homework_5_solutions - MA 265 HOMEWORK ASSIGNMENT #5...

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Unformatted text preview: MA 265 HOMEWORK ASSIGNMENT #5 SOLUTIONS #1. Page 145; Exercise 3. Determine whether each of the following permutations of S = { 1 , 2 , 3 , 4 } is even or odd: (a) 4213 (b) 1243 (c) 1234 Solution: (a) This permutation sends 4213 : 1 7 4 2 7 2 3 7 1 4 7 3 Reading downward, we see that 4 > 2, 4 > 1, 4 > 3, and 2 > 1 . Hence there are 4 inversions, so that 4213 is even. (b) This permutation sends 1243 : 1 7 1 2 7 2 3 7 4 4 7 3 Reading downward, we see that 4 > 3. Hence there is just 1 inversion, so that 1243 is odd. (c) This permutation sends 1234 : 1 7 1 2 7 2 3 7 3 4 7 4 Reading downward, we see that there are no inversions. Hence 1234 is even. #2. Page 145; Exercise 5. Determine the sign associated with each of the following permutations of the column indices of a 5 5 matrix: (a) 25431 (b) 31245 (c) 21345 1 2 MA 265 HOMEWORK ASSIGNMENT #5 SOLUTIONS Solution: (a) We determine the number of inversions. This permutation sends 25431 : 1 7 2 2 7 5 3 7 4 4 7 3 5 7 1 Reading downward, we see that 2 > 1, 5 > 4, 5 > 3, 5 > 1, 4 > 3, 4 > 1, and 3 > 1. Hence there are 7 inversions, so that this permutation is odd. The sign of 25431 is - . (b) Again, we determine the number of inversions. This permutation sends 31245 : 1 7 3 2 7 1 3 7 2 4 7 4 5 7 5 Reading downward, we see that 3 > 1 and 3 > 2. Hence there are 2 inversions, so that this permutation is even. The sign of 31245 is +. (b) Again, we determine the number of inversions. This permutation sends 21345 : 1 7 2 2 7 1 3 7 3 4 7 4 5 7 5 Reading downward, we see that 2 > 1. Hence is just 1 inversion, so that this permutation is odd. The sign of 21345 is - . #3. Page 155; Exercise 2. Compute the following determinants via reduction to triangular form or by citing a particular theorem or corollary: (a) 2- 2 3- 1 (b) 4 2 0- 2 5 0 3 (c) 3 4 2 2 5 0 3 0 0 (d) 4- 3 5 5 2 0 2 0 4 (e) 4 0 0 0- 1 2 0 0 1 2- 3 0 1 5 3 5 MA 265 HOMEWORK ASSIGNMENT #5 SOLUTIONS 3 (f) 2 0 1 4 3 2- 4- 2 2 3- 1 11 8- 4 6 Solution: (a) We use the definition of the determinant: 2- 2 3- 1 = (2)(- 1)- (- 2)(3) =- 2 + 6 = 4 = 2- 2 3- 1 = 4 ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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homework_5_solutions - MA 265 HOMEWORK ASSIGNMENT #5...

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