Therefore x is a linearly independent subset of r4

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Unformatted text preview: namely a = b = c = 0. Therefore X is a linearly independent subset of R4 . Note that any linear combination of the vectors in X will result in a vector whose final entry is zero. Hence X does not span R4 . (Also, dim(R4 ) = 4, and we would need 4 linearly independent vectors to span R4 .) The dimension of Span(X ) is 3. Math 2061: Tutorial 4 (week 5) — Solutions Page 2 Linear Mathematics 1 1 1 5. Let X = Tutorial 4 (week 5) — Solutions 1 1 0 , 1 0 0 , 3 2 0 , Page 3 . 3 a) Show that X spans R . b) Explain why X is not a basis for R3 . c) Find a subset of X which is a basis for R3 . Solution a) X spans R3 if there is a solution, for every x y z 1 1 1 =a +b 1 1 0 +c x y z 1 0 0 ∈ R3 , to the equation +d 3 2 0 1113 1102 1000 = a b c d . The augmented matrix for this system of equations, and a row echelon form is 1113x 1102y 1000z 1113 x 0 1 1 3 x−z 0 0 1 1 x−y Row reduce −−− −−→ So there are infinitely many solutions for any x y z . ∈ R3 , and hence X spans R3 . b) Any basis of R3 contains exactly 3 vectors, and X contains 4 vectors, so it is not a basis. c) Looking at the row echelon form of the matrix in part a), we see that column 4 of the coefficient matrix does not contain a leading one. Removing column 4 from the coeffi111 cient matrix gives us the matrix 1 1 0 , which would reduce to I3 . Hence the subset 100 1 1 1 , 1 1 0 , 1 0 0 of X is linearly independent, and is a basis for R3 . 6. You are given the following data points: x0 2 4 y 7 21 43 Construct a Lagrange basis {p0 , p1 , p2 } of P2 using the x values from the data set. Hence find the unique quadratic p that fits the data exactly. Estimate the value of y when x = 1. Solution By definition, the vectors in the Lagrange basis of P2 are as follows: 1 (x − 2)(x − 4) = (x − 2)(x − 4), (0 − 2)(0 − 4) 8 (x − 0)(x − 4) 1 p 1 ( x) = =...
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This note was uploaded on 06/04/2013 for the course MATH 1002 taught by Professor Cartwright during the One '08 term at University of Sydney.

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