(0510317145024)3(求下列矩阵的秩. . 200007170403171707170

15、4031512311124031) 1 ( :2321131232rrrrrrrrr所以解; 3 000005300042110210111111042110421102101111110421106312121011)2(23432412rrrrrrrrr所以; 2,00000023012100023023012110150101504601210510317145024)3(2321251334122114135107rrrrrrrrrrrrrrr; 3,00000000001123310033810143213381000000000004895014321111140816181004

16、895048950143210722134305143212031214013)4(24232512311514132253rrrrrrrrrrrrrrrrrddccbbaa000000010001010001000000利用分块矩阵计算222222220000000010001,0010001000000:deceeddccbeeaebbaa解.)(0100)(010000)(,)(002222222222222222bdcbdcacaacaebdceaceaeebdceaceaedeceebeeae:均可逆阶矩阵及阶矩阵设bsan；cobaoc1,) 1 (求若.,3221,503512101)2(1cba求时当.00000000:,00000000:,000000) 1 ( :1111111111nsnsnssnnsnsnssnnssnnsabeeabbaab