HWsolutions1 - HOMEWORK ASSIGNMENT 1 SOLUTIONS 1.1 #3.b Let...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
HOMEWORK ASSIGNMENT 1 SOLUTIONS § 1.1 # 3.b Let A = (3 , - 6 , 7), B = ( - 2 , 0 , - 4), and C = (5 , - 9 , - 2) be the three given points. Set u = B - A = ( - 2 , 0 , - 4) - (3 , - 6 , 7) = ( - 5 , 6 , - 11) and v = C - A = (5 , - 9 , - 2) - (3 , - 6 , 7) = (2 , - 3 , - 9) . Following example 2 in the book, the equation for the plane is x = A + su + tv = (3 , - 6 , 7) + s ( - 5 , 6 , - 11) + t (2 , - 3 , - 9) . § 1.1 # 3.d Let A = (1 , 1 , 1), B = (5 , 5 , 5), and C = ( - 6 , 4 , 2) be the three given points. Set u = B - A = (5 , 5 , 5) - (1 , 1 , 1) = (4 , 4 , 4) and v = C - A = ( - 6 , 4 , 2) - (1 , 1 , 1) = ( - 7 , 3 , 1) . Again, following example 2 in the book, the equation for the plane is x = A + su + tv = (1 , 1 , 1) + s (4 , 4 , 4) + t ( - 7 , 3 , 1) . § 1.1 # 7 This was done in class. § 1.2 # 1 (a) True. This is axiom (VS 3) of a vector space. (b) False. The zero vector is unique by Corollary 1, page 11. (c) False. For example, let a = 0, b = 1, and x = ~ 0. Then ax = bx = ~ 0, but a 6 = b . (d) False. For example, consider the vector space R 2 . If a = 0, x = (1 , 1) and y = (1 , 0) then ax = ay = (0 , 0) but x 6 = y . (e) True. Just think of them as column vectors. (f) False. An m × n matrix has m rows and n columns. (g) False. For example, we can add x + 1 and x 2 to get x 2 + x + 1. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 HOMEWORK ASSIGNMENT 1 SOLUTIONS (h) False. For example, let f = x + 1 and g = - x . Then deg( f ) = deg( g ) = 1, but f + g = 1 has degree 0. (i) True. To prove this, let f ( x ) = a n x n + a n - 1 x n - 1 + ··· + a 1 x + a 0 , with a n 6 = 0 (since deg( f ) = n ). Then cf ( x ) = ca n x n + ··· ca 1 x + ca 0 , and if c 6 = 0 then ca n 6 = 0. Therefore deg( cf ( x )) = n . (j) True. Obvious. (k) True. This is by de±nition of equality among functions. See example 3, page 9. § 1.2 # 18 Let V = { ( a 1 , a 2 ) : a 1 , a 2 R } . For ( a 1 , a 2 ) , ( b 1 , b 2 ) V and c R , de±ne ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + 2 b 1 , a 2 + 3 b 2 ) and c ( a 1 , a 2 ) = ( ca 1 , ca 2 ) . The trick to this problem is to notice that the multiplication looks normal, but the addition looks funny. So we should probably check the addition rules, to see if they make sense. Let’s check (VS 1). From our addition rule, we have ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + 2 b 1 , a 2 + 3 b 2 ). Switching the summands, but using the same addition rule, we have ( b 1 , b 2 ) + ( a 1 , a 2 ) = ( b 1 + 2 a 1 , b 2 + 3 a 2 ). These two answers do not look like they are the
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

HWsolutions1 - HOMEWORK ASSIGNMENT 1 SOLUTIONS 1.1 #3.b Let...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online