Chapter 3 Answers. Lesson 3.1

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1 Chapter 3 Answers Lesson 31 1 A 2 5 matrix: any matrix with 2 rows and 5 columns A row matrix with two elements: any matrix with one row of two elements A square matrix: any matrix in which the number of rows equals the number of columns 2 16 elements; 7 elements; mn elements 3 Jacket Shirt Pants Shop Shop Shop a A 21 = $109, A 12 = $1086, A 32 = $389 b A 21 represents the cost of drinks at Vin s A 12 represents the cost of pizza at Toni s A 32 represents the cost of salad at Toni s 5 S 3 represents the cost of pizza at Sal s 6 a 7 8 Cereal Banana N = Milk Toast b N 23 = 26, N 32 = 5, N 42 = 6 Cals g Fat g Carbo mg Chol c N 23 represents the grams of carbohydrates in the banana Pizza Salad N 32 represents the grams of fat in the milk N 42 represents the grams of fat in the toast Vin's Toni's Sal's $1010 $1086 $1065 $369 $389 $385 Vin's Toni's Sal's Add topping $115 $110 $125 Add dressing $000 $045 $050 Chapter 3 Answers 1

2 9 A + B = C = Pizza Salad Vin's Toni's Sal's $1125 $1196 $1190 $369 $434 $ a A 21 = $369, B 21 = $000, C 21 = $ a 12 b A 21 represents the cost of a salad at Vin s B 21 represents the cost of an additional salad dressing at Vin s C 21 represents the cost of a salad with a choice of two dressings at Vin s b It is not possible to add these matrices because their orders are not the same c d It is not possible to add these matrices because their orders are not the same J Reyes R Braun M Kemp At Bats Runs Hits HRs RBIs Avg A decrease in statistics is shown with a negative sign 13 a Matrix representing increases in times: 100-meter race 200-meter race 400-meter race Men Women b The greatest decrease in times is in the 400-meter race for women The smallest decrease is in the 100-meter race for men Chapter 3 Answers 2

3 14 a Answers vary Possible answer: Yes In matrix addition we add the components of the matrices that are in corresponding positions These values are real numbers and since addition of real numbers is both commutative and associative, it follows that addition of matrices is also commutative and associative b Commutative property: A + B = B + A = Associative property: A + (B + C) = = (A + B) + C = = Answers vary Possible answer: No, subtraction of matrices is neither commutative nor associative This follows since the commutative and associative properties do not hold for subtraction of real numbers 16 a To verify that the commutative property does not hold for the given matrices A and B: A B = B A = To verify that the associative property does not hold for the given matrices A and B: A (B C) = = (A B) C = = A + O = = = A O + A = = = A A A = = = O Chapter 3 Answers 3

4 b A + ( A) = = = O ( A) + A = = = O 17 a They are all zeros b Yes, A 11 = A 11, A 12 = A 21, A 13 = A 31, A 14 = A 41, A 15 = A 51, A 22 = A 22, A 23 = A 32, etc c Sample matrix: d No If the matrix is symmetric then the number of rows must equal the number of columns In a matrix that is symmetric the entry Cij will be the same as the entry Cji 18 a The 1s along the diagonal represent the relationship of a variable with itself (Answers may vary) b Answers vary The values are symmetric because they represent the relationship between variables The relationship between ACT math and college GPA, for example, must be the same as the relationship between college GPA and ACT math These values are in row 1 column 6 and in row 6 column 1 c ACT composite d ACT math Lesson 32 1 a T represents the cost of four pizzas with additional toppings and four salads with a choice of two dressings from each of the three pizza houses b $4760 c T 12 represents the cost of four pizzas at Toni s T 21 represents the cost four salads at Vin s 2 a and b J = Pearl Jade c 24 jade necklaces e p n b d J 21 represents the number of jade earrings that the jeweler expects to sell in June J 12 represents the number of pearl pins that the jeweler expects to sell in June Chapter 3 Answers 4

5 3 4 a V = b c Vitamin A Vitamin C Vitamin D Resident Nonresident Resident Nonresident Resident Nonresident % % 25 Vitamin A V = Vitamin C Vitamin D 30 Undergrad Grad $7500 $9925 $20400 $24525 Undergrad Grad $8025 $10620 $21828 $26242 Undergrad Grad $525 $695 $1428 $ a QP is defined Order of QP: Chapter 3 Answers 5 b QP is not defined because the number of entries in the row matrix Q does not equal the number of rows in P c QP is defined Order of QP: 1 4 d QP is not defined because the number of entries in the row matrix Q does not equal the number of rows in P e QP is defined Order of QP: 1 1 Loans Bonds [ ] Rate MA NE CA [ ] = a Two possible answers: Loans Bonds Q = no [ ] P = Cost $300 $850 $950 MA NE CA

6 or P = Cost [$300 $850 $950] no Q = b No Answers vary A possible answer is that the row and column matrices may be written in the two different ways shown in part a c 8 a [10 4 2] 850 = [10(300) + 4(850) + 2(950)] A = b BA = [ ] c BC = [20] d 2C = 4 8 CD CU Bond [ ] Rate = [ ] = [8300] or $8300 CD CU Bonds Amt Invest $10, , 000 = $2, , a The transpose of a row matrix is a column matrix b The transpose of a column matrix is a row matrix c M T = e p n b p j Chapter 3 Answers 6

7 11 a 12 b c 13 a e p n b [ ] Hours Hours [ ] p j Hours [ ] e p n b p j = [ ] d It takes the jeweler 425 hours to make the pearl jewelry and 935 hours to make the jade jewelry trn car pl tk [ ] $ C = Train Car Plane Truck Calendars Mon 10 Tues 15 Wed 20 Thurs 30 Fri 50 b N = [ ] N C = [ ] N S E = [125] N S E [ ] = dollars 1, Chapter 3 Answers 7

8 c A = [1/5 1/5 1/5 1/5 1/5] or A = [ ] A C = [ ] Lesson 33 1 a AB = = [25] b BA is not defined because the number of columns of A is not equal to the number of rows of B c d 2 a 1 2 BC = CB = Mike Liz Kate Mike Liz Kate $261,000 $0 $250 $235,000 $0 $215 $255,000 $0 $325 b The entries in row 1 represent the value to Mike of the items that he, Liz, and Kate received Row 2 represents the values to Liz and row 3 the values to Kate 3 a Matrix Q: Rosa Max Burger Special Potato Fries Shake Chapter 3 Answers 8

9 b Matrix C: Burger Special Potato Fries Shake Cal Fat Chol c The dimensions of Q are 2 by 5, and those of C are 5 by 3 d The dimensions of the product QC are 2 by 3 e The dimensions of C can be described as Foods by Contents, and the dimensions of Q times C can be described as Persons by Contents f R = Rosa Max Cal Fat Chol 1, , g R 12 represents the total grams of fat in Rosa's food R 21 represents the total number of calories in Max's food R 23 represents the mg of cholesterol in Max's food 4 a The number of columns of A must equal the number of rows of B b The number of columns of A must equal the number of rows of B, and the number of columns of B must equal the number of rows of A Chapter 3 Answers 9

10 c Answers vary A possible answer follows A = B = AB = BA = = = AB BA The dimensions of AB are not equal to the dimensions of BA d Students may use trial and error to find a solution Another method is to examine the algebra involved as follows: Suppose we let A = and B = a c a c b d = a c b d b d Then if AB = BA, we have or 3a + c 3b + d 3a + 6b 6a + 5c 6b + 5d = a 3c + 6d + 5b c + 5d Since these two matrices are equal, we have a system of 4 equations in four unknowns 3a + c = 3a + 6b 3b + d = a + 5d 6a + 5c = 3c + 6d 6b + 5d = c + 5d Then AB = BA = Notice that the first and fourth equations are equivalent, leaving three equations in four unknowns There is an infinite number of solutions for the system If we let a = 2 and b = 1, then c = 6 and d = 4 will be one solution = = Students will also discover in Exercise 8 that if A and B are inverses of each other then AB = BA 5 Students should show work for IA = AI = A for any 3 3 matrix A they choose A possible answer follows: Let A = Then AI = and IA = = = Chapter 3 Answers 10

11 6 a Yes A(BC) = (AB)C b A possible answer follows ( ) = A BC = = ( AB)C = = = c Yes Reasons will vary A possible answer follows: The matrix multiplication depends on the associative properties for multiplication and addition of real numbers 7 Answers vary One possibility for matrices A and B follows: A = B = ( A + B) ( A B) = 2 2 A 2 B 2 = = a Students show work for AB = BA = I AB = BA = = = = = b Suppose matrix C does have an inverse Then there exists a matrix D where CD = I Suppose D = x z y w Then x y z w = Performing the matrix multiplication gives 2x + 4y 3x + 6y 2z + 4w 3z + 6w = Since the two matrices are equal, we have a system of four equations in four unknowns Chapter 3 Answers 11

12 2x + 4y = 1 2z + 4w = 0 3x + 6y = 0 3z + 6w = 1 Attempting to solve the system for x and y shows that the first and third equations are inconsistent and that there is no solution Since there is no solution for x and y, matrix D does not have an inverse Another way to show that the system does not have a solution is to rewrite the equations in slope-intercept form and to use the graphing calculator The lines are parallel since their slopes are equal It follows that there is no solution for the system of equations since the lines do not intersect 9 The diagram below shows the polygons plotted for parts a through h b T 1 P = c Polygon A'B'C'D' is the reflection of polygon ABCD in the y-axis Chapter 3 Answers 12

13 d T 2 P = e Polygon A"B"C"D" is the reflection of polygon A'B'C'D' in the x-axis f g h 1 0 T 2 T = R RP = The effect of R on P is to rotate ABCD 180 degrees about the origin T 3 = T 3 P" = T 4 = = T T 3 2 T 4 P' = a T: Bag by Work W: Place by Work P: Place by Bag D: Bag by Month b In this exercise we want the result to be a Work-by-Month matrix We can obtain this result by multiplying a Work-by-Bag matrix times a Bag-by- Month matrix We do this by multiplying the transpose of T times D Cutting T T D = Stitching Finishing May June 880 1,180 1,400 1, ,180 c We want a Cost-by-Place matrix We achieve this by multiplying a matrix representing the hours worked on each bag times a matrix representing the wages paid per hour in each place This translates into multiplying a Bag-by- Work matrix times a Work-by-Place matrix We do this by multiplying T times the transpose of W Chapter 3 Answers 13

14 d Handbag TW T = Standard Roomy May [ ] NY NE CA H S R = May [ $24, 840] e In this exercise, we want a matrix that represents hours of Work by Place We can achieve this by multiplying a Work-by-Bag matrix times a Bag-by-Place matrix One way to do this is to multiply the transpose of T times the transpose of P Cutting T T P T = Stitching Finishing NY NE CA a The figure below shows the result of rotating ABCD through 30, 60, 90, 180, and 90 b = c The polygon A'B'C'D' is the result of rotating polygon ABCD 30 counterclockwise For 180, use the transformation matrix: cos180 sin sin180 cos = Chapter 3 Answers 14

15 d For 60, use the transformation matrix e 12 a cos60 sin60 sin 60 cos = Yes, this transformation matrix has the same effect as applying T 1 twice ( T 1 ) 2 = 90 : 90 : = = = , = These two matrices are inverses of each other The computations for 60 and 60 are as follows = = A = A 2 = A 3 = A 4 = b Answers vary Students may notice, for example, that in any power the entry in the first row third column is the sum of the entries in the first row in the previous power They should also be able to explain why any statements that they make are true c A 5 = A 6 = A 7 = Chapter 3 Answers 15

16 A 8 = A 9 = d A conjecture that can be proved with mathematical induction is that for any natural number n, the nth power of the matrix has the form 1 n n A n = 0 1 n This is clearly true for n = A 1 = Assume true for n = k 1 k k A k = 0 1 k Now, we need to show the statement is true for n = k + 1, that is, that 1 k k + 1 A k +1 = 0 1 k But, A k +1 = AA k = k k 0 1 k k k + k k k + 1 = 0 1 k + 1 = 0 1 k Therefore the conjecture is true for all natural numbers n n 1 Note: Students may also know that k = k =1 2 n( n + 1) So, another conjecture that they could prove using mathematical induction is that 1 1 n 2 n( n + 1) A n = 0 1 n Chapter 3 Answers 16

17 13 a Let A =, B = 0 2 2, C = 0 3 3, and D = Then B 2 = = = = 4A 2 = 2 2 A B 3 = = = = 8A 3 = 2 3 A B 4 = = = = 16A 4 = 2 4 A C 2 = = = = 9A 2 = 3 2 A C 3 = = = = 27A 3 = 3 3 A C 4 = = = = 81A 4 = 3 4 A D 2 = = = = 16A 2 = 4 2 A D 3 = = = = 64A 3 = 4 3 A , 024 2, D 4 = , 024 = = b and c Conjecture: For any natural numbers m and n 4 = 256A 4 = 4 4 A 4 Chapter 3 Answers 17

18 m m m 0 m m 0 0 m n = m n This is clearly true for n = 1 m m m m m = m 0 0 m Assume true for n = k m m m 0 m m 0 0 m k = m k n k Now, we need to show that this is true for n = k + 1, that is, m m m 0 m m 0 0 m k = m k k +1 But m m m 0 m m 0 0 m k +1 m m m = 0 m m 0 0 m m m m 0 m m 0 0 m k = m m k k = m k Therefore, the conjecture is true for all natural numbers m and n Lesson 34 1 a 1897 newborn rats b (166)(06) = 996 move up to the 3 6 age group (9)(09) = 81 move up to the 6 9 age group (81)(09) = 729 move up to the 9 12 age group (117)(08) = 936 move up to the age group (4)(06) = 24 move up to the age group c Distribution after 6 months: 1897, 996, 81, 729, 936, 24; Total 56 rats k +1 Chapter 3 Answers 18

19 d Distribution after 9 months: 1832, 1138, 896, 729, 583, 562; Total 57 rats Distribution after 12 months: 1802, 1099, 1024, 806, 583, 35; Total 57 rats e Answers may vary Possible answer: The population continues to grow The rate of growth seems to have slowed f Answers may vary Possible answer: The population growth may continue to slow or even become constant 2 a 50(0) + 30(08) + 24(17) + 24(17) + 12(08) + 8(04) = 1184 newborn deer b (50)(06) = 300 move up to the 2 4 age group (30)(08) = 240 move up to the 4 6 age group (24)(09) = 216 move up to the 6 8 age group (24)(09) = 216 move up to the 8 10 age group (12)(07) = 84 move up to the age group (8)(0) = 0 No deer lives beyond 12 months 3 c The product is the number of newborn deer after 1 cycle a [ ] b [ ] c [ ] Chapter 3 Answers 19

20 d [ ] e [ ] a Distribution after 3 months: 0, 21, 0, 0, 0, 0; Total 21 rats b Distribution after 3 months: 11, 3, 45, 45, 4, 3; Total 30 rats [ ] ; Total 15 deer 5 a After one cycle: b After two cycles: [ ] ; Total 24 deer c After three cycles: [ ] ; Total 39 deer d After four cycles: [ ]; Total 52 deer Lesson 35 1 a To use the graphing calculator, store P O = [ ] as matrix [A] and store the Leslie matrix for the Rattus norvegicus as matrix [B] P 5 = P 0 L 5 = [A] [B]^5 = [ ] b Let [C] be a 6-by-1 column matrix containing all 1s Then the total population after 15 months (5 cycles) T = [A] [B]^5 [C] = [5935] or T = [ ] [C] = [ 5935 ] = 5935 rats c P 7 = [A] [B]^7 = [ ] Note: Stop and record the distribution and then press [ ] [C] [ENTER] to find the total The calculator will show: T = 6341 rats Chapter 3 Answers 20

21 2 Use trial and error to find the number of cycles for the female population to reach 250 members The number of years equals the number of cycles divided by 4 since there are three months per cycle Cycles Population Years a b c d a Cycle Total Population Growth Rate original % % % % % % b The population appears to decline then increase again c P 25 = , P 26 = , P 27 = The growth rate in each case is 304% 4 a P 25 = 86054, P 26 = 88674, P 27 = a P 25 = 67098, P 26 = 69142, P 27 = P 25 = 54823, P 26 = 56491, P 27 = P 25 = , P 26 = , P 27 = The long-term growth rate of the total population is 304% in each case b The initial population does not affect the long-term growth rate The Leslie matrix L = b P 29 = 371,147427, P 30 = 485, The long-term growth rate is 3076% Chapter 3 Answers 21

22 6 a c P 7 = 1,01315, P 8 = 1,32788 The female herd size reaches 1,250 between 7 and 8 cycles or about 15 years since one cycle equals 2 years d Approximately 3076% of the animals would need to be removed from the herd each year The Leslie matrix L = b P 0 = [0 0 5] c P 16 = [1, ] There will be 1,225 female bugs living in the basement after 16 weeks Note: To use the iteration function on the graphing calculator: Press [A] [ ] [B] and press [ENTER] Press [ ] [B] and press [ENTER] Continue to press [ENTER] while you count the cycles 7 a 1 Week 2 Weeks 3 Weeks P P 22 5, P , P ,8911 P 25 11, P , P ,9714 P 28 23, P , P ,3399 Answers vary Students may observe that only bugs of the same age group are living during each cycle or that the pattern is determined by the growth rate for the previous age group The number of bugs decreases for 2 cycles and then increases dramatically About one third of any population of Chapter 3 Answers 22

23 8 a newborns survives to 3 weeks The population of newborns doubles every fourth week b The population change in each case is 210% 1 Week 2 Weeks 3 Weeks Total % Growth P P 22 4, ,1871 P 23 3,0258 2, , % P 24 1,5129 1,5129 1,5129 4, % P 25 9, , , % P 26 6,3542 4, , % P 27 3,1771 3,1771 3,1771 9, % P 28 19,0627 1,5886 2, , % P 29 13,3439 9,5314 1, , % P 30 6, , , % Note: The percentage growth is for the total population for each cycle Answers vary Students may observe that the number of bugs in each age group is the same every third cycle They may also notice that the total population growth falls into a pattern of 2400, 10486, and times the previous total population b The growth is 110% in each case 9 The statement is true for k = 1 since P 1 = P 0 L by definition Assume that the statement is true for k = n, that is, that P n = P 0 L n is true Then it is necessary to show that P n+1 = P 0 L n+1 But, P n+1 = P n (L) = P 0 L n (L) = P 0 L n+1 It follows by mathematical induction that P k = P 0 L k for all natural numbers k Review 1 Important points covered in this chapter Matrix terminology and definitions including dimensions of matrices, row and column matrices, square matrix, identity matrix, inverse of a matrix, triangular matrices, and transpose of a matrix How to find the sum and difference of two matrices How to find scalar multiples of matrices How to find the product of two matrices How to use mathematical induction to prove conjectures involving matrices How to use matrices to model real-world situations Chapter 3 Answers 23

24 How to use powers of matrices to model population growth How to use the Leslie matrix model for population growth How to make predictions using powers of matrices 2 a 6 elements 3 a b C 12 = 2; C 21 = A + C = b It is not possible to subtract these matrices because their orders are not the same c d 4 a ( A + C) D = A + D = Mex Chips Salsa Drinks [ ] L = b L 2 = number of bags of chips ordered c 5 a L 4 = number of six-packs of drinks Mex Chips Salsa Drinks [ ] Cost = Number Mex Chips Salsa Drinks = [35($450) + 6($197) + 6($210) + 12($289)] = [$21660] Crystal Springs C = Bear Beaver Lodging Food Rec $1300 $2000 $500 $1250 $1950 $750 $2000 $1800 $000 $4000 $000 $000 $450 $197 $210 $289 Chapter 3 Answers 24

25 6 a b C 22 = $1950 C 43 = $000 c C 13 = cost for recreation at Crystal Lodge b c d C 31 = cost for lodging at Bear Lodge Z-Mart Base Z-Mart Base Z-Mart Base 7 a 3 2 System Cartridge Case $3950 $2450 $850 $4990 $2995 $1250 System Cartridge Case $3555 $2205 $765 $3992 $2396 $1000 System Cartridge Case $395 $245 $085 $998 $599 $250 $3555 $2205 $765 4 $3992 $2396 $1000 = Z-Mart Base b 4 3 System Cartridge Case $14220 $8820 $3060 $15968 $9584 $4000 c Not possible The number of columns of Q does not equal the number of rows of S d a Plate Large Small No [ ] b Plate Large Small Ebony Walnut Rose Maple , 200 1, c Ebony Walnut Rose Maple [ 1,450 11,000 9,150 7,200 ] Chapter 3 Answers 25

26 9 d Plate Large Small [ ] No Plate Large Small Total time = = 41 weeks Tennis Golf Soccer [ ] Return Jazz Symp Orch $ [ ] 11 a 12 b 13 a AB = BA = Tennis Golf Soccer Weeks Dollars 50, , , 000 Weeks [ ] = No = [ $16,225] c CA is not defined The number of columns of C does not equal the number of rows of A d DA + E = [12 10] 4 5 A T = b c M = , M 2 = , M 3 = , M 4 = M 5 = M n = 2n 1 2 n 1 2 n 1 2 n 1 where n is a natural number Chapter 3 Answers 26

27 d The conjecture is true for n = 1 since e M 1 = = = Assume true for n = any natural number k, that is, M k = 2k 1 2 k 1 2 k 1 2 k 1 To complete the proof, we need to show that the conjecture is true for n = k + 1 That is, we must show that M k +1 = 2k 2 k 2 k 2 k M k +1 = MM k = k 1 2 k 1 2 k 1 2 k 1 = 2k k 1 2 k k 1 2 k k 1 2 k k 1 = 2 2k k k k 1 = 2k 2 k 2 k 2 k Therefore the conjecture is true for all natural numbers n a b c M = , M 2 = , M = 26 27, M = M 5 = M n = 3 n 1 3 n where n is a natural number d The conjecture is true for n = 1 since 1 0 M 1 = = Assume true for n = any natural number k, that is, 1 0 M k = 3 k 1 3 k To complete the proof, we need to show the conjecture is true for n = k + 1 That is, we must show that Chapter 3 Answers 27

28 1 0 M k +1 = 3 k k +1 M k +1 = MM k = k k k 1 3 k = k k = Therefore the conjecture is true for all natural numbers n 14 If a square matrix A has an inverse A 1, then the product AA 1 = the identity matrix I, where I is a square matrix with ones along the diagonal and zeros elsewhere 15 a Yes AB = BA = I 16 a b Yes AB = BA = I c No These are not square matrices AB BA I b A B C A B C 10th 11th 12th Male Female Male Female Male Female A = B C 10th 11th 12th A = B C Note: There may be a slight difference in the totals matrix due to rounding 17 a 24 months b L = c 10% d After the 17th cycle, or about 35 years Chapter 3 Answers 28