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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) ....
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.
 Spring '10
 FUCKHEAD

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