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# sol2 - SOLUTIONS FOR HOMEWORK 2 Section 1.4 3(b Yes the rst...

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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. Double-checking, 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 = 0 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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