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Unformatted text preview: Problem Set 4 Solutions Economics 6170 TA: Christopher Handy October 10, 2008 Problem 1 We can write the system (1) as Ax = b , where A = 2 4 3 3 2 3 , x = x 1 x 2 , b = 8 9 7 , A b = 2 4 8 3 3 9 2 3 7 (a) To show that Ax = b has a solution, we must show that rank( A ) = rank( A b ). First, rank( A ) = 2: rank( A ) ≤ 2 because A has only two columns, and rank( A ) ≥ 2 because A 1 ,A 2 are linearly independent. Second, rank( A b ) = 2: rank( A b ) ≥ 2 since A 1 b = A 1 and A 2 b = A 2 are linearly independent, and rank( A b ) < 3 since A 3 b = 2 A 1 b + A 2 b , so the three columns of A b are linearly dependent. Because rank( A b ) = rank( A ) = 2, the system Ax = b has a solution by the Existence Theorem from Chapter 3. (b) We can apply the Uniqueness Theorem from Chapter 3, since we have m = 3 ,n = 2. From part (a), rank( A ) = rank( A b ) = 2 = n , so the system Ax = b has a unique solution. Problem 2 [Note: You are free to just solve the system, but the method below may help you solve certain other existence or uniqueness problems.] We can write the system (2) as Ax = b , where A = 3 1 1 1 1 2 1 3 3 , x = x 1 x 2 x 3 , b = t 1 t 1 + t , A b = 3 1 1 t 1 1 2 1 t 1 3 3 1 + t 1 By the Existence Theorem from Chapter 3, the system Ax = b will have a solution when rank( A ) = rank( A b ). Now, rank( A ) = 2 because A 1 ,A 2 are linearly independent, but A 1 = 2 A 2 + A 3 . So we want to find the values of t for which rank( A b ) = 2. Since A 1 ,A 2 are linearly independent, if rank( A b ) = 2 then A 1 ,A 2 will form a basis for the columns of...
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This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell.
 Fall '08
 MITRA
 Economics

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