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Chapter 1-11

MATH136 Chapter 1-11: Wolczuk LinearAlgebra Solutions 136.pdf


Department
Mathematics
Course Code
MATH136
Professor
Dan Wolczuk
Chapter
1-11

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Chapter 1 Solutions
1.1 Problem Solutions
1.1.1 (a) 2
2
1
1
+3
1
0
1
=
1
2
1
(b) 1
3
3
1
2
4
1/2
2/3
=
3
1/6
0
(c) 2
2
0
2
+2
3
6
0
0
=
6
0
2
(d) 3
1
0
1
2
2
3
1
+
7
6
1
=
0
0
0
(e) x
1
0
0
+y
0
1
0
+z
0
0
1
=
x
y
z
(f) 3
1
0
1
+π
0
1
0
=
3
π
3
(g) (a+b+c)
1
1
1
+(a+b)
1
2
3
+(a+2b+c)
1
1
2
=
a
b
c
1.1.2 (a) Since
2
6
2
=2
1
3
1
we get by Theorem 1.1.2 that Span
1
3
1
,
2
6
2
=Span
1
3
1
. Hence,
the set represents the line in R3with vector equation ~
x=t
1
3
1
,tR.
(b) The set only contains two vectors, so the set represents the two points (1,3,1), (2,6,2) in R3.
1

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2Section 1.1 Solutions
(c) Since
1
0
2
,
2
1
1
is linearly independent (neither vector is a scalar multiple of the other), we
get that the set represents the plane in R3with vector equation ~
x=s
1
0
2
+t
2
1
1
,s,tR.
(d) The set only contains the zero vector, so it represents the origin in R3. A vector equation is ~
x=
0
0
0
.
(e) The set is linearly independent (verify this) and so the set represents a hyperplane in R4with vector
equation
~
x=t1
1
0
1
1
+t2
1
0
2
1
+t3
3
1
0
0
,t1,t2,t3R
(f) The set represents the line in R4with vector equation ~
x=t
1
1
1
0
,tR.
1.1.3 (a) Observe that (1)"1
2#+2"1
3#"1
4#="0
0#. Hence, the set is linearly dependent. Solving for the first
vector gives "1
2#=2"1
3#"1
4#.
(b) Since neither vector is a scalar multiple of the other, the set is linearly independent.
(c) Since neither vector is a scalar multiple of the other, the set is linearly independent.
(d) Observe that 2"2
3#+"4
6#="0
0#. Hence, the set is linearly dependent. Solving for the second
vector gives "4
6#=(2)2"2
3#.
(e) The only solution to c
1
2
1
=
0
0
0
is c=0, so the set is linearly independent.
(f) Since the set contains the zero vector, it is linearly dependent by Theorem 1.1.4. Solving for the
zero vector gives
0
0
0
=0
1
3
2
+0
4
6
1
.
(g) Observe that 0
1
1
0
+2
1
2
1
2
4
2
=
0
0
0
. Hence, the set is linearly dependent. Solving for the
second vector gives
1
2
1
=0
1
1
0
1
2
2
4
2
.

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Section 1.1 Solutions 3
(h) Consider
0
0
0
=c1
1
2
1
+c2
2
3
4
+c3
0
1
2
Performing operations on vectors on the right hand side, we get
0
0
0
=
c1+2c2
2c1+3c2c3
c1+4c22c3
Since vectors are equal if and only if there corresponding entries are equal we get the three equa-
tions in three unknowns
c1+2c2=0
2c1+3c2c3=0
c1+4c22c3=0
The first equation implies that c1=2c2. Substituting this into the second equation we get
7c2c3=0. Thus, c3=7c2. Substituting these both into the third equation gives 12c2=0.
Therefore, the only solution is c1=c2=c3=0. Hence, the set is linearly independent.
(i) Observe that
(2)
1
1
2
1
+
2
2
4
2
+0
1
0
1
0
+0
2
1
3
1
=
0
0
0
0
Hence, the set is linearly dependent. Solving for the second vector gives
2
2
4
2
=2
1
1
2
1
+0
1
0
1
0
+0
2
1
3
1
(j) Since neither vector is a scalar multiple of the other, the set is linearly independent.
1.1.4 (a) Clearly "1
0#<Span("3
2#). Thus, ("3
2#)does not span R2and so is not a basis for R2.
(b) Since neither vector is a scalar multiple of the other, the set is linearly independent. Let ~
x="x1
x2#
R2and consider "x1
x2#=c1"2
3#+c2"1
0#="2c1+c2
3c1#
Comparing entries we get
x1=2c1+c2
x2=3c1
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