This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS FOR HOMEWORK 2 Section 1.4 3. (b) Yes , the first vector is a linear combination of the other two. To verify this, we need to find a,b R s.t. (1 , 2 , 3) = a ( 3 , 2 , 1)+ b (2 , 1 , 1). This gives rise to a system of three equations with two unknowns: 3 a + 2 b = 1 2 a b = 2 a b = 3 To solve this system of equations, we eliminate the unknown a from Equations 1 and 2. More precisely, we keep Eq 3, and use Eq 1 + 3 Eq 3 and Eq 2 Eq 3 instead of Eq 1 and Eq 2, respectively. This results in a new system of equations, with the same solution(s) as the old one. b = 8 b = 8 a b = 3 This leads to b = 8, and a = b 3 = 5. Doublechecking, we verify the equality (1 , 2 , 3) = 5( 3 , 2 , 1) + 8(2 , 1 , 1). (d) No , the first vector is not a linear combination of the other two. To verify this, we need to prove that no a,b R satisfy (2 , 1 , 0) = a (1 , 2 3)+ b (1 , 3 , 2). The last equality gives rise to a system of three equations with two unknowns: a + b = 2 2 a 3 b = 1 3 a + 2 b = To solve this system of equations, we eliminate the unknown a from Equations 2 and 3. More precisely, we keep Eq 1, and use Eq 2 2 Eq 1 and Eq 3 + 3 Eq 1 instead of Eq 2 and Eq 3, respectively. This yields a new system of equations, with the same solution(s) as the original one. a + b = 2 5 b = 5 + 7 b = 6 The last two equalities are mutually contradictory, hence the system has no solution. 4. (b) No , 4 x 3 +2 x 2 6 is not a linear combination of x 3 2 x 2 +4 x +1 and 3 x 3 6 x 2 + x +4. Indeed, the first polynomial is a linear combination of the other two iff there exist a,b R for which 4 x 3 + 2 x 2 + 0 x 6 = a ( x 3 2 x 2 + 4 x + 1) + b (3 x 3 6 x 2 + x + 4) ....
View Full
Document
 Spring '10
 FUCKHEAD

Click to edit the document details