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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.
 One '08
 Cartwright
 Linear Algebra, Algebra, Vector Space

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