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专业好文档电大历年试题经济数学基础 线性代数一、 单项选择题:1、设a是mn矩阵,b是st矩阵,且有意义,则c是( )矩阵. a. mt b. tm c. ns d. sn2、设a是可逆矩阵,且a+ab=i,则=( ). a.b b.1+b c.i+b d.3、设a= ,则r(a)=( ). a.0 b.1 c.2 d.34、以下结论或等式正确的是( ). a.若a,b均为零矩阵,则有a=b b.若ab=ac,且ao,则b=cc.对角矩阵是对称矩阵 d.若ao,bo,则abo 5、设a,b均为n阶可逆矩阵,则下列等式成立的是( ). a. b. c. d.ab=ba 6、设a为32矩阵,b为23矩阵,则下列运算中( )可以进行. a.ab b.a+b c. d. 7、设a,b为同阶可逆矩阵,则下列等式成立的是( ). a. b.c. d. ( c. ) 8、设a为34矩阵,b为52矩阵,且乘积矩阵有意义,则c为( )矩阵. a.42 b.24 c.35 d.53 9、设a为34矩阵,b为52矩阵,且乘积矩阵有意义,则c为( )矩阵. a.45 b.53 c.54 d.42 10、设a,b为同阶方阵,则下列命题正确的是( ). a.若ab=o,则必有a=o或b=o b.若abo,则必有ao,且boc.若秩(a)o,秩(b)o,则秩(ab)o d. 11、用消元法解方程组,得到解为( ). a. b. c. d. 12、设线性方程组ax=b的增广矩阵为 ,则此线性方程组的一般解中自由未知量的个数为( ). a.1 b.2 c.3 d.4 13、线性方程组 =的解的情况是( ). a.无解 b.有无穷多解 c.只有0解 d.有唯一解 14、线性方程组解的情况是( ). a. 有无穷多解 b. 只有零解 c. 有唯一解 d. 无解 15、设线性方程组ax=b有唯一解,则相应的齐次方程组ax=o( ). a.无解 b. 有非零解 c. 只有零解 d.解不能确定 16、若线性方程组的增广矩阵为 (或 ),则当=( )时线性方程组无解. a. b.0 c.1 d.2 17、若线性方程组的增广矩阵为 ,则当=( )时线性方程组无解. a.3 b.-3 c.1 d.-1 18、若线性方程组的增广矩阵为 ,则当=( )时线性方程组有无穷多解. a.1 b.4 c.2 d. 19、线性方程组解的情况是( ). a.无解 b. 只有0解 c. 有唯一解 d. 有无穷多解20、设a= ,则r(a)=( ). a.0 b.1 c.2 d.321、设a= ,则r(a)=( ). a.1 b.2 c.3 d.4 二、填空题: 1、矩阵 的秩为 . 2、设a= ,当= 时,a是对称矩阵. 3、设a= ,当= 时,a是对称矩阵. 4、两个矩阵a、b既可相加又可相乘的充分必要条件是 . 5、设矩阵a= ,i为单位矩阵,则 . 6、设a,b均为n阶矩阵,则等式成立的充分必要条件是 . 7、设矩阵a可逆,b是a的逆矩阵,则= . 8、设a= ,则r(a)= . 9、已知齐次线性方程组ax=o中a为35矩阵,且该方程组有非0解,则r(a) . 10、n元齐次线性方程组ax=o有非零解的充分必要条件是r(a) . 11、线性方程组ax=b有解的充分必要条件是 . 12、齐次线性方程组ax=o(a是mn)只有零解的充分必要条件是 . 13、齐次线性方程组ax=o的系数矩阵为a= ,则此方程组的一般解为 .( 或则此方程组的一般解中自由未知量的个数为 .) 14、设齐次线性方程组,且r(a)=rn,则其一般解中的自由未知量的个数等于 . 15、若线性方程组有非零解,则= . 16、若n元线性方程组ax=o满足r(a) n,则该线性方程组 . 17、设齐次线性方程组,且r(a)=2,则方程组一般解中的自由未知量的个数为 . 18、线性方程组ax=b的增广矩阵化成阶梯形矩阵后为 则当d = 时,方程组ax=b有无穷多解. 19.若a为n阶可逆矩阵,则r(a)= . 20.当 时,矩阵a= 可逆.三、计算题: 1、设矩阵a= ,b=,求. 2、已知ax=b,其中a= ,b=(b=),求x. 3、已知ax=b,其中a= ,b=,求x. 4、设矩阵a= ,b= ,求解矩阵方程xa=b. 5、设矩阵a= ,计算.6、设矩阵a= ,计算.7、设矩阵a= ,i是3阶单位矩阵,求.8、设矩阵a= ,b= ,求.9、设矩阵a= ,b= ,i是3阶单位矩阵,求.10、设矩阵a= ,i= ,求.11、设齐次线性方程组,问取何值时有非零解,并求一般解.12、讨论为何值时,齐次线性方程组有非零解,并求一般解.13、求齐次线性方程组的一般解.14、求齐次线性方程组的一般解.15、讨论当为何值时,线性方程组无解,有唯一解,有无穷多解.16、求线性方程组的一般解.17、求线性方程组的一般解.18、当为何值时,线性方程组有解,在有解的情况下求方程的一般解.19、当为何值时,线性方程组有解,在有解的情况下求方程的一般解.参考答案一、 单项选择题:1.d 2.c 3.d 4.c 5.c 6.a 7.c 8.b 9.c 10.b 11.c 12.b 13.d 14.d 15.c 16.a 17.b 18.d 19.a 20.c 21.b 二、填空题:1.2 2.1 3.0 4.a、b为同阶矩阵 5. 6.ab=ba7. 8.1 9.3 10.n 11. 12.r(a)=n13. (或 2 ) 14.n-r 15.-1 16.有非零解 17.318.-5 19.n 20. -3三、计算题: 1.解: , a-ii= ,所以 , =.2.解:ab= ,所以 ( ab= ,所以 )3.解法一:ai= 即 , 所以 = 解法二:ab= ,所以4.解:ai= 即 , = 5.解: , ,所以 6.解: , ,所以 7.解:i-a= ,i-ai= ,所以= 8解: = , ,所以= 9解:前面同第7题 = 10解: , 所以 11解:因为系数矩阵 a= 所以当=4时,方程组有非零解,且一般解为:(其中为自由未知量)(或期末指导p.75三(13)12解:因为系数矩阵 a= 所以当=4时,方程组有非零解,且一般解为: (其中为自由未知量)13解:因为系数矩阵 a= 所以方程组的一般解为:(其中是自由未知量)14解:因为系数矩阵a= 所以方程组的一般解为:(其中是自由未知量)15解:因为增广矩阵= 所以当时,方程组无解;当时,方程组有唯一解;当时,方程组有无穷多解.16.解:因为增广矩阵= ,故方程组的一般解为:(其中是自由未知量)17解:将方程组的增广矩阵化为阶梯矩阵= 由此得方程组的一般解(其中是自由未知量)18解:将方程组的增广矩阵化为阶梯矩阵= ,由此可知当=3时,方程组有解,其一般解为(其中是自由未知量)19解:将方程组的增广矩阵化为阶梯矩阵= 由此可知当=5时,方程组有解,其一般解为(其中是自由未知量) if we dont do that it will go on and go on. we have to stop it; we need the courage to do it.his comments came hours after fifa vice-president jeffrey webb - also in london for the fas celebrations - said he wanted to meet ivory coast international toure to discuss his complaint.cska general director roman babaev says the matter has been exaggerated by the ivorian and the british media.blatter, 77, said: it has been decided by the fifa congress that it is a nonsense for racism to be dealt with with fines. you can always find money from somebody to pay them.it is a nonsense to have matches played without spectators because it is against the spirit of football and against the visiting team. it is all nonsense.we can do something better to fight racism and discrimination.this is one of the villains we have today in our game. but it is only with harsh sanctions that racism and discrimination can be washed out of football.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelsea on sunday.i am going to be at the match tomorrow and i have asked to meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans duringcitys 2-1 win.michel platini, president of european footballs governing body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans of cska.baumgartner the disappointing news: mission aborted.the supersonic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. the balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. as he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic diving platform.below, the earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. it will be like he is diving into the shallow end.skydiver preps for the big jumpwhen he jumps, he is expected to reach the speed of sound - 690 mph (1,110 kph) - in less than 40 seconds. like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to earth. but this will not be enough to stop him completely.if h
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