# 4 04 2 8 p 0 x hence the quadratic that ts the

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Unformatted text preview: x(x − 4), (2 − 0)(2 − 4) 4 (x − 0)(x − 2) 1 p 2 ( x) = = x(x − 2). (4 − 0)(4 − 2) 8 p 0 ( x) = Hence the quadratic that ﬁts the data is deﬁned by p(x) = 7p0 (x) + 21p1 (x) + 43p2 (x). This simpliﬁes to p(x) = x2 + 5x + 7. When x = 1 we estimate that y ≈ p(1) = 13. 7. The matrix J is the reduced row echelon form of the matrix A. A= Math 2061: Tutorial 4 (week 5) — Solutions 1 2 4 8 7 2 5 10 19 17 0 3 7 9 19 4 2 8 14 50 5 1 3 13 18 J= 1 0 0 0 0 0 1 0 0 0 0 40 29 0 −18 −12 14 1 00 0 00 0 Page 3 Linear Mathematics Tutorial 4 (week 5) — Solutions Page 4 a) Find a basis, B , for the column space of A. b) Write each of the columns of A that are not included in B as a linear combination of the vectors in B . c) Find a basis for the null space of A. d) Verify that the sum of the rank of A and the nullity of A is the number of columns of A. Solution a) The ﬁrst three columns of J are a basis for the column space of J , so the ﬁrst three columns of A are a basis for the column space of A. So a basis for the column space of A is 1 2 4 8 7 B= 2 5 10 19 17 , , 0 3 7 9 19 . b) The relationships between the columns of A are the same as the relationships between the columns of J . So we have 4 2 8 14 50 = 40 1 2 4 8 7 − 18 2 5 10 19 17 0 3 7 9 19 +4 5 1 3 13 18 and 1 2 4 8 7 = 29 2 5 10 19 17 − 12 + 0 3 7 9 19 . x1 c) Let x = . . . x5 be in the null space of A. As Null(A) = Null(J ), looking at J we see that x4 and x5 are free variables and that x1 = −40x4 − 29x5 , x2 = 18x4 + 12x5 , x3 = − 4x4 − x5 . Hence x= x1 x2 x3 x4 x5 = −40x4 −29x5 18x4 +12x5 −4x4 −x5 x4 x5 = x4 −40 18 −4 1 0 −29 12 −1 0 1 + x5 , for all x4 , x5 ∈ R. Consequently, a basis for the null space of A is −40 18 −4 1 0 , −29 12 −1 0 1 . d) The rank of A is the dimension of the column space of A, so rank(A) = 3. The nullity of A is the dimension of t...
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