Math 125 – Exam 2 – Version 1
March 28, 2007
60 points possible
1. (5pts) By multiplying out and combining like terms, simplify the matrix expression
[(A − B)(A + B)]A−1 .
Solution: Using the distributive from the left,
[(A − B)A + (A − B)B]A−1 .
By the distributive property from the right,
[AA + AB − BA − BB]A−1 .
Again, using the distributive property from the right,
AAA−1 + ABA−1 − BAA−1 − BBA−1 .
Using the properties of inverses,
A + ABA−1 − B − BBA−1 .
By definition of matrix powers,
A + ABA−1 − B − B 2 A−1 .
(Note: You were not required to state which arithmetic property you were using for full
credit.)
2. The Finest Furniture Factory makes beds, chairs, and couches from the raw materials
time (man-hours), lumber (board feet), and cloth (yards). Let there be a profit of $18 per
bed, $20 per chair, and $32 per couch, and let the supplies be 88 man-hours, 240 feet of
lumber, and 28 yards of cloth. Further, assume that it takes 4 man-hours, 18 feet of lumber,
and 6 yards of cloth to make a bed; 8 man-hours, 12 feet of lumber, and 2 yards of cloth to
make a chair; and 8 man-hours, 24 feet of lumber, and 4 yards of cloth to make a couch.
(a) (7pts) Set up but do not solve a linear program that enables the Factory to maximize
profit. Assume that the Factory will sell everything it produces.
Solution: The above information can be organized in the following table:
time (in man-hours)
lumber (in board feet)
cloth (in yards)
profit in dollars
bed
4
18
6
18
chair couch supplies
8
8
88
12
24
240
2
4
28
20
32
(Note: You did not have to table the information for credit.)
Let x = the number of beds manufactured, y = the number of chairs manufactured, and
z = the number of couches manufactured.
The objective here is to maximize profit. The profit function is P = 18x + 20y + 32z.
Since there is only 88 man-hours available, we get the inequality 4x + 8y + 8z ≤ 88.
Similarly, due to the 240 board feet of lumber available, we get a lumber inequality of
18x + 12y + 24z ≤ 240. Lastly, there is only 28 yards of cloth. This results in the inequality
6x + 2y + 4z ≤ 28.
Lastly, we know that our variables need to be non-negative. (It doesn’t make sense to
manufacture a negative number of items.) That is, x ≥ 0. y ≥ 0, and z ≥ 0.
The above discussion accurately describes the linear program. One could choose to
represent this in its mathematical form:
maximize P = 18x + 20y + 32z
subject to 4x + 8y + 8z ≤ 88
18x + 12y + 24z ≤ 240
6x + 2y + 4z ≤ 28
x ≥ 0, y ≥ 0, z ≥ 0
(Note: You did not have to order the linear program like this for credit.)
(b) (3pts) In class, we learned a restricted version of the simplex algorithm. Clearly explain
whether or not it would be appropriate to use that version to solve this linear program.
Solution: The simplex algorithm we learned requires that the linear program have the
following three characteristics:
1. the program is a maximization
2. all the constraint inequalities are of the form ≤ a positive number
3. all the variables are non-negative
The above linear program satisfies these conditions, hence we can use the simplex algorithm
we learned in class.
3. Consider the linear program:
maximize z = 8x1 + 4x2 − x3
subject to −6x1 + 2x2 − 4x3 ≤ 16
−4x1 + x2 + 8x3 ≤ 10
4x1 − 5x2 + 12x3 ≤ 8
xi ≥ 0 for all i
(a) (2pts) Construct the initial simplex table for this system.
Solution: Moving the objective function into standard form and adding slack variables to
the constraint inequalities, we get:
z − 8x1 − 4x2 + x3
−6x1 + 2x2 − 4x3 + x4
−4x1 + x2 + 8x3 + x5
4x1 − 5x2 + 12x3 + x6
=
=
=
=
0
16
10
8
where x4 , x5 and x6 are slack variables.
Constructing the initial simplex table:

1
−8 −4
1
 0
−6
2 −4

 0
−4
1
8
0
4 −5 12
0
1
0
0
0
0
1
0
0
0
0
1

0
16 
.
10 
8
(b) (2pts) Determine the first pivot point dictated by the simplex algorithm.
Solution: The −8 heading the x1 column is the most negative entry in the top row. Therefore, we look for a pivot point in this column. Recall that we can only pivot on the positive
entries. Since the 4 in the fourth row is the only possible candidate, it is the pivot point.
(c) (3pts) Perform the first pivot dictated by the simplex algorithm and determine the basic
feasible solution for the resulting table.
Solution: The next simplex

1
 0

 0
0
table:
0
−14 25
0 −11/2 14
0
−4 20
1 −5/4 3
0
1
0
0
0
2
0 3/2
1
0
0 1/4

16
28 
.
18 
2
The basic feasible solution for this table is x1 = 2, x2 = 0, x3 = 0, x4 = 28, x5 = 18 and
x6 = 0 or (2, 0, 0, 28, 18, 0).
(d) (3pts) Continue to use the simplex algorithm to attempt to solve the given linear
program. If a solution exist, clearly state the conclusion. If a solution does not exist, explain
how the algorithm demonstrated this and what the geometrical reason for no solution must
be.
Solution: The algorithm is stuck. The third column still starts with a negative entry, but
there are no positive pivot candidates in that column to pivot on. We know that this means
that this linear program does not have a maximum.
The only possible explanation is that the feasibility region must be unbounded and the
value of the objective function can continue to grow unboundedly.
4.
·
α 2
(a) (3pts) Calculate the product
1 β
Solution:
·
¸·
¸
3 2
.
−2 1
¸ ·
¸
α(3) + 2(−2) α(2) + 2(1)
3α − 4 2α + 2
=
.
1(3) + β(−2) 1(2) + β(1)
3 − 2β 2 + β
(b) (2pts) Are there values for α and β such that the following equation is satisfied? Clearly
justify your answer.
·
¸·
¸
·
¸
α 2
3 2
5 1
= −2
.
1 β −2 1
3/2 0
Solution: Using our answer from part (a),
·
¸
·
¸
3α − 4 2α + 2
5 1
= −2
.
3 − 2β 2 + β
3/2 0
Using the definition of scalar multiplication of a matrix,
·
¸ ·
¸
3α − 4 2α + 2
−10 −2
=
.
3 − 2β 2 + β
−3 0
For this matrix equation to be satisfied, we require that the system of equations
3α − 4
2α + 2
3 − 2β
2+β
=
=
=
=
−10
−2
−3
0
must have a single solution of α and β. Although α = −2 satisfies the first two equations,
the third and the fourth equation require that β = 3 and β = −2, respectively. Hence, there
are no values of α and β that will satisfy this matrix equation.
5. (5pts) Compute
¯
¯ x y 0
¯
¯ 0 3 1
¯
¯ 2 0 1
¯
¯
¯
¯.
¯
¯
Solution: We can chose to expand the determinant on across any row or down any column.
For simplicity, I choose to expand across the first row.
¯
¯
¯ x y 0 ¯
¯
¯
¯ 0 3 1 ¯ = a11 A11 + a12 A12 + a13 A13
¯
¯
¯ 2 0 1 ¯
= a11 (−1)1+1 M11 + a12 (−1)1+2 M12 + a13 (−1)1+3 M13
= a11¯ M11 −¯ a12 M¯ 12 + a13
¯
¯ M13¯
¯ 0 3 ¯
¯ 0 1 ¯
¯ 3 1 ¯
¯
¯
¯
¯−y¯
= x ¯¯
¯ 2 1 ¯ + 0¯ 2 0 ¯
0 1 ¯
= x[3(1) − (1)0] − y[0(1) − (1)2] + 0
= 3x + 2y
6. A population is grouped as follows: 50% nonsmokers, 25% smokers of one pack a day or
less per day, and 25% smokers of more than one pack a day. During any given month there
is a 5% probability that a nonsmoker will begin smoking a pack or less per day, and a 2%
probability that a nonsmoker will begin smoking more than a pack per day. For smokers
who smoke a pack or less per day, there is a 5% probability of quitting and a 15% probability
of increasing to more than a pack a day. For smokers who smoke more than a pack per day,
there is a 1% probability of quitting and a 10% probability of dropping to a pack or less per
day.
(a) (2pts) Construct the transition matrix defined by this model. Clearly label the matrix.
nonsmokers

.93
Solution:
 .05
T =
.02
≤ one pack a day
.05
.80
.15
≥ one packa day
.01
.10 
.89
to nonsmokers
to ≤ one pack a day
to ≥ one pack a day
(b) (3pts) What percentage of people will be more-than-one-pack-a-day smokers in 2 years?
(Round your answer to 3 decimal places.)
Solution: First, note that the Markov process above is in the time periods of months. This
question asks for the state of thepopulation
in 2 years. This is equivalent to t = 24. We

.5
have an initial condition of S0 =  .25  . To determine what the population will look like
.25
in 2 years, we use the formula S24 = T 24 S0 .

24 
 

.93 .05 .01
.5
.294
S24 = .05 .80 .10  .25  =  .283  .
.02 .15 .89
.25
.424
The model suggests that in 2 years, 42.4% of the population will be smoking more than one
pack a day.
(c) (5pts) Determine the exact stable vector for this Markov process. (Make sure the stable
vector is in fractions.) In the long run, what percentage of the population will be nonsmokers?
~ such that it satisfies
Solution: To find the exact stable vector, we need to find a vector S
~
the stability requirement ((T − I)S = ~0) and is stochastic.


−.07 .05 .01
~ = ~0
Note that [T − I] =  .05 −.20 .10  . Using the stability requirement [T − I]S
.02
.15 −.1
yields two equations −.07n + .05l + .01m = 0 and .05n − .20l + .10m = 0, where n is the
percentage of nonsmokers at equilibrium, l is the percentage of smokers who smoke a pack
or less per day at equilibrium and m the percentage of more-than-one-pack-a-day smokers at
~ we also have the equation n+l +m = 1.
equilibrium. Using the stochastic requirement for S,


−.07 .05 .01
~ = B
~ where A =  .05 −.20 .10
These three equations form the matrix equation AS
1
1
1
 
0
~ =  0 . Solving for the exact stable vector S,
~
and B
1


7/26
~ = A−1 B
~ =  15/52  .
S
23/52
(Remark: You could have used an augmented matrix to solve this linear system.)
The model suggests that in the long run, approximately 26.9% of the population will be
nonsmokers.
7. Let a small economy contain agriculture, manufacturing, and labor industries. Let $1 of
agricultural production require $0.50 in agriculture, $0.20 in manufacturing, and $1 in labor.
Let $1 of manufacturing production use $0.80 in manufacturing and $0.40 labor, while a $1
in labor takes $0.25 in agriculture and $0.10 in manufacturing.
(a) (3pts) Construct the consumption matrix for this model. Clearly label the matrix.
Solution:
C=
agriculture

.5
 .2
.1
manufacturing
0
.8
.4
labor
.25
.10 
0
agriculture used
manufacturing used
labor used
(b) (2pts) Define what it means for an economy to be productive.
~ there is a proSolution: An economy is productive if given any external demand (D),
~ that can meet the demand. Mathematically, there is an X
~ such that
duction schedule (X)
−1
~ = [I −C] D
~ has all positive entries for any demand vector D.
~ Either answer is acceptable.
X
(c) (2pts) Clearly explain why this economy is a productive economy.
Solution: The only way to show this economy is
positive matrix.

16
−1

[I − C] = 30
28
productive is to show that [I − C]−1 is a

10 5
25 10
20 10
(d) (3pts) Find the production schedule if demand is for $10,000,000 in agriculture,
$50,000,000 in manufacturing, and $70,000,000 in labor.
 
10
~

Solution: We use this information to form the demand vector D = 50 where the units
70
~
on this vector are millions of dollars. We need to solve for the production vector X.


1010
~ = [I − C]−1 D
~ =  2250 
X
1980
The production schedule that meets this demand is for the economy to produce $1.01
billions dollars in agriculture, $2.25 billion dollars in manufacturing, $1.98 billion dollars in
services.
8. (5pts) Determine if the following statement is true or false. For each part, you will receive
1 point for a correct answer, −1 point for an incorrect answer, and 0 points for no answer.
The lowest possible score on this problem is zero.
(a) For any matrix A, both AAt and At A are defined.
Solution: FALSE. This is only true if A is a square matrix.
(b) If A and B are such that AB = 0 and A 6= 0, then B = 0.
Solution: FALSE. This is only true if A is an invertible matrix.
(c) Matrix multiplication is associative.
Solution: TRUE. This is one of the properties of matrix multiplication.
(d) Every matrix A has an additive inverse.
Solution: TRUE. For any matrix A, the matrix (−1)A = −A is the additive inverse.
(A + (−A) = 0 and (−A) + A = 0.)
(e) The product of a 2 × 3 matrix and a 3 × 5 matrix is a matrix that is 5 × 2.
Solution: FALSE. The product of a 2 × 3 matrix and a 3 × 5 matrix is a matrix that is
2 × 5.
(f) If A and B are invertible, then A + B is invertible.
Solution: FALSE. This was a homework question.
(g) The determinant of a matrix can be evaluated using expansion by cofactors in any row
or column.
Solution: TRUE. This is part of the cofactor expansion of the determinant definition.
Math 125 – Exam 2 – Version 2
March 28, 2007
60 points possible
1. Consider the linear system
x − 3y + 2z = −1
−4x + 12y − 7z = 8.
(a) (3pts) Write the above system as a matrix equation.
Solution:
·
 
¸ x
· ¸
1 −3 2  
−1
y =
.
−4 12 −7
8
z
(b) (2pts) Explain why the use of a matrix inverse is not an acceptable way to solve this
system.
Solution: To solve for the variable vector, we would need to multiply (on the left) by the
inverse of the coefficient matrix. But, this is not possible since the coefficient matrix is not
square. Inverses only exist for square matrices.
2. Acme Battery Company produces three kinds of industrial batteries from lead, acid, and
plastic. The 24-month battery is constructed from 4 pounds of lead, 12 ounces of acid, and
10 ounces of plastic. The 36-month battery is constructed from 6 pounds of lead, 14 ounces
of acid, and 12 ounces of plastic. The 48-month battery is constructed from 8 pounds of
lead, 16 ounces of acid, and 16 ounces of plastic. The selling prices are $40 for a 24-month,
$48 for a 36-month, and $64 for a 48-month battery, and there are 7200 pounds of lead,
19,200 ounces of acid, and 28,200 ounces of plastic on hand.
(a) (7pts) Set up but do not solve a linear program that enables the Company to maximize revenue. Assume that the Company will sell everything it produces.
Solution: The above information can be organized in the following table:
lead (in pounds)
acid (in ounces)
plastic (in ounces)
revenue in dollars
24-month
4
12
10
40
36-month 48-months supplies
6
8
7200
14
18
19200
12
6
28200
48
64
(Note: You did not have to table the information for credit.)
Let x = the number of 24-month batteries manufactured, y = the 36-month batteries
manufactured, and z = the number of 48-month batteries manufactured.
The objective here is to maximize revenue. The revenue function is R = 40x + 48y + 64z.
Since there is only 7200 pounds of lead, we get the inequality 4x + 6y + 8z ≤ 7200.
Similarly, due to the 19,200 ounces of acid available, we get an acid inequality of 12x + 14y +
16z ≤ 19200. Lastly, there is only 28,200 ounces of plastic. This results in the inequality
10x + 12y + 16z ≤ 28200.
Lastly, we know that our variables need to be non-negative. (It doesn’t make sense to
manufacture a negative number of items.) That is, x ≥ 0. y ≥ 0, and z ≥ 0.
The above discussion accurately describes the linear program. One could choose to
represent this in its mathematical form:
maximize R = 40x + 48y + 64z
subject to 4x + 6y + 8z ≤ 7200
12x + 14y + 16z ≤ 19200
10x + 12y + 16z ≤ 28200
x ≥ 0, y ≥ 0, z ≥ 0
(Note: You did not have to order the linear program like this for credit.)
(b) (3pts) Assuming that there is a solution to the linear program, give a brief geometric
explanation of how the simplex algorithm works.
Solution: The simplex algorithm always starts at the origin and moves to neighboring corner
points in a way that always increases the value of the objective. The simplex algorithm moves
from corner to corner until it finds the maximum value of the objective function.
3. Consider the linear program:
maximize z = −x1 + 9x2 + 3x3
subject to 3x1 + 9x2 − 6x3 ≤ 420
x1 + 3x2 − 6x3 ≤ 100
x2 + 3x3 ≤ 350
x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
(a) (2pts) Construct the initial simplex table for this system.
Solution: Moving the objective function into standard form and adding slack variables to
the constraint inequalities, we get:
z + x1 − 9x2 − 3x3
3x1 + 9x2 − 6x3 + x4
x1 + 3x2 − 6x3 + x5
x2 + 3x3 + x6
=
=
=
=
0
420
100
350
where x4 , x5 , and x6 are slack variables.
Constructing the initial simplex table:

1
1 −9 −3
 0
3
9 −6

 0
1
3 −6
0
0
1
3
0
1
0
0
0
0
1
0
0
0
0
1

0
420 
.
100 
350
(b) (2pts) Determine the first pivot point dictated by the simplex algorithm.
Solution: The −9 heading the x2 column is the most negative entry in the top row. Therefore, we look for a pivot point in this column. Recall that we can only pivot on the positive
entries.
pivot candidate quotient
9
3
1
420
> 46
9
100
< 34
3
350
= 350
1
Since the 3 in the third row makes the smallest quotient when divided into the entry in the
last column, it is the pivot point.
(c) (3pts) Perform the first pivot dictated by the simplex algorithm and determine the
basic feasible solution for the resulting table.
Solution: The next simplex table:

1
4 0 −21
 0
0 0
12

 0
1/3 1 −2
0
−1/3 0
5
0
3
1
−3
0
1/3
0 −1/3

300
120 
.
100/3 
950/3
0
0
0
1
The basic feasible solution for this table is x1 = 0, x2 = 100/3, x3 = 0, x4 = 120, x5 = 0
and x6 = 950/3 or (0, 100/3, 0, 120, 0, 950/3).
(d) (3pts) Continue to use the simplex algorithm to attempt to solve the given linear program. If a solution exists, clearly state the conclusion. If a solution does not exist, explain
how the algorithm demonstrated this and what the geometrical reason for no solution must
be.
Solution: The next pivot point is
yields the next simplex table.

1
4 0
 0
0 0

 0
1/3 1
0
−1/3 0
the 12 located in the column leading with the −21. This
0
1
0
0
7/4 −9/4
1/12 −1/4
1/6 −1/6
−5/12 11/12
0
0
0
1

510
10 
.
160/3 
800/3
Now the pivot column has moved to the x5 slack variable column. There is only one
positive entry in that column to pivot on. This yields the next, and final, simplex table.


1
35/11 0 0
8/11 0 27/11
12810/11
 0
−1/11 0 1
−1/33 0 3/11
910/11 

.
 0
3/11 1 0
1/11 0 2/11
1120/11 
0
−4/11 0 0
−5/11 1 12/11
3200/11
Now that there are no columns that begin with a negative number, we know that the
12810
simplex algorithm is complete. Using the table, we see that the maximum z value is
11
¶
µ
1120 910
,
.
and occurs at the point 0,
11 11
4. Suppose P is 3 × 2, Q is 2 × 1, R is 1 × 3 and S is 3 × 3. Clearly explain whether the
following expressions exist. If they do exist, determine the size of the resultant matrix.
(a) (3pts) 2P QR.
Solution: Note that the scalar multiple 2 has no bearing on whether the product is sensible
or not. Recall that to multiply two matrices, we need the number of columns of the first
matrix to equal the number of rows of the second. Look at the product P QR.
P QR = [3 × 2][2 × 1][1 × 3]
= [3 × 1][1 × 3]
= 3×3
Therefore, the product makes sense and the resultant matrix is a 3 × 3 matrix.
(b) (2pts) QR − SP Q
Solution: In this problem, the products make sense. QR becomes a 2 × 3 matrix and SP Q
becomes a 3 × 1 matrix. What we can not do is add these matrices together. Recall that
subtraction is just the combination of addition and multiplication by the scalar (−1). In
order to add matrices, they must be the same dimension. Since QR and SP Q are of different
dimensions, this expression does not make sense.

4
1
5. (5pts) Calculate the first iteration of the determinant of A = 
2
0
panding down the second column.

6 2
3
5 0 −1
 by ex4 −3 0 
3 1
5
Solution: Expanding down the second column,
det A =
=
=
=
a12 A12 + a22 A22 + a32 A32 + a42 A42
a12 (−1)1+2 M12 + a22 (−1)2+2 M22 + a32 (−1)3+2 M32 + a32 (−1)4+2 M42
−a12 M12 + a22 M22 − a32 M32 + a42 M42
−6M
¯
¯
¯
¯ 12 + 5M22 −¯4M32¯ + 3M42 ¯
¯ 4
¯ 4 2
¯ 4
¯ 1
2
3
3 ¯¯
2 3 ¯¯
0 −1 ¯¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0 −1
0 ¯ + 5 ¯ 2 −3 0 ¯ − 4 ¯ 1 0 −1 ¯ + 3 ¯ 1
= −6 ¯ 2 −3
¯
¯
¯
¯
¯
¯
¯ 0
2 −3
0
0 1
5
0
1 5
1
5
¯
¯
¯
¯
¯
¯
6. A small community includes a farmer, a baker, and a grocer. Suppose that each dollar’s
worth of farming requires $0.30 of farmed products, $0.10 of bakery goods and $0.30 in
groceries. Each dollar’s worth of bakery goods requires $0.35 of farmed products, $0.20 of
baked goods and $0.10 in groceries. Each dollar’s worth of groceries requires $0.60 of farmed
products, $0.20 of bakery products and $0.35 of groceries.
(a) (3pts) Construct the consumption matrix for this model. Clearly label the matrix.
Solution:
C=
farmer

.30
 .10
.30
baker
.35
.20
.10
grocer

.60
.20 
.35
farm products used
bakery products used
groceries used
(b) (2pts) Define what it means for an industry or company in a Leontiff model to be
profitable.
Solution: A company or industry is profitable if it cost the company less than a dollar to
make a dollar’s worth of its product.
(c) (2pts) Determine if any of the individuals in this economy are profitable. Justify your
answer.
Solution: Summing the columns of the consumption matrix C, we see that it costs $.70 to
make a dollar’s worth of farming goods and $.65 to make a dollar’s worth of baked goods.
Therefore, those individuals are profitable. The grocer is not profitable since it costs $1.15
to produce a dollar’s worth of groceries.
(d) (3pts) If the small economy is currently producing $75,000 in farming, $30,000 in baked
goods, and $45,000 in groceries, determine whether this community is self-sustaining or needs
to purchase additional resources from a neighboring community. Clearly justify your answer.


75000
~ = 30000 and are asked to interpret
Solution: Here, we are given a production schedule X
45000
~
the D in the Leontiff model.

 

75000
15000
[I − C] 30000 =  7500  .
45000
37500
Interpreting this in the context of the problem, given the current production schedule, each
individual is producing a surplus. Hence, the community is self-sustaining.
7. A researcher is modeling college males antiperspirant usage. She dreams of a perfect
world where all college males’ bathe daily and use deodorant. Her research has yielded the
following conclusions. During any given month, 20% of Right-Gard users switch to Speed
Stick and 5% of Right-Gard users switch to Degree. For Speed Stick users, 15% switch to
Right-Gard and 10% switch to Degree. For Degree users, 10% switch to Right-Gard and
20% switch to Speed Stick.
(a) (3pts) Construct the transition matrix defined by this model. Clearly label the matrix.
Right-Gard
Speed Stick Degree


.15
.10
.75
Solution:
 .20
.75
.20 
T =
.10
.70
.05
to Right-Gard
to Speed Stick
to Degree
(b) (3pts) What is the probability that a Degree user chosen at random will still be a
Degree user next month?
Solution: This is an alternative way of interpreting the entries of the transition matrix.
Each month, 70% of Degree users stay Degree users. Therefore, there is a 70% probability
that a Degree user chosen at random will still be a Degree user next month.
(c) (4pts) Find the approximate stable matrix (use 3 decimal place accuracy) and determine
when the researcher’s model predicts that the antiperspirant market will stabilize.
Solution: Looking at powers of the transition matrix, we see that when t = 20,


.349 .349 .349
T 20 = .444 .444 .444 .
.206 .206 .206
This is the first time that all of the columns of T n are the same through three decimal places.
Hence, the model predicts that the antiperspirant market will stabilize in 20 months.
Alternatively, we see that t = 20 is the time that the model stabilizes because it is the
first time period where T 20 = T 21 (through three decimal places).
8. (5pts) Determine if the following statement is true or false. For each part, you will receive
1 point for a correct answer, −1 point for an incorrect answer, and 0 points for no answer.
The lowest possible score on this problem is zero.
(a) The transpose of the sum of matrices is equal to the sum of the transposes of the matrices.
Solution: TRUE. This is one of the properties of transposes.
(b) Matrix addition is commutative.
Solution: TRUE. This is one of the properties of matrix arithmetic. (Remark: It is multiplication that is not commutative.)
(c) If the matrices A, B, and C satisfy AB = AC, then B = C.
Solution: FALSE. This is only true if A is invertible.
(d) If A and B are invertible and of the same dimension, then AB is also invertible.
Solution: TRUE. This was a homework problem.
(e) All n × n matrices are invertible.
Solution: FALSE. Only those matrices who have the property that their reduced row echelon form is the identity matrix are invertible.
(f) The determinant of the 2 × 2 matrix A is a21 a12 − a11 a22 .
·
¸
a11 a12
Solution: FALSE. A general 2 × 2 matrix has the form
. The determinant of this
a21 a22
matrix is a11 a22 − a12 a21 .
(g) The (i, j)–cofactor of a square matrix A is the matrix defined by deleting the ith row
and the jth column.
Solution: FALSE. This is only a part of the definition. The (i, j)–cofactor of a square matrix
A is (−1)i+j times the determinant of the matrix defined by deleting the ith row and the
jth column.