Introduction to Linear Algebra 3rd edition

Introduction to Linear Algebra, Third Edition

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INTRODUCTION TO LINEAR ALGEBRA Third Edition MANUAL FOR INSTRUCTORS Gilbert Strang [email protected] Massachusetts Institute of Technology Wellesley-Cambridge Press Box 812060 Wellesley, Massachusetts 02482
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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 3 v + w = (7 , 5) and v - 3 w = ( - 1 , - 5) and c v + d w = (2 c + d, c + 2 d ). 5 u + v = ( - 2 , 3 , 1) and u + v + w = (0 , 0 , 0) and 2 u +2 v + w = (add first answers) = ( - 2 , 3 , 1). 6 The components of every c v + d w 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 2 v (or 2 w ). 9 The fourth corner can be (4 , 4) or (4 , 0) or ( - 2 , 2). 10 i + j is the diagonal of the base. 11 Five more corners (0 , 0 , 1) , (1 , 1 , 0) , (1 , 0 , 1) , (0 , 1 , 1) , (1 , 1 , 1). The center point is ( 1 2 , 1 2 , 1 2 ). The centers of the six 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 = 16 corners and 2 · 4 = 8 three-dimensional sides and 24 two-dimensional faces and 32 one-dimensional edges. See Worked Example 2.4 A . 13 sum = zero vector; sum = - 4:00 vector; 1:00 is 60 from horizontal = (cos π 3 , sin π 3 ) = ( 1 2 , 3 2 ). 14 Sum = 12 j since j = (0 , 1) is added to every vector. 15 The point 3 4 v + 1 4 w is three-fourths of the way to v starting from w . The vector 1 4 v + 1 4 w is halfway to u = 1 2 v + 1 2 w , and the vector v + w is 2 u (the far corner of the parallelogram). 16 All combinations with c + d = 1 are on the line through v and w . The point V = - v + 2 w is on that line beyond w . 17 The vectors c v + c w fill out the line passing through (0 , 0) and u = 1 2 v + 1 2 w . It continues beyond v + w and (0 , 0). With c 0, half this line is removed and the “ray” starts at (0 , 0). 18 The combinations with 0 c 1 and 0 d 1 fill the parallelogram with sides v and w . 19 With c 0 and d 0 we get the “cone” or “wedge” between v and w . 20 (a) 1 3 u + 1 3 v + 1 3 w is the center of the triangle between u , v and w ; 1 2 u + 1 2 w is the center of the edge between u and w (b) To fill in the triangle keep c 0, d 0, e 0, and c + d + e = 1. 3
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4 21 The sum is ( v - u ) + ( w - v ) + ( u - w ) = zero vector. 22 The vector 1 2 ( u + v + w ) is outside the pyramid because c + d + e = 1 2 + 1 2 + 1 2 > 1. 23 All vectors are combinations of u , v , and w . 24 Vectors c v are in both planes. 25 (a) Choose u = v = w = any nonzero vector (b) Choose u and v in different directions, and w to be a combination like u + v . 26 The solution is c = 2 and d = 4. Then 2(1 , 2) + 4(3 , 1) = (14 , 8). 27 The combinations of (1 , 0 , 0) and (0 , 1 , 0) fill the xy plane in xyz space. 28 An example is ( a, b ) = (3 , 6) and ( c, d ) = (1 , 2). The ratios a/c and b/d are equal. Then ad = bc . Then (divide by bd ) the ratios a/b and c/d are equal! Problem Set 1.2, page 17 1 u · v = 1 . 4, u · w = 0, v · w = 24 = w · v . 2 u = 1 and v = 5 = w . Then 1 . 4 < (1)(5) and 24 < (5)(5). 3 Unit vectors v / v = ( 3 5 , 4 5 ) = ( . 6 , . 8) and w / w = ( 4 5 , 3 5 ) = ( . 8 , . 6). The cosine of θ is v v · w w = 24 25 . The vectors w , u , - w make 0 , 90 , 180 angles with w .
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