Math 2J Lecture B Fall 2009
(Instructor: Professor R. C. Reilly)
Answer Key for Practice Final Exam
(1) Let
A
be the 2by3 matrix whose columns are the given vectors
u
,
v
, and
w
. That is,
A
= [
u v w
] =

3
2

5
1
5
2
.
It was shown in class that the condition for the given vectors to be linearly independent is equivalent to the homoge
neous system
A
x
=
0
having only one solution, namely
x
=
0
. This, in turn, is equivalent to the general solution
of the system having no free variables. However, since there are only two rows in the matrix
A
, the system
A
x
=
0
can have at most 2 lead variables. But the system has 3 variables in all, so there must be at least one free variable,
so the vectors are
not
linearly independent.
To find numbers
a
,
b
, and
c
as required by the problem, let us actually solve the system
A
x
=
0
by GaussJordan
Reduction:
[
A

0
] =

3
2

5
0
1
5
2
0
→
1
5
2
0

3
2

5
0
→
1
5
2
0
0
17
1
0
→
1
5
2
0
0
1
1
17
0
→
1
0
29
17
0
0
1
1
17
0
From this last matrix, which is in reduced rowechelon form, one sees that the general solution of the system has one
free variable, namely
x
3
; and the formula for this general solution is
x
=
x
3

29
/
17

1
/
17
1
NOTE: We are free to use any nonzero value of
x
3
to get the desired
a
,
b
and
c
. To simplify checking our answer,
let us get rid of the fractions by choosing
x
3
= 17, which gives us the special solution
x
=

29

1
17
. That is, we
get
a
=

29,
b
=

1,
c
= 17. Clearly at least one of these numbers is not zero; indeed, this is true for all of them.
Checking Our Answer
: Note that

29
u

v
+ 17
w
= (

29)

3
1
+ (

1)
2
5
+ 17

5
2
=
(

29)
·
(

3) + (

1)
·
2 + 17
·
(

5)
(

29)
·
1 + (

1)
·
5 + 17
·
2
=
0
0
(2) Notice that one has
lim
n
→ ∞
12
n
3

7
n
2
+ 9
n
+ 6
3
n
+ 2
= +
∞
and
lim
n
→ ∞
(4
n
2

5
n
+ 2) = +
∞
.
Thus, trying to compute the limit of the given sequence (
a
1
, a
2
, . . .
) using only the ‘standard limit laws’ leads us to
the indeterminante form
∞  ∞
. However, if we do some preliminary algebra, we can simplify the problem. More
precisely, write
a
n
as a single fraction:
a
n
=
12
n
3

7
n
2
+ 9
n
+ 6
3
n
+ 2

(4
n
2

5
n
+ 2) =
(12
n
3

7
n
2
+ 9
n
+ 6)

(3
n
+ 2)(4
n
2

5
n
+ 2)
3
n
+ 2
=
(12
n
3

7
n
2
+ 9
n
+ 6)

(12
n
3
+ 8
n
2

15
n
2

10
n
+ 6
n
+ 4)
3
n
+ 2
=
13
n
+ 2
3
n
+ 2
=
13 + 2
/n
3 + 2
/n
Thus, one easiy computes that lim
n
→ ∞
a
n
= 13
/
3. In particular, the sequence
does
converge, and its value is 13
/
3.
(3) The coefficient
a
3
is given by
a
3
=
f
000
(

1)
/
3!, so we need only determine
f
000
(

1).
To get
f
000
(
x
) simply
differentiate both sides of the differential equation
f
00
(
x
) + 3
x
2
f
0
(
x
) + 4
f
(
x
) = 0 to get
f
000
(
x
) + 6
xf
0
(
x
) + 3
x
2
f
00
(
x
) + 4
f
0
(
x
) = 0
(
*
)
1
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But
f
(

1) = 2 and
f
0
(

1) =

3 (given facts); and we can substitute these values into the differential equation for
f
to get
f
00
(

1) + 3(

1)
2
·
(

3) + 4
·
2 = 0. Thus,
f
00
(

1) = 9

8 = 1.
Substitute
x
=

1 into the equation (
*
), and use the values of
f
(

1),
f
0
(

1) and
f
00
(

1) obtained above, to get
f
000
(

1)

6
·
(

3) + 3
·
(

1)
2
·
1 + 4
·
(

3) = 0
and thus
f
000
(

1) =

18

3 + 12 =

9.
Thus
a
3
=
f
000
(

1)
/
6 =

9
/
6 =

3
/
2.
Thus the correct answer is
‘(A)

3
/
2’.
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 DONALDSON,NEIL
 Linear Algebra, Maclaurin Series, Power Series, Taylor Series, Eigenvalue, eigenvector and eigenspace

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