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Midterm1-Solutions

Midterm1-Solutions - Name Math 331 Wylie Midterm 1 This is...

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. Name: ...................................................................................................... Math 331 – Wylie – Midterm 1 – September 27, 2011 This is the first midterm exam in Math 331. There are ten questions and each question is worth 10 points. No notes are allowed. You are allowed only a standard scientific calculator (although the exam has been designed so that you don’t need it). You do not need to show any work on the first true/false question. For the other nine problems, show all of your work. Correct answers with no or little justification will receive no points. It is more important that you answer all of the questions that you know how to do well than it is to finish every question on the exam. (DO NOT WRITE BELOW THIS LINE, THIS SECTION IS FOR GRADING PURPOSES ONLY.) ————————————————————————————————————————— 1. ................ 6. ................ 2. ................ 7. ................ 3. ................ 8. ................ 4. ................ 9. ................ 5. ................ 10. ................ Total: ...............................................................

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1. True of False: (a) If two matrices A and B have the same reduced row echelon form, then the equations A x = 0 and B x = 0 have the same solutions. True. (b) There exists a 3 × 2 matrix with 2 pivot columns such that A 1 2 = 0 . False. (c) If A x = b has more than one solution for some vector b , then A x = 0 has more than one solution. True. (d) If A is a 6 × 4 matrix, then there exists a vector b in R 6 such that the system A x = b is inconsistent. True. (e) There is a list of 3 vectors in R 4 which spans R 4 False.
2. Write - 1 1 3 as a linear combination of the vectors 1 1 1 0 2 1 1 1 0 Solution: We solve the linear system 1 0 1 - 1 1 2 1 1 1 1 0 3 II - I ---→ III - I 1 0 1 - 1 0 2 0 2 0 1 - 1 4 II ÷ 2 ---→ 1 0 1 - 1 0 1 0 1 0 1 - 1 4 III - II ----→ 1 0 1 - 1 0 1 0 1 0 0 - 1 3 III ÷ ( - 1) -----→ 1 0 1 - 1 0 1 0 1 0 0 1 - 3 I - III ---→ 1 0 0 2 0 1 0 1 0 0 1 - 3 So we get an answer of x 1 = 1, x 2 = 1, and x 3 = - 2. These are the weights in the linear combination. The solution is then - 1 1 3 = 2 1 1 1 + 1 0 2 1 - 3 1 1 0 Which we can check is true by adding up the right hand side.

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