1. (3pts) State three of the properties of matrix multiplication.

Transcription

1 Math 125 Exam 2 Version 1 October 23, points possible 1. (a) (3pts) State three of the properties of matrix multiplication. Solution: From page 72 of the notes: Theorem: The Properties of Matrix Multiplication. 1. associative: A(BC) = (AB)C 2. distributive property from the left: A(B + C) = AB + AC 3. distributive property from the right: (A + B)C = AC + BC 4. r(ab) = (ra)b = A(rB) 5. I m A = A = AI n (b) (2pts) Simplify the matrix expressions (A B)(A + B) (A + B) 2. Solution: (A B)(A + B) (A + B) 2 = (A B)(A + B) (A + B)(A + B) = [(A B) (A + B)](A + B) = [A B A B](A + B) = 2B(A + B) = 2BA 2BB = 2BA 2B 2

2 2. (a) (3pts) Clearly state the proposition that relates an invertible matrix and the identity matrix. Solution: Proposition: If A is an invertible matrix (has an inverse) then A can be changed through row operations into the identity matrix. Equivalently, one can say that If A is an invertible matrix, then A s reduced row echelon form is the identity matrix I (b) (7pts) By hand, determine the inverse of the matrix A = 2 3 5, if it exists Solution: We need to form the augmented matrix [A I] and move this matrix into reduced row echelon form. I chose to use Gauss-Jordan elimination. (There is more than one way to row reduce this, but since reduced row echelon form is unique there is only one answer.) Augmented Matrix [A I] R 2 = R 2 2R 1 R 3 = R 3 + R 1 R 2 = R 2 R 1 = R 1 + R 2 R 3 = R 3 2R 2 R 1 = R 1 + R 3 R 2 = R 2 R Since the righthand side of the augmented matrix is now the identity matrix, we know that the matrix after the augment is the inverse of A A 1 =

3 3. Consider the following linear system x + 2y + 3z = a 2x + 5y + 3z = b x + 8z = c (a) (5pts) Write the above linear system as a matrix equation x a Solution: y = b z c (b) (2pts) Find the inverse needed to solve the linear system Solution: Using the calculator, = (c) (3pts) Solve the linear system. Leave your answer in terms of the parameters a, b, and c. x a 40a + 16b + 9c Solution: y = b = 13a 5b 3c. z c 5a 2b c

4 4. A researcher is studying the population flows between metropolitan cities, suburbs and rural areas in the United States. Based on collected data, each year 3% of cities dwellers move to the suburbs and 1% move to rural areas; 1% of suburbanites move to the cities and 1% move to rural areas; 1.5% of those living in rural areas move to the city and 0.5% move to the suburbs. (a) (3pts) Construct the transition matrix defined by this model. Clearly label the matrix. Solution: T = cities suburbs rural to cities to suburbs to rural (b) (3pts) According to census data from 2000, 22% of the U.S. population live in the city, 55% live in the suburbs, and 23% live in rural areas. Determine the projected population percentages for the year (Use 3 decimal place accuracy.).22 Solution: According to the given, we have an initial condition of S 0 =.55. To determine.23 what the population will look like in 20 years, we use the formula S 20 = T 20 S S 20 = = The model suggests that in 2020, 22.4% of the population will live in the city, 49.9% will live in the suburbs, and 27.7% will live in rural areas. (c) (4pts) Determine the exact stable vector for this Markov process and interpret the vector in the context of the problem. (Make sure your vector is in fractions.) Solution: To find the exact stable vector, we need to find a vector S such that it satisfies the stability requirement ((T I) S = 0) and is stochastic Note that [T I] = Using the stability requirement [T I] S = yields two equations.04c +.01s +.015r = 0 and.03c.02s +.005r = 0, where c is the percentage of city dwellers at equilibrium, s is the percentage of suburbanites at equilibrium and r the percentage of rural livers at equilibrium. Using the stochastic requirement for S, we also have the equation c + s + r = 1. These threeequations form the matrix equation AS = B where A = and B = 0. Solving for the exact stable vector S, S = A 1 B = 7/30 13/30 1/3..

5 The model suggests that in the long run, 23.3% of the population will live in cities, 43.3% will live in the suburbs, and 33.3% will live in rural areas.

6 a b c g h i 5. (5pts) Assume det d e f = 6. Find the determinant of 2a 3g 2b 3h 2c 3i. g h i 2d 2e 2f Fully justify your answer. a b c Solution: Let A = d e f. g h i g h i R 1 R 3 A 1 = d e f, a b c det A 1 = det A = (6) = 6 g h i R 2 R 3 A 2 = a b c, det A 2 = det A 1 = ( 6) = 6 d e f g h i R 1 = R 1 A 3 = a b c, d e f g h i R 2 = 2R 2 A 4 = 2a 2b 2c, d e f g h i R 3 = 2R 3 A 5 = 2a 2b 2c, 2d 2e 2f g h i R 2 = R 2 + 3R 1 A 6 = 2a 3g 2b 3h 2c 3i, 2d 2e 2f det A 3 = 1 det A 2 = (6) = 6 det A 4 = 2 det A 3 = 2( 6) = 12 det A 5 = 2 det A 4 = 2( 12) = 24 det A 6 = det A 5 = 24

7 6. Consider an economy that is divided into three sectors: agriculture, manufacturing, and services. Suppose that each dollar s worth of agriculture requires $0.10 form the agriculutural sector, $0.15 from the manufacturing sector, and $0.30 from the services sector; each dollar s worth of manufacturing requires $0.20 from the agricultural sector, $0.25 from the manufacturing sector, and $0.30 from the services sector; and each dollar s worth of services requires $0.35 from the agriculutural sector, $0.30 from the manufacturing sector, and $0.40 from the services sector. (a) (3pts) Construct the consumption matrix for this model. Clearly label the matrix. Solution: (b) C = agriculture manufacturing services agriculture used manufacturing used services used (3pts) Are any of the industries profitable? Clearly explain your answer. Solution: The agriculture and manufacturing industries are profitable since both of the columns sum to less than 1. The service industry is not profitable since 1.05 > 1. (c) (3pts) Define what it means for an economy to be productive. Solution: An economy is productive if given any external demand ( D), there is a production schedule ( X) that can meet the demand. Mathematically, there is an X such that X = [I C] 1 D has all positive entries for any demand vector D. Either answer is acceptable. (d) (3pts) Clearly explain why this economy is productive. Solution: The only way to show this economy is productive is to show that [I C] 1 is a positive matrix. (We already showed that the the service industry is not profitable and the last row of C sums to 1.) 80/43 50/43 5/3 [I C] 1 = 40/43 290/129 5/3 60/43 220/129 10/3 (e) (3pts) Determine the production schedule required to satisfy a societal demand of $90 million of agriculture, $72 million of manufacturing, and $75 million of services. Solution: Here we are given a demand vector D 90 = 72 and need to solve for the 75 production vector X X = [I C] 1 D = The production schedule that meets this demand is for the economy to produce $376.2 million dollars in agriculture, $370.6 million dollars in manufacturing, $498.4 million dollars in services.

8 7. (5pts) Determine if the following statement is true or false for matrices A, B, and C. 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) If the sums A + B and B + C exist, then A + C exists. Solution: TRUE. Since A + B exists, A and B are of the same dimensions. Similarly B and C are of the same dimensions. Therefore, A and C are of the same dimensions and their sum exist. (b) If the products AB and BC exist, then AC exists. Solution: FALSE. Since AB exist, we know that the number of columns of A equal the number of rows of B. Similarly, we know that the number of columns of B equal the number of rows of C. But for AC to exist, we need the number of columns of A to equal the number of rows of C. Nothing here guarantees that. In fact, AC exists only if B is square, but that is not guaranteed by the given. (c) AB is never equal to BA. Solution: FALSE. In Homework 4, we found that diagonal square matrices with the exact same entry on the diagonal are commutative with all square matrices of the same size. (Alright, we only proved it for 2 2 matrices, but you can see it would always work. Right?) Another counter-example would be if A or B was the identity matrix. There are more. (d) If the element a ij of a square matrix A lies below the diagonal, the i > j. Solution: TRUE. This is just a notational question. In a ij, i is the row and j is the column. If an entry is on the diagonal, i = j. If an entry is below the diagonal, the row number i must be greater than the row number j. (e) If A and B are matrices such that AB = I n for some n, then both A and B are invertible. Solution: FALSE. This is only true if A and B are square. (f) If B and C are inverses of a matrix A, then B = C. Solution: TRUE. Inverses are unique. (g) If a square matrix has a column consisting of all zeros, then it is not invertible. Solution: TRUE. By theorem, we know that if a square matrix s determinant is zero, then the matrix is not invertible. If a row or column of a square matrix is all zeros, we know that the matrix s determinant is also zero. Hence, the matrix is not invertible.

9 Math 125 Exam 2 Version 2 October 23, points possible a b c 1. (5pts) Assume det d e f = 6. Find the determinant of g h i Fully justify your answer. Solution: Let A = a b c d e f g h i. a b c R 2 R 3 A 1 = g h i, d e f 2a 2b 2c g h i. 2d + 3a 2e + 3b 2f + 3c det A 1 = det A = (6) = 6 a b c R 2 = R 2 A 2 = g h i, det A 2 = det A 1 = ( 6) = 6 d e f a b c R 3 = 2R 3 A 3 = g h i, det A 3 = 2 det A 2 = 2(6) = 12 2d 2e 2f a b c R 3 = R 3 3R 1 A 4 = g h i, det A 4 = det A 3 = 12 2d 3a 2e 3b 2f 3c 2a 2b 2c R 1 = 2R 1 A 5 = g h i, 2d 3a 2e 3b 2f 3c det A 5 = 2 det A 4 = 2(12) = 24

10 Let A = 1 0 c (a) (3pts) Compute the (2, 3)-minor of the matrix A. Solution: M 23 = = 1 1 ( 1) 2 = 3 (b) (2pts) What does the determinant of a matrix tell us about its invertibility? Solution: Theorem: If the determinant of any square matrix A is non-zero, then A 1 exists. Later, we learned that this is actually and if and only if statement. (This is the theorem at the top of page 116 in the notes.) Due to how we motivated the concept of the determinant, it is also correct to say: Theorem: If the determinant of any square matrix A is zero, then A is not invertible. (c) (5pts) For what scalar c is the matrix A not invertible? Solution: Using the theorem from part (b), we need to take the determinant of A. I chose to do the cofactor expansion across the second row in order to take advantage of the zero and our work from part (a). det A = a 21 M 21 + a 22 M 22 a 23 M 23 = ( 1)M 21 + (0)M 22 cm 23 = c = 1(7) 2(1) 3c = 9 3c The det A = 0 when 9 3c = 0, or when c = 3. The matrix A is not invertible when c = 3.

11 3. Consider the following linear system x 2y + 2z = a x + y + 3z = b x y 4z = c (a) (5pts) Write the above linear system as a matrix equation x a Solution: y = b z c (b) (2pts) Find the inverse needed to solve the linear system Solution: Using the calculator, = (c) (3pts) Solve the linear system. Leave your answer in terms of the parameters a, b, and c. x a a 10b 8c Solution: y = b = a 6b 5c. z c b c

12 4. A researcher is studying the population flows between metropolitan areas and nonmetropolitan areas (suburban and rural areas) in the United States. Based on collected data, each year 11% of the metropolitan population moves to the suburbs and rural areas and 8% of the non-metropolitan residents move to the cities. (a) (3pts) Construct the transition matrix defined by this model. Clearly label the matrix. Solution: T = [ metro non-metro ] to metro to non-metro (b) (1pt) What is the probability that someone who lives in the metropolitan area now will live in the metropolitan area next year? Solution: This is another way to interpret the transition matrix. The a 11 entry of T says that the probability that someone who lives in the metropolitan area will stay in the metropolitan area next year is 89%. (c) (3pts) Approximate the stable matrix (to two decimal places) associated with this Markov process and determine how long it will take the population flow to stabilize. 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 2 decimal places) stops changing for higher powers of the transition matrix. [ ] T 23 = [ T 24 = [ T 25 = Hence, it takes 24 years for the market to stabilize. Recall that an alternative way to determine when the market stabilizes is to find the first time period that all of the columns of T k are the same to two decimal places. From the above, we see that this happens when T is raised to the 24 th power. Hence, it takes 24 years for the market to stabilize. (d) (3pts) What will the population equilibrium be when the flow stabilizes? (Again, two decimal place accuracy is all that is needed.) Solution: [ ] Recall that the columns of the stable matrix are the stable vector. Therefore,.42 S =. (You can use the stability requirement (T I) S.58 = 0 and the fact that S is stochastic, but that is a whole lot more work.) Interpreting the stable vector, we see that our model predicts a population equilibrium of 42% of the population living in the metropolitan area and 58% living in the non-metropolitan areas. ]. ].

13 5. Consider an economy that is divided into three sectors: agriculture, manufacturing, and services. Suppose that each dollar s worth of agricultural requires $0.20 form the agricultural sector, $0.25 from the manufacturing sector, and $0.30 from the services sector; each dollar s worth of manufacturing requires $0.35 form the agricultural sector, $0.30 from the manufacturing sector, and $0.40 from the services sector; and each dollar s worth of services requires $0.10 form the agricultural sector, $0.15 from the manufacturing sector, and $0.30 from the services sector. (a) (3pts) Construct the consumption matrix for this model. Clearly label the matrix. Solution: C = agriculture manufacturing services agriculture used manufacturing used services used (b) (2pts) Are any of the industries profitable? Clearly explain your answer. Solution: The agriculture and service industries are profitable since both of the columns sum to less than 1. The manufacturing industry is not profitable since 1.05 > 1. (c) (3pts) Define what it means for an economy to be productive. Solution: An economy is productive if given any external demand ( D), there is a production schedule ( X) that can meet the demand. Mathematically, there is an X such that X = [I C] 1 D has all positive entries for any demand vector D. Either answer is acceptable. (d) (3pts) Clearly explain why this economy is productive. Solution: The only way to show this economy is productive is to show that [I C] 1 is a positive matrix. (We already showed that the the service industry is not profitable and the last row of C sums to 1.) 215/ / /472 [I C] 1 = 55/59 265/ / / / /472 (e) (3pts) If the economy s current production schedule is $90 million of agriculture, $72 million of manufacturing, and $75 million of services, are there any resources leftover for exporting? If so, how much? Solution: Here, we are given a production schedule X 90 = 72 and are asked to interpret 75 the D in the Leontiff model (I C)

14 Interpreting this in the context of the problem, given the current production schedule, the economy is producing a surplus of $39.3 million in agriculture and $16.7 million in manufacturing and a shortfall of $3.3 million of services. Therefore, there is an excess of $39.3 million in agriculture and $16.7 million in manufacturing to export. (f) (1 pt) If the economy s current production schedule is $90 million of agriculture, $72 million of manufacturing, and $75 million of services, is the economy currently self-sustaining or must it purchase external resources? If the latter, how much of each resource must it purchase? Solution: The economy is not currently self-sustaining since, given the current production schedule, there is a shortfall of $3.3 million in services. This economy will need to purchase these services from an external source.

15 6. (a) (3pts) State three of the properties of matrix multiplication. Solution: From page 72 of the notes: Theorem: The Properties of Matrix Multiplication. 1. associative: A(BC) = (AB)C 2. distributive property from the left: A(B + C) = AB + AC 3. distributive property from the right: (A + B)C = AC + BC 4. r(ab) = (ra)b = A(rB) 5. I m A = A = AI n (b) (2pts) Simplify the matrix expression (A + B) 2 2A 2 3AB + A(3B) B 2. Solution: (A + B) 2 2A 2 3AB + A(3B) B 2 = (A + B)(A + B) 2A 2 3AB + A(3B) B 2 = (A + B)(A + B) 2A 2 3AB + 3AB B 2 = AA + AB + BA + BB 2A 2 B 2 = A 2 + AB + BA + B 2 2A 2 B 2 = A 2 + AB + BA

16 7. (5pts) Determine if the following statement is true or false for a square matrix A. 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) If A is invertible, A 2 is invertible. Solution: TRUE. Since A is invertible, A 1 exists. Recall that A 2 = AA. By properties of the inverse, we know that (A 2 ) 1 = (AA) 1 = A 1 A 1. Therefore, the inverse of A 2 exists and is (A 1 ) 2. (b) If A has a zero on the main diagonal, it is not invertible. Solution: FALSE. While it is true that if det A = 0, then A is not invertible, simply having a zero on the diagonal does not imply that det A = 0. (This would not be the case if we knew that A was diagonal or even triangular. If A was triangular and had a zero on the main diagonal, it s determinant would be zero and, hence, not invertible.) (c) Every square matrix is invertible. Solution: FALSE. There are many non-invertible square matrices. For example, a square matrix of all zeros. (d) Invertible matrices are square. Solution: TRUE. Since the definition of an inverse is a matrix such that AB = I and BA = I, to have the both products AB and BA make sense, both A and B have to be square and of the same dimensions. (e) If A and B are matrices such that AB = I n for some n, then both A and B are invertible. Solution: TRUE. From the theorem Properties of the Inverse on page 80 in the notes, we know that if both A and B are square and AB = I n, then BA = I n. Since A is n n and I is n n, the properties of matrix multiplication require that B be n n. Therefore, B is A s inverse and vice versa. (f) If AB = AC, then B = C. Solution: FALSE. Again, from the theorem Properties of the Inverse on page 80 in the notes, we know that this is only true if A is invertible. (g) If A has a row consisting of all zeros, then it is not invertible. Solution: TRUE. By theorem, we know that if a square matrix s determinant is zero, then the matrix is not invertible. If a row or column of a square matrix is all zeros, we know that the matrix s determinant is also zero. Hence, the matrix is not invertible.