MAS 213: Linear Algebra II.
Problem list for Week #2.
Solutions.
.
Tutorial problems:
Problem 1:
(Problems 1.4.3(e), 1.4.4(c) in [FIS].)
In the following parts you are given a vector space
V
, a vector
v
in it, and
a subset
S
of the vector space.
Determine if the given vector
v
lies in the span of the given subset
S
. If
it does, ﬁnd an explicit representation of
v
as a linear combination of the
vectors in
S
.
(1)
V
=
R
3
,
v
= (5
,
1
,
−
5), and
S
=
{
(1
,
−
2
,
−
3)
,
(
−
2
,
3
,
−
4)
}
.
(2)
V
=
P
3
(
R
),
v
=
−
2
x
3
−
11
x
2
+ 3
x
+ 2,
and
S
=
{
x
3
−
2
x
2
+ 3
x
−
1
,
2
x
3
+
x
2
+ 3
x
−
2
}
.
(3)
V
=
M
2
×
2
(
R
),
v
=
[
2
−
1
1
3
]
,
and
S
=
{[
1 1
1 1
]
,
[
0 1
1 1
]
,
[
0 0
1 1
]}
.
Solution to problem 1, part 1.
The vector
v
lies in the span of
S
if and only if there are scalars
a, b
∈
R
such that:
(5
,
1
,
−
5) =
a
(1
,
−
2
,
−
3) +
b
(
−
2
,
3
,
−
4)
.
This is the precise meaning of the statement “
v
lies in the span of
S
”.
Scalars
a
and
b
solve this equation if and only if they solve the system of
three linear equations that we get from the components of these vectors:
a
−
2
b
= 5
,
−
2
a
+ 3
b
= 1
,
−
3
a
−
4
b
=
−
5
.
These equations can be expressed as an augmented matrix as follows:
1
−
2
5
−
2
3
1
−
3
−
4
−
5
.
Now we do the usual row manipulations to put this system in row echelon
form, in order to decide if there are any solutions.
1
−
2
5
−
2
3
1
−
3
−
4
−
5
−→
1
−
2
5
0
−
1
11
0
−
10
10
−→
1 0
3
0 0
10
0 1
−
1
This has
no solutions
because it contains the equation 0
b
= 10, which
cannot be satisﬁed.
1
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So
v
does not lie in the span of
S
.
Solution to problem 1, part 2.
In this case the equation we are solving is:
−
2
x
3
−
11
x
2
+ 3
x
+ 2 =
a
(
x
3
−
2
x
2
+ 3
x
−
1) +
b
(2
x
3
+
x
2
+ 3
x
−
2)
.
This is equivalent to the system of 4 equations that are obtained by equating
coeﬃcients of like powers of
x
:
a
+ 2
b
=
−
2
,
−
2
a
+
b
=
−
11
,
3
a
+ 3
b
= 3
,
−
a
−
2
b
= 2
.
Solving this is the usual way:
1
2
−
2
−
2
1
−
11
3
3
3
−
1
−
2
2
−→
1
2
−
2
0
5
−
15
0
−
3
9
0
0
0
−→
1 0
4
0 0
0
0 1
−
3
0 0
0
.
Thus the system is solved by
a
= 4 and
b
=
−
3. We can check that by
writing out the apparent solution:
−
2
x
3
−
11
x
2
+ 3
x
+ 2 = 4(
x
3
−
2
x
2
+ 3
x
−
1)
−
3(2
x
3
+
x
2
+ 3
x
−
2)
.
This checks out. Thus
v
does
lie in the span of
S
.
Solution to problem 1, part 3.
We want to understand the solutions to the equation:
[
2
−
1
1
3
]
=
a
[
1 1
1 1
]
+
b
[
0 1
1 1
]
+
c
[
0 0
1 1
]
.
Equating entries in the same positions gives us a system of 4 equations in 3
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 Fall '10
 ANDREWKRICKER
 Linear Algebra, Vector Space, 1 W, 1W

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