2009级线性代数与工程应用期末试题解答(b)卷（2009 level linear algebra and engineering application final examination questions (b) volume）.docVIP

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2009级线性代数与工程应用期末试题解答(b)卷（2009 level linear algebra and engineering application final examination questions (b) volume）.doc

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2009级线性代数与工程应用期末试题解答(b)卷（2009 level linear algebra and engineering application final examination questions (b) volume）

2009级线性代数与工程应用期末试题解答(b)卷（2009 level linear algebra and engineering application final examination questions (b) volume） 2009 linear algebra and engineering application of the final answers to questions (volume B) Please note: all the answers will be written on the answer sheet, write invalid in the test paper. First, fill in the blanks (4 points per day, a total of 40 points) 1131. 1. D= 32 Twenty-one Eight Four Zero Three Aij, representative of D row I column J algebraic cofactor element type, then 4125 A+2AAA+=_____0_____. +111 One 11 X. 1 One 2. equation 112 X. 1 0= The solution for the 111. (1) n X.. 12 n.1. _________. _ x=0, x=1, xn2.. .100..100. Three Let A=.. 220.. As with the matrix A *, A*.=_____, 1. 220.. _____. A (1) Ten . . Three hundred and forty-five Three hundred and forty-five . . 4. order n matrix A meet Ak=O, where k is a positive integer, then (. =_).1 Ak.1 +..AE_________. EA + Five Two f=x12.2x1 x2 +2x1 x3.3x22.6x2 x3 ____2_____ rank. Six Let V be a linear space consisting of all matrices of order n, A, V, and A are all commutative matrix A subspace or A? ____ is ______ (fill in yes or no) Seven Let a = (0,1,2,3) 1= (2,2,3,1), beta, beta = 2 (1,2,1,2), beta = 3 (2,1, 1, 2) is Then, alpha... It can be expressed as a linear combination of,,? _____ or _____ (fill in yes or no) Beta Beta One hundred and twenty-three One .1.11. 8. a = D columns, if alpha alpha T=. ..11.1. Then the alpha T alpha =____3____. . 1.11 . The 9. set of 4 order matrix A is similar to B, and the eigenvalue of A is 2, 3, 4, 5, then BE . =___24______. .101. 10. A=.. Twenty K is a positive integer, Ak.2Ak.1 =___0______. And (2). . One hundred and one . Two, find the value of K (8), (=x3) makes the whole coefficient polynomial +kx2 +3x+2 is rational. FX Solution: (=x3) +kx2 integer polynomial +3x+2 with rational root, may only 1, FX + + 2, (4 points) So the solution of F (1) 0, f (1) =0, f.= F (2) =0, k=. (4 points). . (2) 0, 6,2, 4,3. Three, (10) the following defi