MATH311ansHW5

MATH311ansHW5 - Math 311, Homework 5 partial answers and...

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Unformatted text preview: Math 311, Homework 5 partial answers and solutions 3.1.1. 1 2 . 2 4 The columns of A are linearly dependent, and A is not invertible, so f is not one-to-one. A= 3.1.4. Answer: A= Not one-to-one. 3.1.5. 1 0 Thus f (e1 ) = Similarly, f (e2 ) = 3.1.8. Answer: 0 3 2 -1 1 2 . 1 -1 1 = 1 2 1 2 1 1 - 1 2 1 -1 . -1 . 1 = 1 2 1 2 - 1 2 1 . 3 1 , f (e1 ) = 2 -5 0 , f (e2 ) = -1 7 0 . f (e3 ) = 1 3.1.9. |R x| = = = x1 cos - x2 sin x1 sin + x2 cos = (x1 cos - x2 sin )2 + (x1 sin + x2 cos )2 x2 cos2 - 2x1 x2 cos sin + x2 sin2 + x2 sin2 + 2x1 x2 sin cos + x2 cos2 1 2 1 2 x2 + x2 = |x| . 1 2 3.1.10. Answer: the angle is . 3.1.13. The matrix for g f is the product of the matrix for g and the matrix for f . See the textbook for the answer. 1 3.1.19. The formula for Pn is Pn (x) = (x n)n. So 3 Pn (e1 ) = n = 7 6 Pn (e2 ) = n = 7 2 Pn (e3 ) = n = 7 The matrix for Pn has these vectors as columns. 9 18 6 , , 49 49 49 18 36 12 , , 49 49 49 6 12 4 , , 49 49 49 , , . ...
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This note was uploaded on 04/29/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.

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MATH311ansHW5 - Math 311, Homework 5 partial answers and...

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