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Quiz 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.
a b
Question 1: Let V be the vector space over R of the matrices of the following form
,
c 0
1 −3
2 −4
1 −5
where a, b, c ∈ R. Are the vectors A1 =
, A2 =
, A3 =
linearly
2 0
−1 0
7 0
independent in V ?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 A1 + x2 A2 + x3 A3 = 0. Then 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 A1 + x2 A2 + x3 A3 = 0. This
means that the set of vectors {A1 , A2 , A3 } is linearly dependent.
Question 2: Is the set {t2 + 1, t + 1, 1, t2 − t − 1} a basis of the vector space composed of the
real-valued polynomials of degree at most 2, say P2 ? Why?
The set contains 4 vectors, but the dimension of P2 is 3 (recall that {1, t, t2 } is basis of P2 ). As a
result {t2 + 1, t + 1, 1, t2 − t − 1} cannot be a basis of P2 (this set cannot be linearly independent).
2
Quiz 4’, October 4, 2012
Question 3: Express the function v(t) = cos(2t) + 4 cos(t) − 3 as a linear combination of
p1 (t) = cos(2t) − 2 cos(t) + 5, p2 (t) = 2 cos(2t) − 3 cos(t), p3 (t) = cos(t) + 3.
We look for (x1 , x2 , x3 ) ∈ R3 so that v = x1 p1 + x2 p2 + x3 p3 . This means that the following holds
for all t ∈ R:
cos(2t) + 4 cos(t) − 3 = x1 (cos(2t) − 2 cos(t) + 5) + x2 (2 cos(2t) − 3 cos(t)) + x3 (cos(t) + 3)
= cos(2t)(x1 + 2x2 ) + cos(t)(−2x1 − 3x2 + x3 ) + 1(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
 
0 1
1
1 4  ∼ 0
3 −3
0
2
1
−10
 
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
This means that v = −3p1 + 2p2 + 4p3 .
Question 4: Is the set S := {(1, 2, 3), (1, 0, 1)} a basis of R3 ? Why?
R3 is three-dimensional. The bases of R3 have three vectors. S contains two vectors only. This
cannot be a basis. (Actually it cannot be a spanning set).