Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Introduction to Linear...

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INTRODUCTION TO LINEAR ALGEBRA Fourth Edition MANUAL FOR INSTRUCTORS Gilbert Strang Massachusetts Institute of Technology math.mit.edu/linearalgebra web.mit.edu/18.06 video lectures: ocw.mit.edu math.mit.edu/ ± gs www.wellesleycambridge.com email: [email protected] Wellesley - Cambridge Press Box 812060 Wellesley, Massachusetts 02482

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2 Solutions to Exercises Problem Set 1.1, page 8 1 The combinations give (a) a line in R 3 (b) a plane in R 3 (c) all of R 3 . 2 v C w D .2; 3/ and v NUL w D .6; NUL 1/ will be the diagonals of the parallelogram with v and w as two sides going out from .0; 0/ . 3 This problem gives the diagonals v C w and v NUL w of the parallelogram and asks for the sides: The opposite of Problem 2. In this example v D .3; 3/ and w D .2; NUL 2/ . 4 3 v C w D .7; 5/ and c v C d w D .2c C d; c C 2d/ . 5 u C v D . NUL 2; 3; 1/ and u C v C w D .0; 0; 0/ and 2 u C 2 v C w D . add first answers / D . NUL 2; 3; 1/ . The vectors u ; v ; w are in the same plane because a combination gives .0; 0; 0/ . Stated another way: u D NUL v NUL w is in the plane of v and w . 6 The components of every c v C d w add to zero. c D 3 and d D 9 give .3; 3; NUL 6/ . 7 The nine combinations c.2; 1/ C d.0; 1/ with c D 0; 1; 2 and d D .0; 1; 2/ will lie on a lattice. If we took all whole numbers c and d , the lattice would lie over the whole plane. 8 The other diagonal is v NUL w (or else w NUL v ). Adding diagonals gives 2 v (or 2 w ). 9 The fourth corner can be .4; 4/ or .4; 0/ or . NUL 2; 2/ . Three possible parallelograms! 10 i NUL j D .1; 1; 0/ is in the base ( x - y plane). i C j C k D .1; 1; 1/ is the opposite corner from .0; 0; 0/ . Points in the cube have 0 ² x ² 1 , 0 ² y ² 1 , 0 ² z ² 1 . 11 Four more corners .1; 1; 0/; .1; 0; 1/; .0; 1; 1/; .1; 1; 1/ . The center point is . 1 2 ; 1 2 ; 1 2 / . Centers of faces are . 1 2 ; 1 2 ; 0/; . 1 2 ; 1 2 ; 1/ and .0; 1 2 ; 1 2 /; .1; 1 2 ; 1 2 / and . 1 2 ; 0; 1 2 /; . 1 2 ; 1; 1 2 / . 12 A four-dimensional cube has 2 4 D 16 corners and 2 ± 4 D 8 three-dimensional faces and 24 two-dimensional faces and 32 edges in Worked Example 2.4 A . 13 Sum D zero vector. Sum D NUL 2 : 00 vector D 8 : 00 vector. 2 : 00 is 30 ı from horizontal D . cos ± 6 ; sin ± 6 / D . p 3=2; 1=2/ . 14 Moving the origin to 6 : 00 adds j D .0; 1/ to every vector. So the sum of twelve vectors changes from 0 to 12 j D .0; 12/ . 15 The point 3 4 v C 1 4 w is three-fourths of the way to v starting from w . The vector 1 4 v C 1 4 w is halfway to u D 1 2 v C 1 2 w . The vector v C w is 2 u (the far corner of the parallelogram). 16 All combinations with c C d D 1 are on the line that passes through v and w . The point V D NUL v C 2 w is on that line but it is beyond w . 17 All vectors c v C c w are on the line passing through .0; 0/ and u D 1 2 v C 1 2 w . That line continues out beyond v C w and back beyond .0; 0/ . With c ³ 0 , half of this line is removed, leaving a ray that starts at .0; 0/ . 18 The combinations c v C d w with 0 ² c ² 1 and 0 ² d ² 1 fill the parallelogram with sides v and w . For example, if v D .1; 0/ and w D .0; 1/ then c v C d w fills the unit square.
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