ps4_solution - TA: Yankun Wang Fall 2006 Economics 617...

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TA: Yankun Wang Fall 2006 Economics 617 Problem Set 4 Suggested Solution 1. [Properties of Determinants] Suggested Proof: Since we are free to use any properties listed in the lecture notes, there are many ways to prove this problem. However, you are not allowed to use Property (D.6) on page 26 in the lecture notes, since otherwise there is nothing to prove. Proof (1). Let B = a 11 a 12 ... a 1 n a 11 a 12 ... a 1 n . . . . . . . . . . . . a n 1 a n 2 ... a nn , i.e., the f rst two rows of B are both the same as A 1 , the B 3 , ...B n are the same as A 3 , ...A n . Then expanding by the second row, we will have | B | = a 11 A 21 + a 12 A 22 + ... + a 1 n A 2 n . Property (D.2), which has been proved in class, tells us that | B | = | B | , and this could only be true if | B | =0 . Hence a 11 A 21 + a 12 A 22 + ... + a 1 n A 2 n =0 . Proof (2). The above proof is the most basic proof, and it’s derived directly from the de f nition of determinant. Thus is also the desired proof. However, after noticing | B | = a 11 A 21 + a 12 A 22 + ... + a 1 n A 2 n , we can use either Property (D.5) or Property (D.4) to get | B | =0 . Essentially (D.5) and (D.4) are con- sequences of our result in Problem 1, but there is nothing wrong with proving Problem 1 using these derived properties. 2. [Test of Linear Dependence of Vectors] Suggested Proof: Since this is an "if and only if" statement, we have two parts to prove; (i) If S = { x 1 ,x 2 , .., x m } is linearly dependent, then | G | =0 ,where G = x 1 x 1 x 1 x 2 ... x 1 x m ... ... ... ... x
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ps4_solution - TA: Yankun Wang Fall 2006 Economics 617...

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