Practice Workbook Answers

Practice Workbook Answersb. n 126. a. 1, 30 c.5. a. 2z 2w 32y 2w 12y 2z 0i 1(continued)b. 2z 2w 1 2y 2w 32y 2z 01c. y 12 , z 12 , w 1d. x 0, y 12 , z 12 , w 1 i6. a 2, b 1, c 37. There is no solution to the system.The equations represent two linesthat are parallel and therefore neverintersect.7. a. The magnitude of each solutionis 1. The directions goingcounterclockwise are 60 , 120 ,180 , 240 , 300 , and 360 .b. {1, {i, { a 12 1 "23 b , { a 12 2 "23 bLesson 4.3 Additional Practice1. a8.i221 22 28 x ba b a b2 21 27 y332. a22 1026 x ba b a b26 y30213x21 3 11 y 21 3. 5 226 22 22 16z211182161x224. 1 23 22 3 216 y 2 7510z4 i5. 7, 3, and 11Chapter 46. 2x 3y 6z 5x 3y z 06x 2y z 2Lesson 4.2 Additional Practice1. x 2, y 02117. 332. x 2, y 2, z 33. x 0, y 0, z 04. x 1, y 4, z 2, w 3CME Project Algebra 2 Teaching Resources Pearson Education, Inc. All rights reserved.165

Practice Workbook Answers8. a. The reduced form is3q328 4r3 Sa04 2211111 21114. a. 11 211111 218 4 b.0 0The matrix on the right siderepresents this system.3x 8y 40 0Since 0 0 is always true, thesolutions to the system are allordered pairs (x, y) on the line3x 8y 4. There are an infinitenumber of values that solve it.211bb. a1 212111c. 1 211 11 215. z 23x6. z 14xIn terms of Algebra 1, the twooriginal equations are the sameline and therefore have an infinitenumber of solutions.b. The reduced form is7. 48. a. 938b. The calculation is exactly parallelto Exercise 7. Let z spending indollars and y1 value in dollars ofthe pound, y2 value in dollars ofthe euro, and y3 value in dollarsof the Swiss franc.Then z 200y1 320y2 275y3and y1 1.75, y2 1.15, andy3 0.80. Solve for total spendingin dollars by substituting y1, y2,and y3 into the equation for z.226 100 11 bS a ab.1 23 51 235The matrix on the right siderepresents this system.0 11x 3y 5Since 0 11 is always false, thesystem has no solutions.In terms of Algebra 1, the twooriginal equations are parallellines. Since they do not intersect,there are no solutions.Lessons 4.6 and 4.7Additional PracticeLesson 4.5 Additional Practice1. a. 2 3b. 3c. 6d. 5e. 23 42. a. 9 10 19 20b. a3. a(continued)1. a. 44c. x ze. 0b. undefinedd. 83f. 02. a. ( 2, 3)c. (1, 0, 1)b. ( a, b)d. (0, 0, 0, 0)3. a. Let A Aa1, a 2, . . . , an B andB Ab1, b2, . . . , bnB .A B a1b1 a 2b2 · · · anbnB A b1a1 b2a 2 · · · bnan3 4 5 6b9 10 11 12Since real-number multiplicationcommutes, a1b1 b1a1,a 2b2 b2a 2, and so on. Therefore,A B B A.1 2 3 4b1 4 9 16CME Project Algebra 2 Teaching Resources Pearson Education, Inc. All rights reserved.166

Practice Workbook Answersb. Let A Aa1, a 2, . . . , an B ,B Ab1, b2, . . . , bn B , andC Ac1, c 2, . . . , cn B .A (B C) Aa1, a 2, . . . , an B Ab1 c1, b2 c2,.,b c Bnn a1 Ab1 c1B a 2 Ab2 c 2B . . . an Abn cn B Aa1b1 a1c1B Aa 2b2 a 2c 2B . . . Aanbn ancn B [Distr. Prop.for real numbers] A B A CSimilarly,(A B) C Aa1 b1, a 2 b2, . . . , an bn B Ac1, c 2, . . . , cn B Aa1 b1B c1 Aa 2 b2B c 2 . . . Aa n bn B cn Aa1c1 b1c1B Aa 2c 2 b2c 2B . . . Aa ncn bncn B[Distr. Prop. for real numbers] A C B C21 04. a. ab23 226 6c. ab26 6e. ( 15 15)b. a(continued)b. A: 47 students, B: 68 students,C: 64 students, D: 22 students,F: 6 students;(1 1 1)M (47 68 64 22 6)c. 207 students; adding the resultsin part (a) will give the totalenrollment. The same is true forthe results in part (b).68 71 68 207 or47 68 64 22 6 207.In matrix form this is1111((1 1 1)M) 1 (1 1 1)(M) 1 .1111Lessons 4.8 and 4.9Additional Practice1. a. Graph the system of equations2x 3y 8 .x 2y 35 26b3 24y2d. ( 12 10) 4O246x 215295. AB 212 , AC 10 ,129 415 212AD 2128 14The graphs intersect at the point(1, 2), so the solution is x 1and y 2.b. Solving by Gaussian Eliminationgives6. a. 68 sixth grade students, 71 seventhgrade students, 68 eighth gradestudents;a1168M 1 71 16812 23 81 bSa12 2300 1b.1 22The solution is x 1 and y 2.CME Project Algebra 2 Teaching Resources Pearson Education, Inc. All rights reserved.167

Practice Workbook Answersc.2 37 7The inverse is M 1 2 .27 7Lesson 4.11 Additional Practice1. a. The map dilates the plane by thefactor 2 in the x-direction and thefactor 4 in the y-direction.b. The map dilates the plane by thefactor 12 in the x-direction andthe factor 14 in the y-direction;Solve by the inverse method.a2 23 x8ba b a b1 2 y2327377723777 782 23 xb a b 1 2 a b232 y2 1 2 a12(continued)12 1A 0 .0 141 0 x1aba b a b0 1 y22The solution is x 1 and y 2.221.52. ab1 20.5y221.53. ab1 20.5X14X24. x 1 2, x 2 1, x 3 32c2c 2d5. a. A a bb. A abcc d 42A 1X2 2O 26. Yes; by definition of a matrix sum, theij entry of A A A is aij aij aij.By real-number algebra, this is 3aij.By definition of scalar multiplication,3aij is the ij entry of 3A. SoA A A 3A, since the two sidesagree entry by entry. Alternately,A A A 1A 1A 1A (1 1 1)A 3A,using the basic rules of matrices.A 1X142x6A 1X3 4X32. a. A maps (x, y) ] ( x, y). It rotatesevery point 180 counterclockwisearound the origin.b. B maps (x, y) ] (0, y). It projectsevery point horizontally onto they-axis.c. C maps (x, y) ] (0.25x, 0.25y).It scales every point by the factor0.25.7. Yes; expand the left side of theequation.(X I)2 (X I )(X I)[def. (X – Y )2] X(X I ) I(X I)[Distr. Prop.] (XX XI) ( IX II) [Distr. Prop.] X 2 XI IX I 2[def. X 2]22 X (XI IX) I[Assoc. Prop.]22 X 2XI I[I commutes.]3. a. image of L1: x 1, image ofL 2: x 0, image of L 3: y 2,image of L 4: y 3, image ofL 5: y x, image of L 6: y x 1b. image of L1: x 0, image ofL 2: x 0, image of L 3: (0, 2), imageof L 4: (0, 3), image of L 5: x 0,image of L 6: x 0c. image of L1–L 6: y x1 0b0 1a bb. Any matrix of the form abc dcommutes under multiplication1b.with a18. a. yes, a4. a. Q 13 , 1Rb. ( 10, 0)x2 1 x2ba b a b5. a b ] ay5 0 y1CME Project Algebra 2 Teaching Resources Pearson Education, Inc. All rights reserved.168c. (8, 6)

Practice Workbook Answers6. a. 360 rotation counterclockwiseb. (x, y) ] ( 3x, 3y)c. (x, y) ] (x, y)d. (x, y) ] (3x, 3y)(continued)a(n)b. Let T(n) b(n) M nT(0) andc(n)50T(0) 50 . After two weeks,50Lessons 4.12 and 4.13Additional Practice53.25T(2) 43.875 . After three weeks,52.875200,000rqif n 0200,0001. Let P(n) ,MP(n 1) if n 053.5T(3) 43.65 . In the long run,52.8553.63the distribution settles at 43.63 .52.73NSN 0.96 0.06b.where M aS 0.04 0.940.6 0.6In the long run, M ab, and0.4 0.43. a. Answers may vary. Checkstudents’ work.b. X (312.5, 187.5)0.6 0.6 200,000P( ) abab0.4 0.4 200,000240,000b. This means that a160,000the population distribution becomesstable at 240,000 people living northof downtown and 160,000 peopleliving south of downtown. The peoplewho live in each area may change, butthe population will remain stable.MM 04. G J 12E 1212JE000 121a. 1261 0.016b. Q 12 R 642. a. At the end of each week, of thetrucks that start at location A,60% return there, 30% are at B,and 10% are at C. Of the trucksthat start at location B, 25% areat A, 45% return to B, and 30%are at C. Of the trucks that start atlocation C, 20% are at A, 15% areat B, and 65% return to C;ABC1 0.031c. 325. a. 181b. 241c. 966. a. 3b. 24c. 72d. 131e. 24Chapter 5A 0.6 0.25 0.2M B 0.3 0.45 0.15 .C 0.1 0.3 0.65Lessons 5.2 and 5.3Additional Practice1. a. 10,00064c. 125CME Project Algebra 2 Teaching Resources Pearson Education, Inc. All rights reserved.169b. 60,466,176d. 1

L2: x 0, image of L3: y 2, image of L4: y 3, image of L5: y x, image of L6: y x 1 b. image of L1: x 0, image of L2: x 0, image of L3: (0, 2), image of L4: (0, 3), image of L5: x 0, image of L6: x 0 c. image of L1– 6: y x 4. a. Q1 3, 1R b. ( 10, 0) c. (8, 6) 5. a x y b] a 21 50 ba x b a 2 1 b 4 2 O 46 2 4 2 2 4 y x A 1X2 A 1X1 A 1X 3 X1 X2 X3