线性代数王定江第3章答案.pdf

-1-习习题题三三1.分别写出下列矩阵的行阶梯形、行最简形和标准形矩阵：（1）1234（2）123456789（3）111032212411115.解：（1）行阶梯形：31312123402rr行最简形：132213122121210340201rrrrr也是标准形（2）行阶梯形：3121212471231231234560360367890612000rrrrrr行最简形：212132123123101456036012789000000rrrLL标准形：31322123101100456012010.789000000cccc+LL（3）行阶梯形：32213122111031110311103221240012200122111150021200036rrrrrr行最简形：123132313-22111031110311021221240012200122111150003600012110050010200012rrrrrrr+LL标准形：-2-2151535423345221110311005100002212400102001001111500012000101000001000.00100cccccccccccc+L2.利用矩阵的初等行变换判断下列矩阵是否可逆若可逆求其逆矩阵.（1）121342541（2）231123415（3）31411110.20111120解：（1）对矩阵()AE做初等行变换即2131351211001211003420100213105410010146501rrrr2312327100210100210021310020136100116710011671rrrrrr+23121002100101323120011671rr所以矩阵可逆且112121042013421323121361.2541167132142=（2）对矩阵()AE做初等行变换即12231100123010123010231100415001415001rr-3-213124123010077120077041rrrr3210127370077120000121rr所以矩阵不可逆.（3）对矩阵()AE做初等行变换即123141100011100100111001003141100020110010201100101120000111200001rr2131141332421110010044400400047113000471130002310210046204200030010100300101rrrrrrrr+123240311100047113000011112000300101rrrr340311100047113000011112000300101r1323433734004226004086101400011112000033461rrrrrr+1233331200126618001202418304200033336000033461rrr-4-1424344812000610640120062680030010100033461rrrrrr+1234111212113310001256121301001216122300100130130001143213rrrr所以矩阵可逆且1314112561213353211101216122331341.201101301302026112014321368122=3.利用矩阵的初等变换解下列矩阵方程.（1）0122311415.21036=X解：对矩阵()AB做初等行变换即12012231141511415012232103621036rr312114150122303814rr123231021801223002713rrrr+1232310065010916002713rrrr+3121006501091600172132r所以-5-65916.72132X=（2）200200130030.521001=X解：方法一：对矩阵()TTAB做初等行变换即132152006315600032030032030001001001001r13122313260136306006313032030030032001001001001rrrrrr+1216131001121360100123001001rr所以112136100600101231210=360.60011362311346TX=方法二：（特殊方法）因为11|6|1|6BBABAA=所以1BA为可逆矩阵.又因为()()()11BAAEBBABX=所以对矩阵()AE做一系列初等行变换将A变为B则E变为1XBA=即原问题的解.对矩阵()AE做初等行变换即()213125220010020010013001003012105210010215201rrrrAE+=-6-()32232001000301210001136231rrBX+=所以10060011210=360.61362311346X=4.求下列矩阵的秩.（1）101011.111（2）12340112.1231（3）31325323.13507514（4）1211124311.1213300252解：（1）对矩阵做初等行变换化为行阶梯形得3132101101101011011011111012003rrrrA+=所以()3.RA=（2）对矩阵做初等行变换化为行阶梯形得31123412340112011212310005rrA=所以()3.RA=（3）对矩阵做初等行变换化为行阶梯形得-7-2131314153731321350135053235323012273135031320818275147514016364rrrrrrrrA=233242411321413501350049104910491000004910000rrrrrrr所以()2.RA=（4）对矩阵做初等行变换化为行阶梯形得213121211112111243110011112133002420025200252rrrrA+=3324243126271211112111001110011100060000100007000000rrrrrrr所以()3.RA=5.设矩阵111111111111kkkk=A且()3AR=求k的值.解：对矩阵A做初等行变换化为行阶梯形得213114412111111111111111010111111100111111110111rrrrrrrkrkkkkkkkAkkkkkkkkk=43421111110101010100110011001(1)(2)000(1)(3)rrrrkkkkkkkkkkkkkkk+因为()3RA=所以10k(1)(3)0kk+=求解得3.k=-8-6.设矩阵122343123119a=A问a为何值时()3RB6.设A为n阶可逆矩阵则（B）（A）若=ABCB则=AC.（B）A总可以经过初等行变换化成E.（C）对矩阵()AE施行若干次初等变换当A变为E时相应地E变为1A.（D）以上都不对.7.设A是mn矩阵()Rr=AC是n阶满秩矩阵1()rr=BACB则（C）（A）1rr.（B）1rr时必有|0AB.（B）当mn时必有|0=AB.（C）当nm时必有|0AB.（D）当nm时必有|0=AB.9.设(3)nn阶方阵1111aaaaaaaaaaaa=ALLLMMMML若()1rn=A则a必为（B）（A）1.（B）11n.（C）1.（D）11n.解：对A做初等变换111111001011010001iirraaaaaaaaaaaAaaaaaaaaa+=LLLLLLMMMMMMMMLL-18-1111(1)(1)(2)010000100001jjccjnnananaaaaa+=+LLLLMMMML因为()1rAn=所以1(1)0na+=即11an=.10.设AB均为非零n阶矩阵且=ABO则A与B的秩（D）（A）必有一个为零.（B）一个小于n一个等于n.（C）都等于n.（D）都小于n.证明：因为AB均为非零n阶矩阵所以()()RAnRBn.假设()RAn=那么A为可逆矩阵进而由ABO=得BO=与已知条件矛盾.因此必有()RAn所以AX=也可能无解.16.设A为mn矩阵mn非齐次线性方程组=AX对应齐次线性方程组=AX0下面结论正确的是（B）（A）=AX有无穷多解.（B）=AX0有非零解.（C）=AX0只有零解.（D）=AX无解.17.若n阶矩阵A满足2+=AAO证明()()RRn+=AAE.证明：因为2AAO+=即()AAEO+=所以()()RARAEn+.另一方面()()()()()RARAERARAEREn+=+=.综上()()RARAEn+=.18.若2AE=证明()().RRn+=AEAE证明：因为2AE=即()()AEAEO+=所以()()RAERAEn+.另一方面()()()()(2)()RAERAERAEREAREREn+=+=.所以()()RAERAEn+=.19.设A为(2)nn阶矩阵A为A的伴随矩阵则()()1()1()1AAAAnRnRRnRn=若若0，若证明：当()RAn=时A为可逆矩阵从而1A也为可逆矩阵进而由1AAA=得A也为可逆矩阵所以().RAn=-21-当()1RAn=时|0A=由行列式展开定理得AAAEO=所以()()RARAn+即()()1RAnRA=.另一方面由()1RAn=知存在一个非零的1n阶子式即A有非零元素（一阶子式）所以()1