ma253
, Fall
2007
— Problem Set
2
Solutions
1.
Problems from the textbook:
a. Section
1
.
1
, problem *
46
The equations are
x
1
+
x
2
+
x
3
=
1000
.2x
1
+
.5x
2
+
2x
3
=
1000.
Rowreducing and solving leads to
⎡
⎣
x
1
x
2
x
3
⎤
⎦
=
⎡
⎣
5a

5000
3

6a
+
800
3
a
⎤
⎦
,
where
a
is arbitrary. Unfortunately for the metermaids, whenever
a
is a whole
number, the expressions for
x
1
and
x
2
are
not
whole numbers! Since we can’t have
fractions of coins, no one will ever get the Ferrari.
Unless they figure this out and sue, of course.
b. Section
1
.
2
, problem
50
, *
53
.
50.
Write the polynomial
f
(
t
)
as
f
(
t
) =
a
+
bt
+
ct
2
+
dt
3
. The first condition says
that
f
(
0
) =
3
; plugging in, we get
a
=
3
. The second gives
f
(
1
) =
a
+
b
+
c
+
d
=
2
,
f
(
2
) =
a
+
2b
+
4c
+
8d
=
0
, and the final one says
2
0
f
(
t
)
dt
=
2a
+
2b
+
8
3
c
+
4d
=
4.
That gives us a system of equations in the four unknown
a
,
b
,
c
,
d
. Setting up the
rowreduction, we see that the system is inconsistent: there is no solution, hence
there is no such polynomial!
The reason is this: for a cubic polynomial Simpson’s Rule gives the
exact
value of
the integral, and the information that goes into Simpson’s rule is exactly the values
f
(
0
)
,
f
(
1
)
and
f
(
2
)
. Using that, we can see that for the given values the integral
must be equal to
11/3
. Hence, asking that the integral be
4
is asking for something
that is impossible.
53.
The (only) hard part of this is extracting a system of equations from the given
data. Let’s start by letting
x
1
be the number of students who start out as liberals and
x
2
be the number of students who start out as conservatives. So we know that
x
1
+
The first piece of luggage out of the chute doesn’t belong to anyone, ever.
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x
2
=
260
. At the end, 30% of the liberals, i.e.
.3x
1
have become conservatives; on
the other hand,
.4x
2
are now liberals. So how many conservatives are there at the
end? Well,
x
2

.4x
2
+
.3x
1
. But the final number of conservatives is equal to the
initial number of liberals, so this expression is equal to
x
1
. So the second equation is
x
1
=
x
2

.4x
2
+
.3x
1
, which shakes out to
.7x
1

.6x
2
=
0
. That gives us the other
equation. Sovling, we see that
x
1
=
120
and
x
2
=
140
. So we started with 120
liberals and 140 conservatives and ended with 140 liberals and 120 conservatives.
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 Spring '07
 GHITZA
 Linear Algebra, Algebra, Equations, left hand, linear space

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