© Scarborough MATH 141 Week In Review 3 Key Spring 2013 1
1.
The monthly supply and demand functions for a metallic Texas A&M car emblem are given below, where x is the quantity and p is the price per emblem in dollars. Determine the equilibrium price and quantity for the car emblems. Be sure to state your answer in terms of the appropriate units.
Supply:
3 x

1 0 0 p 6 0 0
Demand:
5 x

2 5 0 p

3 5 0 0
  x

1 0 0 p
  
5
 
6 0 0
 x

2 5 0 p
   
OR rref



3
5

100
250

600
3500







1
0
0
1
160
10 .
8




15 x

500 p

3000
15 x

750 p

10500
0 x

1250 p

13500 p

1 0 .8
3 x
    
6 0 0
3 x

4 8 0 x

1 6 0
The equilibrium price is $10.80 and the equilibrium quantity is 160 car emblems.
2.
Given A

6


2 n
0

1
3 p


T



5

0
4 1 2

 t

1 r




. a.
Fully simplify matrix A .
A

6



2 n
0

1
3 p



T



5

0
4 1 2

 t

1 r





6




3
2 0 n

1



 p



5

0
4 1 2

 t

1 r
 





 
1 2
6 n

0
6
 
 

2 0 0
4 8 r
1 8 6 p
 
 

4 t 4









6 n

8

4 8 r
0

6
1 8

4 t 6 p

4



 b.
a
3 ,1
 a
3 1

1 8

4 t

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 2
3.
A snowman accessory company has monthly fixed costs of $67005. If 10,000 snowman accessories are produced and sold, each for $20, there is a loss of $17005. a.
Find the linear cost function.
C ( x ) = cx + F
C ( x ) = cx + 67005
R ( x ) = sx
R ( x ) = 20 x
P ( x ) = R ( x ) – C ( x )
P ( x ) = 20 x – cx – 67005
P (10000) = 20(10000) – c (10000) – 67005 = –17005
10000 c = 150000 c = 15
Therefore C ( x ) = 15 x + 67005 dollars cost per x snowmen accessories produced b.
What is the revenue break-even point? Include units in your point.
C ( x ) = R ( x )
15 x + 67005 = 20 x
5 x = 67005 x = 13401
R (13401) = 268020
Therefore (13401 snowman accessories, $268020) c.
Compute the profit or loss when 1420 snowmen accessories are produced and sold.
       

2 0 x
 
1 5 x

6 7 0 0 5


5 x

6 7 0 0 5
P

1 4 2 0
  
5 9 9 0 5
Loss of $59,905

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 3
4.
The table shows the average SAT score for certain years. Find the best fitting line for the data where t is the number of years after 1995 and y is the SAT score. t
Year
SAT Score
10
2005
1028
9
2004
1026
7
2002
1020
5
2000
1019
3
1998
1017
L1
L2 a.
What is the calculator command you are to write down that shows your work when doing linear regression?
LinReg L
1
, L
2
, Y
1 b.
What is the best fitting line?
y = 1.585365854
t + 1011.219512 average SAT score for t years after 1995 c.
Using the line of best fit, in what year would you expect an average SAT score to be 1030?
Y
2
= 1030 (an appropriate window would be [0,15] by [1010, 1040])
Intersect (11.846154, 1030)
1995 + 11.846154 = 2006.846154 so the year would be 2006 d.
Using the line of best fit, what would you predict the average SAT score to be in the year 1996?
Y
1
(1996 – 1995) or Y
1
(1) = 1012.804878
In 1996, the best fitting line would predict an average SAT score of 1013 . e.
What is the value of the correlation coefficient? Interpret.
r = 0.957073226
There is a strong positive correlation so the linear model is a good model and the line has positive slope.

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 4
5.
The demand for eReaders is 4560 units when the price is $310. If the price drops down to $278 then
5328 are sold. The manufacturer will not market eReaders if the price drops to $92. For each $33 increase in price, the manufacturer will supply an additional 6552 eReaders. Find and interpret the equilibrium point.
Demand
( x , p )
(4560, 310)
(5328, 278) m

2 7 8


3 1 0


3 2


1
5 3 2 8 4 5 6 0 7 6 8 2 4 p

3 1 0


1  x

4 5 6 0
 p

3 1 0

2 4

1 x

1 9 0
2 4 p


1 x
2 4

5 0 0
Demand = Supply at the equilibrium point

1 x

5 0 0

1 1
2 4

1 7 x
 
4 0 8
3 6 4 x

8 7 3 6
2 1 8 4 x

9 2 p

8736
 
136
Supply
( x , p )
(0, 92) m

3 3

1 1
6 5 5 2 2 1 8 4 p
 m x
 b p

1 1
2 1 8 4 x

9 2
When 8736 eReaders are produced and sold at a price of $136, both producers and consumers are satisfied.

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 5
6.
An industrial paper shredder is originally purchased for $24,000. After 8 years, its scrap value is $4200. a. Find a linear equation for the value, V , of the paper shredder as a function of time, t , in years.

,
    
Could find the slope and use the point-slope form of a line or can use linear regression.
To use linear regression, put the x -coordinates in L1 and the y -coordinates in L2.
LinReg L1, L2, Y1
   
2475 t

24000 dollars in value after t years. b. What is the rate of depreciation of the paper shredder?
Since m
 
2 4 7 5 , the depreciation rate is $2475 per year . c. What is the paper shredder worth after 6 years?
Y1(6) = 9150
V
      
24000

$9150
7.
The matrix




1 0 8 1
0 2 4 1 0
0 0 6 0




represents a system of equations. Perform Gauss-Jordan row operations to solve the system of equations.




1 0 8 1
0 2 4 1 0
0 0 6 0




1
2
R
2


R
2




1 0 8 1
0 1 2 5
0 0 6 0




1
6
R
3


R
3




1 0 8 1
0 1 2 5
0 0 1 0



 
8 R
3

R
1


R
1




1 0 0 1
0 1 2 5
0 0 1 0





2 R
2

R
2


R
2




1 0 0 1
0 1 0 5
0 0 1 0




Therefore the solution is (1, 5, 0) .
8.
A farmer has x acres of corn, y acres of peanuts and z acres of soybeans planted. Write an equation that represents that the number of acres of soybeans is three times the number of acres of corn and peanuts combined.
z = 3( x + y ) OR z = 3 x + 3 y OR
–
3 x
–
3 y + z = 0 OR
1
3 z
  y

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 6
9.
It costs the CoGo Company $164 per cow to grow it to maturity and $36 per goat to grow it to maturity.
CoGo wants to have half as many cows as goats. If the entire budget of $145,848 is to be used, how many of each animal should they have? a.
Define your variables and set up the system of equations.
Let x = number of cows
Let y = number of goats
1 6 4 x

3 6 y

1 4 5 8 4 8
2 x
 y b.
How many of each animal should CoGo have? rref
 1 6 4 3 6 1 4 5 8 4 8
2

1 0





1 0 6 1 8
0 1 1 2 3 6


Therefore CoGo should have 618 cows and 1236 goats.
10.
The matrix




1 4 0 0

1 0

0 0 1 3 8


0 0 0 0 0
represents a system of equations. Give the solution. If the solution is parametric, give two particular solutions.
1 w

4 x
1 y

3 z

 
10
8
means w y
 
4 x

1 0
8 3 z
so let x
 r and z
 s
The solution is
 
4 r
 r
 s s

, where r and s are any real numbers.
Two particular solutions are (–10, 0, 8, 0) and (–14, 1, 2, 2).
11.
If A




2
0 n v



and B




1 p
5
0



, find AB and BA
T
.
A B



2
0 n v
 
 
1 p
5
0



  n p v p
B A
T 



1 5
 
2 0 p 0
 
  n v



 
 
2

5 n
2 p
5 v
0


 n v
0
0



 
2
 vp n p 1 0
0



© Scarborough MATH 141 Week In Review 3 Key Spring 2013
12.
Solve the matrix equation 5 X

2 B

X C

4 D
Method I (solving for X on the left):
5 X

2 B

X C

4 D
for X .
5 X

X C

2 B

4 D
5 X I

X C

2 B

4 D
X 5 I

X C

2 B

4 D
X

5 I

C
 
2 B

4 D
X

5 I

C
 
5 I

C
 
1  
2 B

4 D
 
5 I

C
 
1
X I
 
2 B

4 D
 
5 I

C
 
1
X
 
2 B

4 D
 
5 I

C
 
1
Method II(solving for X on the right):
5 X

2 B

X C

4 D
4 D

2 B

X C

5 X
4 D

2 B

X C

X 5
4 D

2 B

X C

X 5 I
4 D

2 B
  
5 I


4 D

2 B
 
C

5 I
 
1 

4 D

2 B
 
C

5 I
 
1 
X I
 
5 I
 
C

5 I
 
1

4 D

2 B
 
C

5 I
 
1 
X
13.
If C


5
1
2 n



and D n p

0
1



, find 2 C + D
T
. n p

1
0

T

10 4


 2 2 n
 n p
1 0





10
1
 n
C

D
T
2 = 2


5 2
1 n
4

2 n p


7

14.
A

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 m

1




0
2
1 r p

3
4 t n q




8 a.
What are the dimensions of matrix A ?
3

4 b.
What are all the possible dimensions of matrix B such that AB is defined?
4
 n , where n is any natural number. c.
a
3 , 2
 a
3 2
 r d.
Give matrix C such that C is an identity matrix and where CA is defined.
C
 



1 0 0
0 1 0
0 0 1



 e.
A
T f.
If B








0 m n
2 p q
1 t r

1

3 4








4


0
1 1

2
3 r m p


5 n
2
9 q
6




, find A – 2 B .
A

2 B





0
2
1 m r p



3 m 2 n

1
4
1
3





2
8

2 1
7
 p r

1 3
8 n q t




 


4
2 0

2
3

 1 1 r m p t n


 q
1 8
1 2






5 n
2
9 q
6
 





 
0
2
1 m p r

1

3
4 t n q
 
 

8
0

2 2

4 m
6 p

2 r

2
4 n
1 0

1 8


2 q
1 2




 g.
If D







0
2

1 0
0
1
1
2






, find AD.
A D





0
2
1 m p r

1

3
4 n q t











0 1
2 1
1 0
0 2











2
2
2 m p r



1
3
4
2
1 m

 p
2
 n
2
2 t q





15.
If
© Scarborough MATH 141 Week In Review 3 Key Spring 2013
A

1
3
 6

4

2


, find A .
In calculator, let B

A

1 , then
B

1 

1

A
. [Use math frac.]
9
A
 




1 2

9
1
9
1




3 6
16.
The table gives the dimensions and characteristics of five matrices.
Matrix
A
B
C
D
E
, non-singular
, singular
Dimensions
5  6
7  7
6  5
6  6
7  5
Which of the following matrix operations are defined? If defined, what is the dimension of the resulting matrix? a.
T
Is defined and the resulting matrix is a 5 x 6.
5 x 5 | 5 x 6 | 6 x 6 b.
2 A
T 
6 C
Is defined and the resulting matrix is 6 x 5.
6 x 5 | 6 x 5 c.
D

1 
3 I
6 d.
e.
Is NOT defined since D is a singular matrix and has no inverse.

1
4 B E
Is defined and the resulting matrix is 7 x 5.
7 x 7 | 7 x 5

1 T
D D A
Is NOT defined since D is a singular matrix and has no inverse.

© Scarborough MATH 141 Week In Review 3 Key Spring 2013
17.
Solve the system of equations. If the solution is parametric, give two particular solutions. x

8 z

1 y

2 z
 
0 x

1 2 z 4 2 y rref




1 0 8 1
0 1 2 5
1 2 1 2 4
 





 
1 0 8 0
0 1 2 0
0 0 0 1




There is NO SOLUTION since 0

1 .
10
18.
The Berry Bucket has 105 acres of land allotted for cultivating berries. The cost of cultivating blackberries is $300 per acre, of blueberries is $200 per acre, and of strawberries is $420 per acre. The
Berry Bucket will invest exactly $30,100 for the cultivation of all 105 acres. For every 11 acres of blueberries planted, 9 acres of blackberries will be planted. How many acres of each crop should be planted? Then find the product of the number of acres planted in blackberries and strawberries.
Let x = acres of blackberries
Let y = acres of blueberries
Let z = acres of strawberries x y z 105 acres
300 x

200 y

420 z

30100 $ cultivation cost
11 x
 y
To double check the ratio equation
1 1 x

9 y
, let x = 9, and solve for y . Since y = 11, it is set up correctly. rref




1
3 0 0 2 0 0 4 2 0 3 0 1 0 0
1 1

1
9
1
0
1 0 5
0
 





 
1 0 0 3 6
0 1 0 4 4
0 0 1 2 5




Therefore 36 acres of blackberries, 44 acres of blueberries, and 25 acres of strawberries should be planted. The product of the number of acres planted in blackberries and strawberries is xz
     
900 .

© Scarborough MATH 141 Week In Review 3 Key Spring 2013
19.
Given the system of equations:
2 x

5 y
3 x

4 y


3 6
2 0
. a.
Represent the system of equations as a matrix equation. Solve the system of equations by first solving a matrix equation.
A




2

3 4
5 


, X

 
, B

 3 6 

 2 0


Coefficient Variable Constant
Matrix Matrix Matrix
A X

B
 
1
X A B
X
 
1
A B




2

5
3 4

1
  3 6 
 
2 0

 


   
2 4 4

2 3
6 8




11
Therefore the solution is


2 3
2 4 4
,

6 8
2 3 2 3


. b.
Solve the system of equations by transforming an augmented matrix into a row-reduced form. rref



2
3

5
4
36
20









1
0
0
1
244

23
68
23





Therefore the solution is


2 4 4
,

6 8
2 3 2 3


.

© Scarborough MATH 141 Week In Review 3 Key Spring 2013 12
20.
In a zoo 65% of the mammals are male and 70% of the non-mammals are female, as shown in the matrix
Z below. Find a matrix Y that when multiplied by Z
T
will give a matrix containing the number of male and female animals that live in temperate zones and live in non-temperate zones, given there are 100 mammals live in temperate zones, 400 non-mammals that live in temperate zones, 200 mammals that live in non-temperate zones, and 500 non-mammals that live in non-temperate zones. m a le fem a le
Z
 m a m m a l n o n
 m a m m a l



0 .6 5 0 .3 5
0 .3
0 .7


 m a m m a l n o n -m a m m a l
Z
T  m a le fem a le


0 .6 5 0 .3
0 .3 5 0 .7

 m a m m a l n o n -m a m m a l tem p era te n o n -tem p era te
 T
X Z Y
 m a le fem a le


0 .6 5 0 .3
0 .3 5 0 .7

 m a m m a l n o n
 m a m m a l

1 0 0 2 0 0
4 0 0 5 0 0


tem p era te n o n -tem p e ra te
 m a le fem a le
 1 8 5 2 8 0

 3 1 5 4 2 0



Therefore d is the correct answer. a.
b.
Y
Y

1 0 0 4 0 0
2 0 0 5 0 0



, X = YZ
T

1 0 0 4 0 0
2 0 0 5 0 0



, X = Z
T
Y c.
Y

1 0 0 2 0 0
4 0 0 5 0 0



, X = YZ
T d.
Y

1 0 0 2 0 0
4 0 0 5 0 0



, X = Z
T
Y e.
None of these