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exam1_review - Exam 1 review Page 1 of 10 Review for...

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Exam: 1 - review Page: 1 of 10 Review for Midterm Exam 1 General stuff : The exam will cover the following sections in the book: 1.1, 1.2, 1.3, 1.4, 1.5, 1.7 and 1.8. : The questions are going to be similar to hw problems and to problems done in class. They might be phrased as multiple-choice questions, though. : No textbooks, notebooks, calculators, laptops, cell-phones (or any other communication devices) will be allowed. : You will have 45 min for the exam (the exam will start on 10:05am promptly). : Do not cheat. NOTE: I might have made simple mistakes (such as 2+3 = 6) in the solutions. Please recheck all the computations (please e-mail me if you find any). On the other hand, the methods of solutions are correct. Course: M340L Instructor: G. ˇ Zitkovi´ c Semester: Fall, 2005
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Exam: 1 - review Page: 2 of 10 Section 1.1 You should: : know what a solution to a system of linear equations means. : know the terminology (you will not be asked to define terms, but they will appear in problems). : be familiar with consistency (existence) and uniqueness issues. Problem 1. Here is a system of linear equations: 3 x 1 + 2 x 2 + 2 x 4 - 6 x 5 = 0 2 x 1 + 8 x 2 - 2 x 3 + 3 x 4 - 2 x 5 + x 6 = 5 - 4 x 1 + 16 x 2 + x 4 + 7 x 5 + 2 x 6 = 0 32 x 1 + 3 x 2 - x 4 + 27 x 5 - 7 x 6 = 0 (a) Find a solution to this system. (b) Are there any other solutions? Solution: (a) Inspection of the system above reveals that x 1 = x 2 = x 4 = x 5 = x 6 = 0, x 3 = - 5 / 2 is a solution. (b) Once we know (from (a)) that the system admits at least one solutions, the fact that the reduced echelon form will necessarily have pivot-less columns (more columns than rows) implies that the description of the solution set will involve free variables. Presence of free variables in a consistent system indicates that there are infinitely many solutions. Therefore, the answer is yes. Problem 2. For what values of h and k does the following system have infinitely many solutions? 4 x + 6 y = h - 2 x - 3 y = k Solution: First we reduce the augmented matrix of the system to echelon form (no need for reduced echelon form): 4 6 h - 2 - 3 k 4 6 h 0 0 k + 1 2 h The system will be consistent only if there are no rows of the form [0 0 . . . 0 a ], a = 0. In this case, if and only if k + 1 2 h = 0. In that case, x 2 is a free variable and the system has infinitely many solutions. (When k + 1 2 h = 0, then the system has no solutions. For no combination of h and k will the system have a unique solution.) Course: M340L Instructor: G.
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