Name: Section Registered In: Math 125 – Exam 2 – Version 1 March 27, 2006 

     α 2 2 1 −2 7 , B =  , and C =  . 1. Let A =  2 4 3 2 5 −1 (a) (1pt) Compute AB.

Solution:



      α 2 α·2+2·3 α·1+2·2 2α + 6 α + 4 2 1  = = . AB =  2 4 3 2 2·2+4·3 2·1+4·2 16 10

(b) (1pt) Compute AC.

Solution:        α 2 −2 7 α · −2 + 2 · 5 α · 7 + 2 · −1 −2α + 10 7α − 2  = = . AC =  2 4 5 −1 2 · −2 + 4 · 5 2 · 7 + 4 · −1 16 10

(c) (2pt) Solve for α such that AB = AC.

Solution: Recall that for two matrices to be equal, we need the entries in each matrix to be the same. For AB = AC, we need 2α + 6 = −2α + 10 α + 4 = 7α − 2 16 = 16 10 = 10 The third and fourth equations are obviously satisfied. Solving for α in the first equation, we see that for equality in the a11 position, α = 1. Solving the second equation for α, we get

the same solution: α = 1. Therefore, AB = AC when α = 1.

(d) (1pts) Note that here AB = AC but B 6= C. What property is A lacking that AB = AC does not require that B = C? Briefly explain your answer. Solution: When A is invertible, we can left multiply the equation AB = AC by A−1 . This results in the equation B = C.

2. (a) (3pts) Define the multiplicative inverse of a matrix A.

Solution: A matrix B is the multiplicative inverse of a matrix A if AB = I and BA = I.

(b)  (7pts) 2 1   A = −4 −1  −2 2

Usethe proposition for formally calculating inverses to find the inverse of −4   6 , if it exists.  −2

Solution: We need to form [A|I] and move the augmented matrix into reduced row echelon form.



 2

1 −4

  [A|I] =  −4 −1 6  −2 2 −2

1 0 0

  0 1 0 .  0 0 1

Using the calculator to move this to reduced row echelon form, we get the matrix   0 −1/5 −1/10 1 0 −1      0 1 −2 0 −1/5 2/5  .   0 0 0 1 3/5 −1/5 Since the reduced row echelon form of A is not the identity matrix, we know that no inverse exists for matrix A.

3. A small town has three primary industries: a copper mine, a railroad, and an electric utility. To produce $1 of copper, the copper mine uses $0.20 of copper, $0.10 of transportation, and $0.20 of electric power. To provide a $1 of transportation, the railroad uses $0.10 of copper, $0.10 of transportation, and $0.40 of electric power. To provide a $1 of electricity, the electric utility uses $0.20 of copper, $0.20 of transportation, and $0.30 of electric power. (a) (2pts) Form the consumption matrix for the above economic model.

Solution:

C=

copper  .20    .10  .20

railroad .10 .10 .40

electric  .20   .20   .30

copper used railroad used electricity used

(You did not have to label the matrix for full-credit.)

(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) ~ = [I −C]−1 D ~ has all positive entries for any demand vector D. ~ Either answer is acceptable. X

(c) (3pts) Determine if the small town’s economy is productive.

Solution: There are multiple solution to this question. First, we can note that every industry is profitable (since every column sums to less than 1), hence the economy is productive. Another solution would be to recognize that every row sums to less than 1, hence the economy is productive. Lastly, we can show that [I − C]−1 is non-negative.   10/7 30/77 40/77     [I − C]−1 =  2/7 104/77 36/77  .   4/7 68/77 142/77 Since no entry in the inverse is negative, the economy is productive.

(d) (3pts) Suppose that during the year there is an outside demand of 1.2 million dollars for copper, 0.8 million dollars for transportation, and 1.5 million dollars for electric power. How much should each industry produce to satisfy the demands?   1.2    ~ = 0.8 Solution: We are given that D  (where the units on each entry is in millions of   1.5 −1 ~ ~ ~ dollars).  are asked to solve for X such that X = [I − C] D. Using the calculator,  We 2.81     ~ X ≈ 2.12. Therefore the copper industry should produce 2.81 million dollars of copper,   4.16 the railroad should produce 2.12 million dollars of transportation, and the electricity industry should produce 4.16 million dollars in electricity.

4. (a) (3pts) Let A be a square matrix and let B be the matrix obtained from A after a stated row operation. Using A and B, describe the effects each row operation has on the value of the determinant.

Solution: If Ri ↔ Rj , then det(B) = − det (A). If Ri = kRi , then det(B) = k det (A). If Ri = Ri + kRj , then det(B) = det (A).   2 4 6     (b) (7pts) Find the determinant of A = 4 3 2 by using row operations to make   3 −2 5 the matrix upper triangular. Solution:  R20 = R2 − 2R1 , 3 R30 = R3 − R1 ; 2



2 4 6     A1 = 0 −5 −10 ,   0 −8 −4 



2 4 6   1   0 A 2 = 0 1 R2 = − R2 ; 2 ,   5 0 −8 −4   2 4 6     0 A =  R3 = R3 + 8R2 ; 0 1 2 , 3   0 0 12

det A1 = det A

1 det A2 = − det A1 5 . 1 = − det A 5 det A3 = det A2 1 = − det A 5

.

Since A3 is upper triangular, the determinant of A3 is 2 · 1 · 12 or det A3 = 24. Using 1 det A3 = − det A, we see that det A = 24 · −5 = −120. 5

5. Let A be an invertible n × n matrix. Determine if the following are always true. Fully justify your answer. (a) (3pts) (I + A)(I + A−1 ) = 2I + A + A−1

Solution: This is always true. Using the properties of matrix arithmetic: (I + A)(I + A−1 ) = II + IA−1 + AI + AA−1 = I + A−1 + A + AA−1 by properties of I = I + A−1 + A + I by properties of the inverse = 2I + A−1 + A since addition is commutative

(b) (2pts) If A2 is invertible then (A2 )−1 = (A−1 )2 .

Solution: This is always true. (A2 )(A−1 )2 = (AA)(A−1 A−1 ) = AAA−1 A−1 = AIA−1 = AA−1 = I Therefore (A2 )−1 = (A−1 )2 .

(Technically, we should also do the same calculation for

(A−1 )2 (A2 ) = I. But since A2 is a square matrix, we are guaranteed that, if we have the right inverse, it is the left inverse as well.)

6. Consider the linear system x + 2y + 3z = 5 2x + 5y + 3z = 3 x + 8z = 17. (a) (2pts) Write this linear system as a matrix equation of the form A~x = ~b.      5 x 1 2 3           Solution: 2 5 3 y  =  3       17 1 0 8 z (b) (1pt) Find A−1 .

Solution: The calculator says that A−1

  −40 16 9     =  13 −5 −3 .   5 −2 −1

(c) (2pts) Use A−1 to solve for the variable vector. Be sure to show work. Solution: Since A~x = ~b and A−1 inverse exist, we can solve for ~x by the equation ~x = A−1~b.      −40 16 9 5 1           −1~ ~x = A b =  13 −5 −3  3  = −1 .      5 −2 −1 17 2

7. A new mass transit system has just gone into operation. The transit authority has made studies that predict the percentage of commuters who will change to mass transit (M) or continue driving their automobile (A). Data suggests that 20% of mass transit users will switch to their auto next year and 30% of auto commuters will switch to mass transit. (a) (3pts) Construct the transition matrix for this model. Be sure to label your matrix.

 Solution:

M

A

.8

.3

.2

.7

T = 

 

to M to A

(b) (2pts) Suppose that the population of the area remains constant and that initially 30% of commuters are using mass transit. What percentage of commuters will be using mass transit after 4 years?  Solution: Writing the initial state as a vector, we get S0 =  S4 = T 4 S0 = 



 .3



.7

M A

.

 .58125 .41875

.

Mass transit has roughly 58.1% of the market and auto commuters are 41.9% of the market after four years.

(c) (5pts) Using four decimal places of accuracy, determine how long it takes the market to stabilize and determine what percentage of commuters will be using mass transit in the long run. (Be sure to clearly explain how you determined your answer.)

Solution: To determine when a market stabilizes, we look at powers of the transition matrix and look for the first time period k where the transition matrix (with entries rounded to 4 decimal places) stops changing for higher powers of the transition matrix. 

 T 13 = 

0.6000 0.5999 0.4000 0.4001

.

 T 14 = 

 0.6000 0.6000 0.4000 0.4000

 T 15 = 

. 

0.6000 0.6000 0.4000 0.4000

.

Hence, it takes 14 years for the market to stabilize.

An alternative way of finding the time period that the market stabilizes is to look for the first power of T such that, when rounding, all of the column vectors are the same vector. From the above matrices, we see that the first time this happens is when t = 14. In the long run, 60% of commuters will be using mass transit.

8. (1pt each) Answer the following either True or False and justify your answer. (a) If A and B are matrices, then both AB and BA are defined if and only if A and B are square matrices.

Solution: False. To multiply matrices, you need the number of columns of the first matrix to equal the number of rows of the second. To have both AB and BA, this requires that A be n × k and B be k × n. Note that if n does not equal k, then A and B are not square but the multiplication is still defined.

(b) Every square matrix is invertible.

Solution: False. We have encountered many square matrices that do not have a multiplicative inverse.

(c) If A and B are matrices such that AB = I, then both A and B are invertible.

Solution: False. The definition of multiplicative inverses is that AB = I and BA = I. If A and B are not square, we are not even guaranteed that AB even exists, let alone the identity matrix.

(d) If A is a 3 × 2 matrix and B is a 2 × 4 matrix, than AB is a 3 × 4 matrix.

Solution: True. Since the number of columns of A equal the number of rows of B, we can calculate AB. By construction, the resulting matrix will have the dimensions: the numbers of rows of the first by the number of columns of the second. Here, that gives a 3 × 4 matrix.

(e) The determinant of an n × n matrix can be evaluated using a cofactor expansion along any row.

Solution: True. The cofactor expansion definition of the determinant allows us to expand along any row or down any column.

Name: Section Registered In: Math 125 – Exam 2 – Version 2 March 27, 2006

1. Consider the linear system x + 2y + 3z = 6 2x − 3y + 2z = 14 3x + y − z = −2. (a) (2pts) Write this linear system as a matrix equation of the form A~x = ~b.      1 2 3 x 6           Solution: 2 −3 2  y  =  14       3 1 −1 z −2 (b) (1pt) Find A−1 .  Solution: The calculator says that A−1



1/50 1/10 13/50     =  4/25 −1/5 2/25  .   11/50 1/10 −7/50

(c) (2pts) Use A−1 to solve for the variable vector. Be sure to show work. Solution: Since A~x = ~b and A−1 inverse exist, we can solve for ~x by the equation ~x = A−1~b.      1 6 /50 1/10 13/50           ~x = A−1~b =  4/25 −1/5 2/25   14  = −2 .      3 −2 11/50 1/10 −7/50

2. (a) (3pts) How is it possible to use the determinant to determine if a matrix is invertible?

Solution: If the determinant of a matrix is nonzero then it is invertible. 

 k

−k

3

    (b) (7pts) Find all values of k for which the matrix A = 0 k + 1 1  is invertible.   k −8 k − 1 Solution: We want to use part (a) and take this determinant. To compute the determinant, I will choose to do a cofactor down the first column since there is a zero in it. ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −k 3 ¯ ¯ −k ¯ k+1 3 ¯ 1 ¯ ¯ ¯+k¯ ¯ − 0¯ det A = k ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ k+1 1 ¯ ¯ −8 k − 1 ¯ ¯ −8 k − 1 ¯ = k[(k + 1)(k − 1) + 8] + k[−k − 3(k + 1)] = k[k 2 − 1 + 8] + k[−4k − 3] = k 3 + 7k − 4k 2 − 3k = k 3 − 4k 2 + 4k = k(k 2 − 4k + 4) = k(k − 2)2 When the determinant is not equal to zero, then A is invertible. We want k(k − 2)2 6= 0. Therefore, the matrix is invertible when k 6= 0 and k 6= 2.

3. An island’s economy is divided into three sectors – tourism, transportation, and services. Suppose that to produce a dollar’s worth of tourism requires that inputs of $0.30 from the tourism sector, $0.10 from the transportation sector, and $0.30 from the services sector; each dollar’s worth of transportation requries $0.20 from the tourism sector, $0.40 from the transportation sector, and $0.20 from the services sector; and each dollar’s worth of services requires $0.05 from the tourism sector, $0.05 from the transportation sector, and $0.15 from the services sector. (a) (2pts) Form the consumption matrix for this economy.

Solution:

C=

tourism transportation  .30 .20    .10 .40  .30 .20

services  .05   .05   .15

tourism used transportation used services used

(You did not have to label the matrix for full-credit.)

(b) (2pts) Is the consumption matrix from part (a) a stochastic matrix? Clearly explain your answer.

Solution: C is not stochastic since the columns of C do not sum to 1.

(c) (2pts) Are any of the industries profitable? Clearly explain your answer.

Solution: All of the industries are profitable since every column sums to less than 1.

(d) (2pts) Is the economy productive? Clearly explain your answer.

Solution: Since every industry is profitable, we know that the economy is productive.

Another solution would be to recognize that every row sums to less than 1, hence the economy is productive.

Lastly, we can show that [I − C]−1 is non-negative. Since no entry in the inverse is negative, the economy is productive.

(e) (2pts) The current gross production for this economy is $70 million in tourism, $20 million in transportation, and $10 million in services. Is the economy self-sustaining in each area? Solution: Here we are given a production schedule and are asked if the is producing enough to sustain its current production.     70 44.5         [I − C] 20 =  4.5  .     −16.5 10 The economy is not self-sustaining since there is a shortage of 16.5 million dollars of services.

 4. Let A = 

 0 1



, B = 

β 2 (a) (1pt) Compute AB.

 1 1 3 4



, and C = 

 2 5 3 4

.

Solution:  AB = 













0 1

1 1 0·1+1·3 0·1+1·4 3 4  = = . β 2 3 4 β·1+2·3 β·1+2·4 β+6 β+8

(b) (1pt) Compute AC.

Solution:  AC = 

 0 1





2 5





0·2+1·3 0·5+1·4

 3

4

 = = . β 2 3 4 β·2+2·3 β·5+2·4 2β + 6 5β + 8

(c) (2pts) Solve for β such that AB = AC.

Solution: Recall that for two matrices to be equal, we need the entries in each matrix to be the same. For AB = AC, we need 3 = 3 4 = 4 β + 6 = 2β + 6 β + 8 = 5β + 8 The first and the second equations are obviously satisfied. Solving for β in the third equation, we see that for equality in the a21 position, β = 0. Solving the fourth equation for β, we get the same solution: β = 0. Therefore, AB = AC when β = 0.

(d) (1pts) Note that here AB = AC but B 6= C. What property is A lacking that AB = AC does not require that B = C? Briefly explain your answer. Solution: When A is invertible, we can left multiply the equation AB = AC by A−1 . This results in the equation B = C.

5. Let A and B be two invertible n × n matrices. Determine if the following are always true. Fully justify your answer. (a) (3pts) (A + B)2 = A2 + 2AB + B 2 .

Solution: This is false. Using the properties of matrix arithmetic: (A + B)2 = (A + B)(A + B) = AA + AB + BA + BB since multiplication is distributive = A2 + AB + BA + B 2 Recall that matrix multiplication is usually not commutative. Therefore, (A + B)2 = A2 + 2AB + B 2 only if AB = BA. (b) (2pts) (ABA−1 )2 = AB 2 A−1 .

Solution: This is true. (ABA−1 )2 = (ABA−1 )(ABA−1 ) = ABA−1 ABA−1 since multiplication is associative = ABIBA−1 by properties of inverses = ABBA−1 by properties of the identity = AB 2 A−1

6. A new mass transit system has just gone into operation. The transit authority has made studies that predict the percentage of commuters who will change to mass transit (M) or continue driving their automobile (A). Data suggests that 30% of mass transit users will switch to their auto next year and 20% of auto commuters will switch to mass transit. (a) (3pts) Construct the transition matrix for this model. Be sure to label your matrix.

 Solution:

M

A

.7

.2

.3

.8

T = 

 

to M to A

(b) (2pts) Suppose that the population of the area remains constant and that initially 30% of commuters are using mass transit. What percentage of commuters will be using mass transit after 3 years?  Solution: Writing the initial state as a vector, we get S0 =  S3 = T 3 S0 = 



 .3 .7



M A

.

 .3875 .6125

.

Mass transit has roughly 38.8% of the market and auto commuters are 61.3% of the market after three years.

(c) (5pts) Find the exact stable vector and determine what percentage of commuters will be using mass transit in the long run. 

 −.3

.2

~ = ~0  . Using the stability requirement [T − I]S .3 −.2 yields only one equation −.3m + .2a = 0, where m is the percentage of mass transit users

Solution: Note that [T − I] = 

at equilibrium and a the percentage of auto commuters at equilibrium. Using the stochastic ~ we also have the equation m + a = 1. These two equations form the requirement for S,     0 −.3 .2 ~ =  . Solving for the exact ~ =B ~ where A =   and B matrix equation AS 1 1 1

~ stable vector S,

 ~ = A−1 B ~ = S

 .4 .6

.

In the long run, 40% of commuters will be using the mass transit system.

7. (1pt each) Answer the following either True or False and justify your answer. (a) If A and B are m × n matrices, then AB is defined and is an m × n matrix.

Solution: False. For AB to be defined, we require that the number of columns of A equal the number of rows of B. Here, m 6= n.

(b) If A is any invertible matrix, then det A = 0.

Solution: False. This is the exact opposite of how we developed the determinant. If A is invertible, then det A 6= 0.

(c) Invertible matrices are square.

Solution: True. The definition of matrix inverses requires AB = BA = I. By default, these equations require that A and B be square matrices. Therefore, if a matrix has an inverse, it must be square.

(d) If a square matrix has a column consisting of all zeros, then it is not invertible.

Solution: True. If we expand the determinant down the column of all zeros, then the det equals 0. Hence, the matrix is not invertible. (e) For any square matrix A, det At = − det A.

Solution: False. Since the definition of the determinant calculates the exact same value regardless of which row or column of the matrix we choose to expand on, the definition of the determinant is unable to tell the difference between A and At . Hence, det At = det A.

8. Consider the linear program: maximize

z = x + 2y

subject to −3x + y ≤ 3 x−y ≤2 x ≥ 0, y ≥ 0 (a) (6pts) Use the simplex algorithm to show that there is no upper bound on the objective function value z. Be sure to explain how the simplex algorithm shows that there is no upper bound.

Solution: First we construct the initial simplex table.  1 −1 −2 0 0    0 −3 1 1 0  1 −1 0 1 0

 0

  3 .  2

We choose the variable column with the most negative number as the pivot column. Here, we pivot on the column starting with the −2. Recall that one can only pivot on positive numbers. Therefore, the 1 is the first pivot point. Pivoting on the 1 yields   −7 0 2 0 6 1      0 −3 1 1 0 3 .   −2 0 1 1 5 0 The algorithm dictates that we find the next variable column that starts with a negative number and choose the next pivot point. Here, we need to pivot in the column that starts with the −7. Unfortunately, there are no positive entries in the new pivot column to pivot on. Therefore, the algorithm is stuck and is unable to determine a maximum. This tells us that no maximum exists for this linear program.

(b) (4pts) Sketch the feasibility region, draw the lines of constancy z = 0 and z = 6 and explain why z can become arbitrarily large.

Solution: We need to find the feasibility region by looking for the intersection of the halfplanes −3x + y ≤ 3, x − y ≤ 2, x ≥ 0, and y ≥ 0. In doing so, we see that the feasibility region is unbounded.

y 12 10 8 6 4 2 2

4

6

8

10

12

14

x

-2 -4

The question then asks us to graph the lines of constancy for z = 0 and z = 6. Solving x z the objective function for y, we see that y = − + . When z = 0, the line of constancy 2 2 x x is y = − . When z = 6, the line of constancy is y = − + 3. Graphing these over the 2 2 feasibility region, y 12 10 8 6 4 2 -2.5

z=6 z=0 2.5

5

7.5

10 12.5

x

-2 -4

Notice that as z increases, the lines are moving upward. Since the feasibility region is x z unbounded, for any positive z, y = − + will intersect the feasibility region. Hence, z can 2 2 grow unbounded and will never obtain a maximum.