FEBRUARY 2015 MATH1231 Page 5 Use a separate book clearly marked Question 3 3. a

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FEBRUARY 2015 MATH1231 Page 5 Use a separate book clearly marked Question 3 3. a) Suppose S-(Pi (2), P2(x), Pa(z), P4(r), ps(r), where P2(z) = 1-1-13 i) Without using the Maple output below or doing any row operations, explain why S is a linearly dependent set ii) For the remainder of this subquestion, you may use the Maple output below, or any other method a) Write one of the polynomials in S as a linear combination of the others Find conditions, if any, on a general polynomial p(x) = ao + a a2x2 + a3z? e P3(R) so that p(x) E span(S) ) )What is the dimension of span(S)? Give a reason 6) Find a basis for P (R) containing as many polynomials from S as possible > with(LinearAlgebra) > 1d4:IdentityMatrix(4): > A1 : = ReducedRowEchelonForm (AI); 1 0 2 -1 -3 00 -1 0 01 3 23 00 -2 -1 0 0 0 0 0 1 0 31 0 0 0 0 0 0 -3 -1 Please see over .. .

Explanation / Answer

3 a) (i) There are 5 polynomials in 3 unknowns x1, x2 and x3 and so we get 5x3 matrix

Hence the system will be linearly dependent

(ii) a(1-x-x2+2x3 )+b(1-x-x3) +c(5-5x-2x2+x3)+d(1-x+x2-4x3)+e(3x2-9x3)=0

=> a+b+5c+d=0, -a-b-5c-d=0, -a-2c+d+3e=0, 2a-b+c-4d-9e=0

c,d,e are free using Maple output

a= 2c+d+3e; b = -3c-2d-3e

Let c=1,d=0,e=0

=> a=2, b =-3

Hence (a,b,c,d,e) = (2,-3,1,0,0)

So we have 2p1(x)-3p2(x)+p3(x)=0

=> p3(x) = -2p1(x)+3p2(x)

There is no condition

Span(S) = {p1(x), p2(x), -2p1(x)+3p2(x), p4(x) ,p5(x)}

Dimension Span(S) = 3

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