Final practice _Solutions - Ohio State University MATH 2568

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Math 2568 Name Final Exam Instructor 12/7/18 Time Limit: xx Minutes The exam contains 10 questions on 16 pages, including the cover page. Circle your final answers, and indicate clearly if they ... are on the back of the page. You may not use your books, notes, phones, calculators, or any other external resource. Problem Points Score 1 22 2 16 3 20 4 22 5 20 6 18 7 18 8 16 9 24 10 24 Total: 200 Math 2568 Final Exam - Page 2 of 16 12/7/18 1. Consider the matrix A and vector b given below. A =      1 0 −1 −2 0 2 0 1 −2 0 0 0      b =      y1 y2 y3 y4      ∈ R 4 (a) (8 points) Describe the sets of vectors ~y ∈ R 4 so that the system of equations Ax = b has (i) no solution, (ii) one solution, and (iii) infinitely many solutions. Solution: We row reduce the corresponding augmented matrix      1 0 −1 y1 −2 0 2 y2 0 1 −2 y3 0 0 0 y4      →      1 0 −1 y1 0 0 0 2y1 + y2 0 1 −2 y3 0 0 0 y4      →      1 0 −1 y1 0 1 −2 y3 0 0 0 2y1 + y2 0 0 0 y4      (i) There are no solutions if y1 + 2y2 6= 0 or y4 6= 0. (ii) rank(A) = 2 < 3, so the system cannot have a unique solution. (iii) There is a unique solution if y1 + 2y2 = 0 and y4 = 0. Math 2568 Final Exam - Page 3 of 16 12/7/18 (b) (8 points) Which (if any) of the sets you found in part (a) are subspaces of R 4 ? Explain your answer. Solution: (i) The set of vectors y ∈ R 4 so that Ax = y is inconsistent does not contain 0, so it is not a subspace of R 4 . (ii) The empty set is not a subspace of R 4 , so the set of vectors y so that Ax = y has a unique solution is not a subspace of R 4 . (iii) The set of vectors y ∈ R 4 so that Ax = y has infinitely many solutions is a subspace of R 4 ; it equals the range of LA : R 4 → R 4 , where LA is given by LA(x) = Ax. (c) (6 points) Is the vector y given below a linear combination of the columns of A? If so, express it as one. y =      2 −4 5 0      Solution: Plugging y1 = 2, y2 = −4, y3 = 5, and y4 = 0 into the reduced augmented matrix in part (a) gives:      1 0 −1 2 0 1 −2 5 0 0 0 0 0 0 0 0      So [ 2 5 0 0 ] T is a solution to Ax = y, and 2      1 −2 0 0      + 5      0 0 1 0      + 0      −1 2 −2 0      =      2 −4 5 0      Therefore, y is a linear combination of the columns of A. Math 2568 Final Exam - Page 4 of 16 12/7/18 2. Let u =    3 2 −5   , v =    −6 1 7   , w =    0 5 −3    (a) (8 points) Determine whether the sets {u, v}, {u, w}, {v, w} are linearly independent. Solution: We know that a set of two vectors is linearly dependent if and only if one is a scalar multiple of the other. We can see that none of the vectors is a scalar multiple of another. Thus, {u, v}, {u, w}, {v, w} are linearly independent. (b) (8 points) Is the set {u, v, w} linearly independent? Solution: To tell whether {u, v, w} is linearly independent, we solve the equation xu + yv + zw = 0: x    3 2 −5    + y    −6 1 7    + z    0 5 −3    =    0 0 0    We row reduce the coefficient matrix of this system:    3 −6 0 2 1 5 −5 7 −3    →    1 −2 0 2 1 5 −5 7 −3    →    1 −2 0 0 5 5 0 −3 −3    →    1 −2 0 0 1 1 −5 1 −1    →    1 −2 0 0 1 1 0 0 0    →    1 0 2 0 1 1 0 0 0    Therefore, there are inf [Show More]

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