ma253
, Spring
2007
— Problem Set
1
Solutions
1.
Problems from the textbook:
a. Section
1
.
1
, problems
5
,
7
,
9
, *
20
, *
28
.
Problems 5, 7, and 9 should be straightforward; the answers are in the back of the
book. Let’s do the other two.
20.
The total demand for the product of A is
1000
(from consumers) plus
0.1b
(from B). So we must have
a
=
1000
+
0.1b
. Similarly,
b
=
780
+
0.2b
. This leads
to the system
a

0.1b
=
1000

0.2a
+
b
=
780
The unique solution is
a
=
1100
,
b
=
1000
.
28.
Thermal equilibrium requires that
T
1
=
T
2
+
200
+
0
+
0
4
T
2
=
T
1
+
T
3
+
200
+
0
4
T
3
=
T
2
+
400
+
0
+
0
4
Rewriting as a linear system and solving gives
(
T
1
, T
2
, T
3
) = (
75, 100, 125
)
.
b. Section
1
.
2
, problems
5
,
11
,
13
, *
38
, *
44
, *
61
.
Again, 5, 11, and 13 should be routine; the answers are in the back of the book.
You should be able to do 5 by hand, but getting 11 right by hand is challenging.
38.
This is in the same genre as problem 20 in section 1.1, but is a little closer to
what would happen in real economic modeling.
For (a), the vectors are
v
1
=
⎡
⎣
0
0.1
0.2
⎤
⎦
v
2
=
⎡
⎣
0.2
0
0.5
⎤
⎦
v
3
=
⎡
⎣
0.3
0.4
0
⎤
⎦
b
=
⎡
⎣
320
90
150
⎤
⎦
.
For (b), notice first
x
j
is the output of
I
j
and the
i
th entry
a
ij
in
v
j
is the perunit
demand of
I
j
for the goods made by
I
i
. So
x
j
a
ij
is the actual demand of
I
j
for
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the goods made by
I
i
. That’s the
i
th entry in
x
j
v
j
, so the vector
x
j
v
j
gives the
demands made by
I
j
for the goods of each of the other industries.
Now if we add all of those up,
x
1
v
1
+
· · ·
+
x
n
v
n
+
b
we get a vector whose
i
th entry represents the total demand for the goods pro
duced by
I
i
. So this could be called the total demand vector.
Finally, the equation
x
1
v
1
+
· · ·
+
x
n
v
n
+
b
=
x
says that the total demand should equal the total production.
44.
Kyle’s predicament boils down to the following system of equations:
x
1
+
x
2
+
x
3
=
24
3x
1
+
2x
2
+
1
2
x
3
=
24
Two equations and three unknowns, so we expect one degree of freedom. Solving
this (and taking
x
3
=
2a
to avoid denominators) gives
⎡
⎣
x
1
x
2
x
3
⎤
⎦
=
⎡
⎣
3a

24

5a
+
48
2a
⎤
⎦
.
All three numbers need to be positive, so
a
≥
8
(from the value of
x
1
) and
a
≤
9.6
(from the value of
x
2
). Since
a
needs to be a whole number, either
a
=
8
or
a
=
9
,
which leads to two possible solutions:
⎡
⎣
x
1
x
2
x
3
⎤
⎦
=
⎡
⎣
0
8
16
⎤
⎦
or
⎡
⎣
3
3
18
⎤
⎦
.
Since Kate loves lillies, the second choice is better: Kyle buys 3 lillies, 3 roses, and
18 daisies.
61.
This is very similar to the previous one, except that the handling of the inequal
ities is much messier and there are many more solutions.
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 Spring '07
 GHITZA
 Linear Algebra, Algebra, Equations, Quadratic equation, Elementary algebra, Leonhard Euler

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