M346 Midterm Exam 1 Solutions
I.
Let
L
:
R
2
→
R
2
be de ned by
L
(
x
) = (15
x
1
+ 24
x
2
,
−
8
x
1
−
13
x
2
)
T
, where
x
= (
x
1
, x
2
)
T
.
Let
E
be the standard basis of
R
2
and
B
=
{
(2
,
−
1)
T
,
(
−
3
,
2)
T
}
be an alternate basis.
1. (10 points) Find the changeofbasis matrices
P
EB
and
P
BE
.
Answers:
P
EB
=
[
2
−
3
−
1
2
]
and
P
BE
=
P
−
1
EB
=
[
2
3
1
2
]
2. (5 points) If
v
= (1
,
−
1)
T
, nd
[
v
]
B
.
Answers:
[
v
]
B
=
P
BE
[
v
]
E
=
[
2
3
1
2
][
1
−
1
]
=
[
−
1
−
1
]
3. (10 points) Find the matrices
[
L
]
E
and
[
L
]
B
.
Answers:
Since
L
(
e
1
) = (15
,
−
8)
T
= 15
e
1
−
8
e
2
and
L
(
e
2
) = (24
,
−
13)
T
= 24
e
1
−
13
e
2
,
we have
[
L
]
E
=
[
15
24
−
8
−
13
]
Since
L
(
b
1
) = (6
,
−
3)
T
= 3
b
1
and
L
(
b
2
) = (3
,
−
2)
T
=
−
b
2
, we have
[
L
]
B
=
[
3
0
0
−
1
]
4. (5 points) Let
x
(
n
)
∈
R
2
be the sequence of vectors de ned by
x
(
n
+ 1) =
L
(
x
(
n
))
and
x
(0) = (2
,
1)
T
. De ne
y
(
n
) = [
x
(
n
)]
B
. Find
y
(3)
.
Answers:
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 Spring '08
 RAdin
 Linear Algebra, Algebra, Matrices, basis, Standard basis, Pxy

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