164(Summer10)_PracticeMidterm_Solutions

164(Summer10)_PracticeMidterm_Solutions - MATH 164 Lecture...

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MATH 164 - Lecture 1 - Summer 2010 Practice Midterm Solutions - July 9, 2010 NAME: STUDENT ID #: This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 6 problems for a total of 100 points. POINTS: 1. 2. 3. 4. 5. 6. TOTAL: 1
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2 1. (20 points) Suppose x R n is an extreme point of the set { x : Ax = b, x 0 } where A is an m × n matrix and b R m . Show that x is a basic feasible solution. Solution. Since x is an extreme point , x S = { x : Ax = b, x 0 } and thus feasible. We need to show that x is a basic solution. Suppose x has k nonzero components. We can reorder the variables so that zero components of x are the last n - k . Denote the first k components of x by x B ( x B R k ), and the first k columns of A by B ( B is m × k matrix). Then Ax = Bx B . If the columns of B are linearly independent then we are done. Suppose that it is not the case. Then there exist a vector p R k , p 6 = 0 such that Bp = 0. Since x B > 0, we can find a small ² such that x B + ²p > 0 and x B - ²p > 0. Denote by y a vector in R n with the first k coordinates x B + ²p and the rest zero and by z a vector in R n with the first k coordinates x B - ²p and the rest zero. Then y, z 0 and Ay = B ( x B + ²p ) = Bx B + ²Bp = b, Az = B ( x B - ²p ) = Bx B - ²Bp = b So, y, z S . Since x = 1 2 y + 1 2 z
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