2014-2015(2)期末考试试卷(A)(线性代数)

20142学期《线性代数》考试A 卷答案及评分标准一、选择题(每题2分，共计20分)1-5 D C C A C 6-10 C A A A C二、填空题(每题2分，共计20分)1、-122、1或33、10123321015432176543---⎛⎫⎪-⎪ ⎪⎪⎝⎭4、1815、28a6、11121321111222231331323322()2a a a a a a a a a aa a -⎛⎫⎪--- ⎪ ⎪-⎝⎭7、R （A ）=R （A ，b ）或线性方程组系数矩阵的秩与线性方程组增广矩阵的秩相等。

8、21,αα3,α 9、无 10、16三、证明题(每题10分，共计20分)1、证明：线性方程组的系数矩阵为A=1100011000111001-⎛⎫⎪-⎪ ⎪-⎪-⎝⎭； （1分） 线性方程组的增广矩阵为12341100011000111001a a A a a -⎛⎫⎪-⎪= ⎪-⎪-⎝⎭； （1分） 又线性方程组有解的充分必要条件为R （A ）=R （A ）， （2分）12341100011000111001a a a a -⎛⎫⎪- ⎪ ⎪-⎪-⎝⎭~12314110001100011011a a a a a -⎛⎫ ⎪- ⎪ ⎪- ⎪-+⎝⎭~123214110001100011011a a a a a a -⎛⎫ ⎪- ⎪ ⎪- ⎪-++⎝⎭~12332141100011000110a a a a a a a -⎛⎫ ⎪- ⎪ ⎪- ⎪+++⎝⎭（4分）∴3214a a a a +++=0 （2分） 证毕。

2、证明：假设存在一组数12,r k k k ，使得02211=+++r r k k k βββ 成立， （2分）即++++++++++p r p r r k k k k k k ααα)()()(2211 0=+r r a k 因向量组r a a a ,,,21 线性无关，所以⎪⎪⎩⎪⎪⎨⎧==++=+++000221r r r k k k k k k ⇔⎪⎪⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎪⎪⎭⎫⎝⎛00010011011121 r k k k ，因为01100110111≠= ,(6分） 故方程组只有零解，即当且仅当021====r k k k ，故r βββ,,,21 线性无关. （2分）四、计算题（共计40分）1、解：将第2，3…n 列都加到第一列得：(3分)()()()()1111a n bb b b a n b a bb a n b b a b a n b bba+-+-+-+-D =[]11(1)1b b b a b b a n b b a b =+-(4分)(1分)2、解：由 B AX X +=2，得 B X A E =-)2(. 因为032110111|2|≠=--=-A E ，所以矩阵A E -2可逆， (2分) B A E A E B A E X |2|*)2()2(1--=-=- 求出1(2)E A --得(4分)或者(2)E A E -=110100101010102001-⎛⎫ ⎪- ⎪ ⎪⎝⎭~1((2))E E A --=10002/31/301012/31/300101/31/3⎛⎫ ⎪- ⎪⎪-⎝⎭,即1(2)E A --=02/31/312/31/301/31/3⎛⎫⎪- ⎪ ⎪-⎝⎭ X = 02112211321303330110311--⎛⎫⎛⎫⎛⎫⎪⎪ ⎪--=-⎪⎪ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭⎝⎭(2分) 3、解：非齐次线性方程组的增广矩阵为()⎪⎪⎪⎭⎫ ⎝⎛-----==b a A B 1223131121β⎪⎪⎪⎭⎫ ⎝⎛-+---225050501121~b a []10011101201j c bc a b a (n )b a b j ,,nab--======+--=-[]1(1)().n a n b a b -=+--⎪⎪⎪⎭⎫ ⎝⎛++---320010101121~b a (2分) 所以（1）当3,2-≠-=b a 时，()()B R A R ≠，非齐次线性方程组无解； (2分)（2）当2-≠a 时，()()3==B R A R ，非齐次线性方程组有唯一解； (2分)（3）当3,2-=-=b a 时，()()3<=B R A R ，非齐次线性方程组有无穷多解，(2分)当3,2-=-=b a 时，⎪⎪⎪⎭⎫⎝⎛---⎪⎪⎪⎭⎫ ⎝⎛---000010101101~000010101121~B =R (4分) 矩阵R 对应的线性方程组为1321,1.x x x -=⎧⎪=-⎨⎪⎩把3x 看成自由未知数，取3x =k,k 为任意实数得1231,1.x k x x k=+⎧⎪=-⎨⎪=⎩所以，其通解为123111*********x k k x x k x k k +⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪--- ⎪⎪ ⎪ ⎪ ⎪ ⎪===+=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭（其中k 为任意实数.） 4、解 (1) A 的特征多项式为|A -λE|=λλλ---111011002=(1-λ)2(2-λ)所以A 的特征值为λ1=2, λ2=λ3=1. (4分)当λ1=2时,解线性方程组(A-2E)x =0.由A-2E=⎪⎪⎪⎭⎫ ⎝⎛--111011000∽⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛--00021102101得基础解系x 1=(1/2，1/2，1)T所以对应于λ1=2的所有特征向量为k 1 x 1 （k 1≠0）当λ2=λ 3 =1时,解线性方程组(A-E)x =0.由 (4分)A- E=⎪⎪⎪⎭⎫ ⎝⎛011001001∽⎪⎪⎪⎭⎫ ⎝⎛000010001得基础解系x 2=(0，0，1)T所以对应于λ2=λ 3 =1的所有特征向量为k 2 x 2 （k 2≠0） (4分)。

西南大学课程考核《 线性代数 》课程试题 【A 】卷3. A =⎪⎪⎪⎭⎫ ⎝⎛300020001, then A -1 is ( )A. ⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛10002100031 B. ⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛31000210001 C. ⎪⎪⎪⎪⎪⎭⎫ ⎝⎛21000100031 D.⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛100031000214. A , B , C are square matrices and AB =AC , we can conclude ( ):A. A is a zero matrixB. A is a zero matrix if B ≠CC. B =C if A is not zero matrixD. B =C if det A ≠05. Rewrite ⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛-=2121211422),(),(x x x x x x f as a standard quadratic form x T A x , A = ( ) A. ⎪⎪⎭⎫ ⎝⎛1-332B. ⎪⎪⎭⎫ ⎝⎛1-422C. ⎪⎪⎭⎫⎝⎛1-212 D. ⎪⎪⎭⎫⎝⎛1001III. Solve the following problem (15points)Given a homogenous linear system: ⎪⎩⎪⎨⎧-=+++=+++=+++2321321321)1()1(0)1(λλλλλx x x x x x x x xFind the value of λ when the system has:a) Exactly one solution; b) No solution;c) Infinitely many solutions.In case c), describe all possible solutions as a linear combination of some vectors. Answer:西南大学课程考核(试题【A】卷) ——————————————密————————————封————————————线——————————————《 线性代数 》课程试题 【A 】卷IV . Calculate the determinant of matrix A (9points)⎪⎪⎪⎪⎪⎭⎫⎝⎛--+---+---=1111111111111111x x x x A Answer:西南大学课程考核(试题【A】卷)——————————————密————————————封————————————线——————————————V. Calculation (10 points)Matrix A and B satisfy AB=A+2B,⎪⎪⎪⎭⎫⎝⎛=321-11324A, Calculate matrix B. Answer:《线性代数》课程试题【A】卷VI. (12points)A linear transformation in 2 is defined as the composition of the following steps:(1)Rotate θ counterclockwise (逆时针旋转θ), where θ=tan-1(0.75); (0.75的反正切)(2)Scale by the factor 2 (放大到原来的2倍);(3)Reflection through x1-axis (绕x1轴翻转).Determine: the standard matrix of this linear transformation.Is this linear transformation a one to one linear transformation? Explain the reason. Answer:西南大学课程考核(试题【A】卷)——————————————密————————————封————————————线——————————————VII. System analysis (12points)The Internet service supply is a duopoly (两家垄断) in a city. Investigation shows that in customer who currently contracted with company A, 10% will contract with company B in the next year. On the other hand, in the customer who currently contracted with company B, 15% will move to company A in the next year.a)Establish the model which describes the relationship between the market share of currentyear and the market share of the next year.b)Estimate the vector which describes the long-term market share (长期以后的市场份额) ofthese two companies. Explain why it is a stable status and how to converge to this status.(解释为何此份额是稳定的，以及是怎样收敛到此数值的).Answer:《线性代数》课程试题【A】卷VIII. Prove the following statement (7points)A is a 3×3 matrix and A T=-A.Prove: A is not invertible.Answer:。