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

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



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

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_editionINTRODUCTION TO LINEAR ALGEBRA Third Edition
MANUAL FOR INSTRUCTORS

Gilbert Strang
gs@math.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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