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# homework_5_solutions - MA 265 HOMEWORK ASSIGNMENT#5...

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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) 4 2 1 3 (b) 1 2 4 3 (c) 1 2 3 4 Solution: (a) This permutation sends 4 2 1 3 : 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 4 2 1 3 is even. (b) This permutation sends 1 2 4 3 : 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 1 2 4 3 is odd. (c) This permutation sends 1 2 3 4 : 1 7→ 1 2 7→ 2 3 7→ 3 4 7→ 4 Reading downward, we see that there are no inversions. Hence 1 2 3 4 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) 2 5 4 3 1 (b) 3 1 2 4 5 (c) 2 1 3 4 5 1

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2 MA 265 HOMEWORK ASSIGNMENT #5 SOLUTIONS Solution: (a) We determine the number of inversions. This permutation sends 2 5 4 3 1 : 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 2 5 4 3 1 is “ - ”. (b) Again, we determine the number of inversions. This permutation sends 3 1 2 4 5 : 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 3 1 2 4 5 is “+”. (b) Again, we determine the number of inversions. This permutation sends 2 1 3 4 5 : 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 2 1 3 4 5 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 0 - 2 5 0 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 0 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 . (b) The matrix is upper triangular, so we use Theorem 3.7 on page 149 of the text: 4 2 0 0 - 2 5 0 0 3 = (4) ( - 2) (3) = - 24 = 4 2 0 0 - 2 5 0 0 3 = - 24. (c) Upon interchanging the first and third columns, we see that the matrix will be upper triangular. We use Theorems 3.2 on page 147 and 3.7 on page 149 of the text: 3 4 2 2 5 0 3 0 0 = ( - 1) 2 4 3 0 5 2 0 0 3 = ( - 1) · (2) (5) (3) = - 30 = 3 4 2 2 5 0 3 0 0 = - 30.

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homework_5_solutions - MA 265 HOMEWORK ASSIGNMENT#5...

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