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1
Quizz 4 (Notes, books, and calculators are not authorized)
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded.
Question 1: Are the polynomials p1 = 1 − 3t + 2t2 , p2 = 2 − 4t − t2 , p3 = 1 − 5t + 7t2 linearly
independent in P2 (t)?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 p1 + x2 p2 + x3 p3 = 0. Then (x1 + 2x2 + x3 ) + (−3x1 −
4x2 − 5x3 )t + (2x1 − x2 + 7x3 )t2 = 0. This is equivalent to saying that X solves


1
2
1
−3 −4 −5 X = 0.
2 −1 7
Let us reduce the matrix of the linear system in echelon form

 


1
2
1
1 2
1
1
−3 −4 −5 ∼ 0 2 −2 ∼= 0
2 −1 7
0 −5 5
0
2
2
0

1
−2 .
0
There are only two pivots. There is one free variable. This means that the solution set of the above
linear system is not {0}. There is some nonzero vector X so that x1 p1 + x2 p2 + x3 p3 = 0. This
means that the polynomials p1 , p2 , p3 are linearly dependent.
Question 2: Express the vector v = (1, 4, −3) as a linear combination of v1 = (1, −2, 5),
v2 = (2, −3, 0), v3 = (0, 1, 3).
We look for (x1 , x2 , x3 ) ∈ R3 so that v = x1 v1 + x2 v2 + x3 v3 . For all t ∈ R the following holds
(1, 4, −3) = x1 (1, −2, 5) + x2 (2, −3, 0) + x3 (0, 1, 3) = (x1 + 2x2 , −2x1 − 3x2 + x3 , 5x1 + 3x3 ).
This implies
x1 + 2x2 = 1,
We can write this
form:

1
2
−2 −3
5
0
−2x1 − 3x2 + x3 = 4
5x1 + 3x3 = −3.
system in the form of an augmented matrix and compute the reduced echelon
 
1
0 1
1 4  ∼ 0
0
3 −3
2
1
−10
This means that v = −3v1 + 2v2 + 4v3 .
 
0 1
1
1 6  ∼ 0
3 −8
0
2
1
0
0
1
13
 
1
1
6  ∼ 0
52
0
0
1
0
0
0
1

−3
2
4
2
Quizz 4, September 29, 2011
Question 3: Is W = {at2 + bt + c : a, b, c ∈ R, a + b + c = 0} a subspace of P2 (t) over the field
R? Why?
(i) Clearly p = 0 is a member of W . (ii) Let p1 = a1 t2 + b1 t + c1 and p2 = a2 t2 + b2 t + c2 be two
members of W and let λ, µ ∈ R. Then
λp1 + µp2 = λ(a1 t2 + b1 t + c1 ) + µ(a2 t2 + b2 t + c2 ) = (λa1 + µa2 )t2 + (λb1 + µb2 )t + (λc1 + µc2 ).
Now let us verfy if λp1 + µp2 is a member of W by summing the coefficients. The sum of the
coefficients of λp1 + µp2 is
(λa1 + µa2 ) + (λb1 + µb2 ) + (λc1 + µc2 ) = λ(a1 + b1 + c1 ) + µ(a2 + b2 + c2 ) = 0,
thereby proving that λp1 + µp2 is a member of W . Items (i) and (ii) prove that W is a subspace
of P2 (t).
Question 4: Does S := {(1, 1, 1), (1, 0, 1), (0, 0, 1), (0, 1, 1)} form a basis of R3 ?
R3 is three-dimensional. The bases of R3 have three vectors. S contains four vectors, i.e., one too
many. This cannot be a basis. (Actually it cannot be a linearly independent set.)