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선형대수학 introduction to linear algebra gilbert strang 3rd edition 3판 솔루션

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선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition
선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition
선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition
선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition
선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition
선형대수학_introduction_to_linear_algebra_gilbert_strang_3rd_edition

INTRODUCTION TO LINEAR ALGEBRA Third Edition
MANUAL FOR INSTRUCTORS

Gilbert Strangxxxx@xxxx.mit.edu

Massachusetts Institute of Technology
http://web.mit.edu/18.06/www http://math.mit.edu/?gs http://www.wellesleycambridge.com

Wellesley-Cambridge Press Box 812060 Wellesley, Massachusetts 02482

Solutions to Exercises
Problem Set 1.1, page 6
1 Line through (1, 1, 1); plane; same plane! 3 v = (2, 2) and w = (1, ?1). 4 3v + w = (7, 5) and v ? 3w = (?1, ?5) and cv + dw = (2c + d, c + 2d). 5 u + v = (?2, 3, 1) and u + v + w = (0, 0, 0) and 2u + 2v + w = (add ?rst answers) = (?2, 3, 1). 6 The components of every cv + dw add to zero. Choose c = 4 and d = 10 to get (4, 2, ?6). 8 The other diagonal is v ? w (or else w ? v ). Adding diagonals gives 2v (or 2w ). 9 The fourth corner can be (4, 4) or (4, 0) or (?2, 2). 10 i + j is the diagonal of the base.
1 11 Five more corners (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1). The center point is ( 2 , 1 , 1 ). The 2 2 1 centers

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