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Mathematics for Economists

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ANSWERS PAMPHLET 51 b ) x ( s, t ) 5 0 3 2 1 3 0 2 1 t 1 2 2 2 2 s , x 1 2 y 1 3 z 5 12. 10.37 a ) Rewriting the symmetric equations, y 5 y 0 2 ( b 6 a ) x 0 1 ( b 6 a ) x and z 5 z 0 2 ( c 6 a ) x 0 1 ( c 6 a ) x . Then x ( t ) y ( t ) z ( t ) 5 x 0 y 0 z 0 1 1 b 6 a c 6 a t or x ( t ) y ( t ) z ( t ) 5 x 0 y 0 z 0 1 a b c t. b ) The two planes are described by any two distinct equalities in system (20). For example, x 2 x 0 a 5 y 2 y 0 b and y 2 y 0 b 5 z 2 z 0 c . In other words, 2 bx 1 ay 5 2 bx 0 1 ay 0 and cy 2 bz 5 cy 0 2 bz 0 . c ) i ) x 1 2 2 2 1 5 x 2 2 3 4 5 x 3 2 1 5 , ii ) x 1 2 1 4 5 x 2 2 2 5 5 x 3 2 3 6 . d ) i ) 4 x 1 1 x 2 5 11 and 5 x 2 2 4 x 3 5 11. ii ) 5 x 1 2 4 x 2 5 2 3 and 6 x 2 2 5 x 3 5 2 3 . 10.38 a ) Normals (1 , 2 , 2 3) and (1 , 3 , 2 2) do not line up, so planes intersect. b ) Normals (1 , 2 , 2 3) and ( 2 2 , 2 4 , 6) do line up, so planes do not intersect. 10.39 a ) ( x 2 1 , y 2 2 , z 2 3) ? ( 2 1 , 1 , 0) 5 0, so x 2 y 5 2 1. b ) The line runs through the points (4 , 2 , 6) and (1 , 3 , 11), so the plane must be orthogonal to the difference vector ( 2 3 , 1 , 5). Thus ( x 2 1 , y 2 1 , z 1 1) ? ( 2 3 , 1 , 5) 5 0, or 2 3 x 1 y 1 5 z 5 2 7. c ) The general equation for the plane is a x 1 b y 1 g z 5 d . The equations to be satisfied are a 5 d 6 a , b 5 d 6 b , and c 5 d 6 g . A solution is a 5 1 6 a , b 5 1 6 b . g 5 1 6 c and d 5 1, so (1 6 a ) x 1 (1 6 b ) y 1 (1 6 c ) z 5 1. 10.40 Plug x 5 3 1 t , y 5 1 2 7 t , and z 5 3 2 3 t into the equation x 1 y 1 z 5 1 of the plane and solve for t : (3 1 t ) 1 (1 2 7 t ) 1 (3 2 3 t ) 5 1 = t 5 2 6 3 . The point is ( 11 6 3 , 2 11 6 3 , 1 ) . 10.41 1 1 2 1 . . . 4 1 2 1 . . . 3 = 1 0 2 3 . . . 5 0 1 2 . . . 2 1 = x 5 5 1 3 z y 5 2 1 2 2 z. Taking z 5 t , write the line as x y z 5 5 2 1 0 1 t 3 2 2 1 .
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52 MATHEMATICS FOR ECONOMISTS 10.42 IS : [1 2 c 1 (1 2 t 1 ) 2 a 0 ] Y 1 ( a 1 c 2 ) r 5 c 0 2 c 1 t 0 1 I p 1 G LM : mY 2 hr 5 M s 2 M p . I p rises = IS moves up = Y p and r p increase. M s rises = LM moves up = Y p decreases and r p increases. m rises = LM becomes steeper = Y p decreases and r p increases. h rises = LM flatter = Y p increases and r p decreases. a 0 rises = IS flatter (with same r -intercept) = r p and Y p rise. c 0 rises = IS moves up = Y p and r p increase. t 1 or t 2 rises = IS steeper (with same r -intercept) = r p and Y p decrease. Chapter 11 11.1 Suppose v 1 5 r 2 v 2 . Then 1 v 1 2 r 2 v 2 5 r 2 v 2 2 r 2 v 2 5 0 . Suppose c 1 v 1 1 c 2 v 2 5 0 . Suppose that c i 0. Then c i v i 5 2 c j v j , so v i 5 ( c j 6 c i ) v j . 11.2 a ) Condition (3) gives the equation system 2 c 1 1 c 2 5 0 c 1 1 2 c 2 5 0. The only solution is c 1 5 c 2 5 0, so these vectors are independent. b ) Condition (3) gives the equation system 2 c 1 1 c 2 5 0 2 4 c 1 2 2 c 2 5 0. One solution is c 1 5 2 2, c 2 5 1, so these vectors are dependent.
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