hw13.solutions

# At a 3 3 3 11 1 56 10 8 8 12 24 2 4 3 and at

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Unformatted text preview: −4, −4, −4)|| = 64 = 8. Section 6.5, #2. Compute AT A = 12 8 8 10 12 8 8 , and AT b = 10 x y = −24 −2 so solve −24 −2 Since the matrix is invertible this is easy x y #4. AT A = 3 3 3 11 = 1 56 10 −8 −8 12 −24 −2 = −4 3 and AT b = (6, 14), so solve 33 3 11 x y = 6 14 The solution is (1, 1). #6. The equaion AT Ax = AT b is (augmented matrix form) 6 3 3 27 3 3 0 12 3 12 3 15 By row reduction the solution is x = (5, −1, 0) + x3 (−1, 1, 1) 1 #8. The error is ||Ax − b||, with A = 1 ˆ 1 √ from Problem 4. The answer is 6. 3 −1, x = (1, 1) and b = (5, 1, 0) ˆ 1 #10. The projection is ˆ = (4, −1, 4) and x = (3, 1/2). b ˆ 2...
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## This note was uploaded on 12/15/2013 for the course MATH 461 taught by Professor Jake during the Fall '08 term at Maryland.

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