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fm241sol3

# fm241sol3 - M241 Matrix Algebra Homework Assignment 3...

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M241: Matrix Algebra Homework Assignment 3 Answers Graded Problems: Section 2.1 Problems 14, 22; Section 2.2 Problems 2, 5, 30, 48. Section 1.6. Problem 52. If A = 0 1 0 0 , then A 2 = 0 0 0 0 . However for any A = a b c d , we have that A T A = a c b d a b c d = a 2 + c 2 ab + cd ab + cd b 2 + d 2 . If A T A = 0, then all the entries in A T A must be zero. In particular a 2 + c 2 = 0 and b 2 + d 2 = 0, which implies a = 0, b = 0, c = 0 and d = 0. Thus A T A = 0 implies that A = 0. Q.e.d. Problem 60. The A = LDL T factorizations: Let A = 1 3 3 2 . Gaussian elimination gives the upper triangular form U 0 = 1 3 0 - 7 = 1 0 0 - 7 1 3 0 1 . Since L = 1 3 0 1 T , we have the following A = LDL T factorization: A = 1 3 3 2 = 1 0 3 1 1 0 0 - 7 1 3 0 1 . Let A = 1 b b c . Gaussian elimination gives the upper triangular form U 0 = 1 b 0 c - b 2 = 1 0 0 c - b 2 1 b 0 1 .

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Since L = 1 b 0 1 T , we have the following A = LDL T factorization: A = 1 b b c = 1 0 b 1 1 0 0 c - b 2 1 b 0 1 . Let A = 2 - 1 0 - 1 2 - 1 0 - 1 2 . Gaussian elimination gives the upper triangular form U 0 = 2 - 1 0 0 3 / 2 - 1 0 0 4 / 3 = 2 0 0 0 3 / 2 0 0 0 4 / 3 1 - 1 / 2 0 0 1 - 2 / 3 0 0 1 . Since L = 1 - 1 / 2 0 0 1 - 2 / 3 0 0 1 T , we have the following A = LDL T factorization A = 2 - 1 0 - 1 2 - 1 0 - 1 2 = 1 0 0 - 1 / 2 1 0 0 - 2 / 3 1 2 0 0 0 3 / 2 0 0 0 4 / 3 1 - 1 / 2 0 0 1 - 2 / 3 0 0 1 . Section 2.1. Problem 2. (a) The set of vectors of form (0 , b 2 , b 3 ) is a subspace of R 3 because 1) The vector (0 , 0 , 0) is in this set; 2) (0 , b 2 , b 3 ) + (0 , b 0 2 , b 0 3
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fm241sol3 - M241 Matrix Algebra Homework Assignment 3...

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