1、 第三章 矩陣的初等變換與線性方程組一、計算題1解：112111211121112110001231 0112 0105 0105 01052343010501120013001310001000 0105 010000130010（行最簡）（標準型）2解：321112331233123312330221 0221 0221 0221 022112333211041080254003510141014101411 0221 011 0112200355001037100313 0106550100133100001000010（行最簡）（標準型）3解：121100121100121100342

2、010 021310 02131054100101465010011671100210100210131 0201361 01032200116710011671A4.解：（1）531105311053110( ,)17221 10121725015555020515104132172501725A E 1325P104017PA(2) 523111TA52100521005210013(,)31010010013505511001071057100155TAE52100505100101200135001350013500020355002035500471120103500147100TQ

3、QA5解： 111213112112221323212223212223313233313233110110010010001001aaaaaaaaaAaaaaaaaaaaaa111213111211132122232122212331323331323133110110010010001001aaaaaaaAaaaaaaaaaaaaaa6解：1112131112132122232122233132333132330000010010001001kkaaakakakaAaaaaaaaaaaaa111213111213212223212223313233313233000001001000100

4、1kaaakkaaaAaaakaaaaaakaaa7解：111213313233212223212223313233111213001001010010100100aaaaaaAaaaaaaaaaaaa111213131211212223232221313233333231001001010010100100aaaaaaAaaaaaaaaaaaa8解:123200101001010013122212327120017124120101120101121121004121314123032222A9解:121100121100121100342010 021310 021310541001014

5、65010011671100210100210131 0201361 010322001167100116712101313221671逆矩陣是10解：（方法同 8，過程略） X131341351321311137131133131013413513213413113213613 11解：1412131022212215331131124X 12解：*1*11121(3 )(3 )(3 )( (3 )(3 )31926262631993126931262626193193262626AE XAXAEAAEA AA A AEAAA 13解: ,R=2 4329101111113201761513

6、204 0111111141211100000是最高階非零子式321314.解：=2AE3001001001402 0101200030010011001001001002)120010020110001001001001AE E（，1001001101002200100112 )EAE（，）故=12 )AE（1001102200115.解：1111100111100)121010010110113001002101AE（，51100110121022010110010110( , )1111001000102222E A故=，A5112211011022511221110211022A 因此

7、=*A-1（）2AA5112211011022521220101注：若已知，求，則需利用（）經過初等變換（）AA-1,A E1,EA若已知，求，則需利用（）經過初等變換（）A-1A1,AE,EA16.解：1000000000rsrsrrAECBECBAECBEAEAE1111000srEC BACEA故1111100AC BACCBA17解：，R=21110243102351121 012413440000是最高階非零子式311118解：323111132(1) (2)011Rkkkkkkk21kk 且19解：121234124310873564 01654523000087651001Axx

8、ccxx通解20解：121234111111011113 00121123000011100201Axxccxx 通解：21解：710105123114()31532 010052122300001Ab無解22解：123411001111102()11131 010001111230012221120012210Abxxcxx 通解23解: 2222121222231111234(4)(1)11414(21)4(1)(2)34434124(1)(2)3443444(1)24(1)(4)(1)(4)kAkkkkkkkkkkkkkkkkxkkkxA bkkkkkxkkkkkkk （1）且時，解唯一

9、。，123(2)4,11441030()1124 0114141160000034101kAbxxcx 時123(3)1,31002111410307()1124 0114 010211110000001132721kAbxxx時24解:232432(1)3(1)3(1)3111111(3)111xyzxyzxyzA3243223243224332221(1)03233(2)()133(2)()133(2)01110() 00000000110 xyzAbxycz 且時，有唯一解。時，2101(3)321101010() 1210 011011200000111cAbxycz 時，二、證明題1

10、.證明:00000()000000()( )TTTTTTTTTAXA AXAXA AXA AXXA AXAXAXAXA AXAXAXA AXR A AR A只需要證明與同解即可。顯然，的解是的解,若，則（），因此的解是的解與同解2.證明：對分塊矩陣作初等變換，得000AABABBBBB注意到是分塊矩陣的子塊，而子塊的秩不會大于該矩陣的秩，AB所以：0()( )( )0ABBAr ABrrr Ar BBBB3.證明：設則存在可逆矩陣，使得( ),r Ar r Bt），,PQ，則000rEPAQ，令=1110()00rEPABPA QQBPAQQ BQ B（）1Q BC則，上式為1112200,00000rrCCEECCCCC其中故11()()()( )()( )( )( )0Cr ABr PABrr Cr Csrr crsr Br As 即，()( )( )r ABr Br As4. 證明：由上題可知，而，則，故有:()( )( )r ABr Ar Bn0AB ()0r AB ,即0( )( )r Ar Bn( )( )r Ar Bn5.分析：注意到 ()(),r AEr EA()(),r AEr EA故要證 ()()()()r AEr AEr EAr EA