SYSTEMS OF LINEAR EQUATIONS AND MATRICES
2.1 Solution of Linear Systems by the Echelon Method
Your Turn 1
2312() 341() xy1 xy2 += -=
Use row transformations to eliminate x in equation (2).
2312 3R(2)RR1734 xy y += +-=
Now make the coefficient of the first term in each equation equal to 1.
Back-substitute to solve for
The solution of the system is (3,2).
Your Turn 2
Let x = number of shares of Kohl’s stock y = number of shares of Best Buy stock 188() 52278601() xy1 xy2 += += 212
Substitue 47 y = in equation (1).
47188 141 x x += =
Olinda owns 141 shares of Kohl’s and 47 shares of Best Buy.
Your Turn 3
Since there are 60 utensils with a total weight of 189.6 pounds, the equations of Example 5 become 60() 3.93.63.0189.6() xyz1 xyz2 ++= ++= where x = the number of knives, y = the number of forks, and z = the number of spoons.
Solve for y in terms of the parameter z: 1483 yz =−
Substitute this expression into equation (1). (1483)60xzz +−+=
Then solve for x. 288xz=−
Since both x and y must be nonnegative, we have: 148 3148or,so49 3 288or44 zzz zz ≤≤≤ ≥≥
The largest number of spoons is thus 49 ; the corresponding number of knives is (2)(49)8810 −= , and the number of forks is 148(3)(49)1. −=
2.1 Exercises
In Exercises 1–16 and 19–28, check each solution by substituting it in the original equations of the system. 1. 5() 222() xy1 xy2 += -=
To eliminate x in equation (2), multiply equation (1) by 2 - and add the result to equation (2). The new system is 122 5() 2RRR48.() xy1 y3 += -+-=-
Now make the coefficient of the first term in each row equal 1. To accomplish this, multiply equation (3) by 1 4 .1 22 4 5() RR2() xy1 y4 += -=
Substitute 2 for y in equation (1). 25 3 x y += =
The solution is (3, 2).
2. 49() 35() xy1 xy2 += -=
First use transformation 3 to eliminate the x-term from equation (2). Multiply equation (1) by 3 and add the result to 4 - times equation (2).
122 49() 3R(4)RR77() xy1 y3 += +-=
Now use transformation 2 to make the coefficient of the first term in each equation equal to 1. 1 11 4 1 22 7 19 RR() 44 RR1() xy4 y5 += =
Complete the solution by back-substitution. Substitute 1 for y in equation (4) to get 19 (1) 44 8 2. 4 x x += ==
The solution is (2, 1).
3. 323() 52() xy1 xy2 -=-=
To eliminate x in equation (2), multiply equation (1) by 5- and equation (2) by 3. Add the results. The new system is
122 323() 5R3RR721.() xy1 y3 -=-+=
Now make the coefficient of the first term in each row equal 1. To accomplish this, multiply equation (1) by 1 3 and equation (3) by 1 7 . 1 11 3 1 22 7 2 RR1() 3 RR3() xy4 y5 -==
Back-substitution of 3 for y in equation (4) gives 2 (3)1 3 21 1. x x x -=-==
The solution is (1, 3).
4. 278() 2312() xy1 xy2 +=-+=-
122 278() RRR1020() xy1 y3 +=+=-
Make each leading coefficient equal 1. 1 11 2 1 22 10 7 RR4() 2 RR2() xy4 y5 +==-
Substitute 2 - for y in equation (4). 7 (2)4 2 74 3 x x x +-=-==
The solution is (3, 2 - ).
5. 326() 5210() xy1 xy2 +=-=-
Eliminate x in equation (2) to get the system
122 326() 5R(3)RR160.() xy1 y3 +=+-=
Make the coefficient of the first term in each equation equal 1. 1 11 3 1 22 16 2 RR2() 3 RR0() xy4 y5 +==
Substitute 0 for y in equation (4) to get 2. x =The solution is (2,0).6. 34() 224() xy1 xy2 -+= -=-
Eliminate x in equation (2).
122 34() 2R3RR44() xy1 y3 -+= +-=-
Make the coefficient of the first term in each row equal 1. 1 11 3 1 22 4 14 RR() 33 RR1() xy4 y5 --=-=
Back-substitution of 1 for y in equation (4) gives 14 (1) 33 14 33 1. x x x -=-==-
The solution is ( 1,- 1).
7. 624() 348() xy1 xy2 -=+=
Eliminate x in equation (2). 122 624() 1R2RR1020() xy1 y3 -=-+=
Make the coefficient of the first term in each row equal 1.
Substitute 2 for y in equation (4) to get x = 0. The solution is (0, 2).
8. 431 253 mn mn +=+= 122 431 R(2)RR77 mn n +=+--=-
Make each leading coefficient equal 1. 1 11 4 1
Back-substitution
The solution is (1,1). -
9. 5117() 3825() pq1 pq2 +=-=
Eliminate p in equation (2).
122 5117() 3R5RR73146() pq1 q3 +=-+-=
Make the coefficient of the first term in each row equal 1. 1 11 5 1 22 73 117 RR() 55 RR1() pq4 q5 +=-=-
Substitute 2 - for q in equation (4) to get p = 3. The solution is (3, 2 - ).
10. 1259 3818 st st -= -=122 1 11 12 1 22 27 1259 R(4)RR2781 53 RR 124 RR3 st t st t -= +-= -= =
Back substitution gives 53 (3) 124 53 44 8 2. 4 s s s -= -= ==
The solution is (2, 3).
11. 672() 7626() xy1 xy2 +=-=
Eliminate x in equation (2).
122 672() 7R(6)RR85170() xy1 y3 +=+-=-
Make the coefficient of the first term in each equation equal 1. 1 11 6 1 22 85 71 RR() 63 () RR2 xy4 y5 +==-
Substitute 2 - for y in equation (4) to get x = 2. The solution is (2, 2 - ).
12. 3814() 22() ab1 ab2 -= -=
Eliminate a in equation (2).
122 3814() R(3)RR28() ab1 b3 -= +--=
Make the coefficient of the first term in each row equal 1. 1 11 3 1 22 2 814 RR() 33 RR4() ab4 b5 -= -=-
Back substitute 4 - for b in equation (4) to get a = 6.- The solution is (6,4).
13. 325() 648() xy1 xy2 += +=
Eliminate x in equation (2).
122 325() 2RRR02() xy1 3 += -+=-
Equation (3) is a false statement. The system is inconsistent and has no solution.
14. 951 18101 xy xy -= -+=
122 951 2RRR03 xy-= +=
The equation 03 = is a false statement, which indicates that the system is inconsistent and has no solution.
15. 324() 648() xy1 xy2 -=-+=
Eliminate x in equation (2).
122 324() 2RRR00() xy1 3 -=+=
The true statement in equation (3) indicates that there are an infinite number of solutions for the system. Solve equation (1) for x. 324()
324 24 () 3 xy1 xy y x4 -==-=
For each value of y, equation (4) indicates that 24 3 , xy= and all ordered pairs of the form () 24 3 , yy - are solutions.
16. 3520 91560 xy xy ++= ++=
Begin by rewriting the equations in standard form, 122 352 9156 352 3R(1)RR00 xy xy xy +=+=+=+-=
The true statement, 00, = shows that the two equations have the same graph, which means that there are an infinite number of solutions for the system. All ordered pairs that satisfy the equation 352 xy+=- are solutions. Solve this equation for x.
352 52 3 xy y x =-=
The general solution is the set of all ordered pairs of the form
52 ,, 3 y y æö ÷ ç ÷ ç ÷ ÷ ç èø where y is any real number.
17. 35 () 22 42 6() 33 y x1 xy2 -= +=
Rewrite the equations without fractions. 11 22 2RR235() 3RR4218() xy3 xy4 -= +=
Eliminate x in equation (4). 122 235() 2RRR88() xy3 y5 -= -+=
Make the coefficient of the first term in each equation equal 1. 1 11 2 1 22 8 35 RR() 22 RR1() xy6 y7 -= =
Substitute 1 for y in equation (6) to get 4. x = The solution is (4, 1).
18. 331 5 28 5 x y y x += -=
Multiply each equation by 5 to eliminate fractions. 122 1 22 151 15155 1040 15155 10R(1)RR1511510 15155 RR10 xy xy xy y xy y += -= += +-= += =
Back-substitution gives 15(10)155 5. x x += =
The solution is (5, 10).
19. 3 () 22 1 () 33 x y1 x y2 += +=
Rewrite the equations without fractions. 11 22 2RR23() 3RR31() xy3 xy4 += +=
Eliminate x in equation (4).
122 23() 1RRR2() xy3 y5 += -+=-
Substitute 2 - for y in equation (3) to get 7. x =
The solution is (7, 2 - ).
20. 1 () 963 82 2() 55 xy1 y x2 += +=
Rewrite the equations without fractions.
11 22 18RR236() 5RR1082() xy3 xy4 += +=
Eliminate x in equation (4).
122 236() 5RRR728() xy3 y5 += -+-=-
Make the coefficient of the first term in each row equal 1. 1 11 2 1 22 7 3 RR3() 2 RR4() xy6 y7 += -=
Substitute 4 for y in equation (6) to get 3. x =-
The solution is ( 3,- 4).
21. An inconsistent system has no solutions.
22. The solution of a system with two dependent equations in two variables is aninfinitesetof orderedpairs.
25. 231() 353() xyz1 xyz2 +-= ++=
Eliminate x in equation (2).
122 231() 3R2RR53() xyz1 yz3 +-= -++=
Since there are only two equations, it is not possible to continue with the echelon method as in the previous exercises involving systems with three equations and three variables. To complete the solution, make the coefficient of the first term in the each equation equal 1. 1 11 2 311 RR() 222 53() xyz4 yz3 ++= +=
Solve equation (3) for y in terms of the parameter z. 53 35 yz yz += =-
Substitute this expression for y in equation (4) to solve for x in terms of the parameter z.
311 (35) 222 91511 2222 84 84 xzz xzz xz xz +--= +--= -==-
The solution is (84, 35, ). zzz
26. 30() 237() xyz1 xyz2 +-= -+=-
Eliminate x in equation (2). 122 30() 2R(3)RR51121() xyz1 yz3 +-= +--=
Make the coefficient of the first term in each equation equal 1. 1 11 3 1 22 5 11 RR0() 33 1121 RR() 55 xyz4 yz5 +-= -=
Solve equation (5) for y in terms of z. 1121 55yz=+
Substitute this expression for y in equation (4), and solve the equation for x. 111211 0 3553 1171 0 553 27 55 27 55 xzz xzz xz xz æö ÷ ç ++-= ÷ ç ÷ ÷ ç èø ++-= +==--
The solution is 271121 ,,or 5555 271121 55,,. zzz zz z æö ÷ ç --+ ÷ ç ÷ ÷ ç èø æö --+ ÷ ç ÷ ç ÷ ÷ ç èø
27. 2311() 22() xyz1 xyz2 ++= -+=
122 2311() 2RRR5520() xyz1 yz3 ++= -+--=1 22 5 2311() RR4() xyz1 yz4 ++= -+=
Since there are only two equations, it is not possible to continue with the echelon method. To complete
the solution, solve equation (4) for y in terms of the parameter z. 4 yz =-
Now substitute 4 z - for y in equation (1) and solve for x in terms of z. 2(4)311 82311 3 xzz xzz xz +-+= +-+= =-
The solution is ( 3, z - 4, z - z)
28. 7() 237() xyz1 xyz2 -+-=++=
Eliminate x in equation (2).
Make the coefficient of the first term in each equation equal 1. 11 1 22 5 1RR7() 17 RR() 55 xyz4 yz5 --+= -=-
Solve equation (5) for y in terms of z. 17 55yz=-
Substitute this expression for y in equation (4), and solve the equation for x. 17 7 55 17 7 55 428 55 428 55 xzz xzz xz xz
The solution of the system is 42817 ,,or 5555 4287 55,,. zzz zz z æö
29. 2390() 3436() xyz1 yz2 ++= +=
Let z be the parameter and solve equation (2) for y in terms of z. 3436 3364 4 12 3 yz yz yz += ==-
Substitute this expression for y in equation (1) to solve for x in terms of z.
Thus the solutions are ( ) 14 33 66,12,, zzz where z is any real number. Since the solutions have to be nonnegative integers, set 1 3 660. z -³
Solving for z gives 198. z £
Since y must be nonnegative, we have 4 3 120. z -³ Solving for z gives 9. z £
Since z must be a multiple of 3 for x and y to be integers, the permissible values of z are 0, 3, 6, and 9, which gives 4 solutions.
30. 7475() 2760() xyz1 yz2 -+= +=
Let z be the parameter and solve equation (2) for y in terms of z 2760 2607 7 30 2 yz yz yz +=
Substitute this expression for y in equation (1) to solve for x in terms of z
Thus the solutions are () 577 22 285,30,, zzz where z is any real number. Since the solutions have to be nonnegative integers, set 57 2 2850. z -³ Solving for z gives 10. z £
Since y must be nonnegative, we have 7 2 300. z -³ Solving for z gives 8.57. z £
Since z must be a multiple of 2 for x and y to be integers, the permissible values of z are 0, 2, 4, 6, and 8, which gives 5 solutions.
31. 32480() 310() xyz1 yz2 ++= -=
Let z be the parameter and solve equation (2) for y in terms of z. 310 310 yz yz -= =+
Substitute this expression for y in equation (1) to solve for x in terms of z.
Thus the solutions are () 10 3 20,310,, zzz-+ where z is any real number. Since the solutions have to be nonnegative integers, set 10 3 200. z -³ Solving for z gives 6. z £
Since y must be nonnegative, we have 3100. z +³ Solving for z gives 1 3 10. z ³-
Since z must be a multiple of 3 for x to be an integer, the permissible values of z are 0, 3, and, 6, which gives 3 solutions.
32. 42372() 2312() xyz1 yz2 ++= -=
Let z be the parameter and solve equation (2) for y in terms of z 2312 2312 3 6 2 yz yz yz -= =+ =+
Substitute this expression for y in equation (1) to solve for x in terms of z 3
have to be nonnegative integers, set 3 2 150. z -³ Solving for z gives 10. z £
Since y must be nonnegative, we have 3 2 60. z +³ Solving for z gives 4. z £-
Since 410 z -££ and z must be a multiple of 2 for x and y to be integers, the permissible values of z are 0, 2, 4, 6, 8, and 10, which gives 6 solutions.
33. 2 ()() ()()() nbxmy1 xbxmxy2 +å=å å+å=å
Multiply equation (1) by 1 n 2 () ()()() xy bm3 nn xbxmxy2 åå += å+å=å
Eliminate b from equation (2). () xy bm3 nn +=åå 122 2 2 ()RRR ()()() () x xxyxmxy4 nn -å+ éùå-åå êú -+å=+å êú ê ú ë û
Multiply equation (4) by 2 2 1 . () x x n å -+å () xy bm3 nn +=åå 2 2 ()()1 () () xy mxy5 nx x n éù êú êú éù -åå êú =+åêú êú êú å ëû êú -+å êú ëû
Simplify the right side of equation (5). 22 22 ()() ()() ()() ()() xynxyn m nxnx nxyxy m nxx éù éù -åå+å êú êú = êú êú -å+å ëû êú ëû å-åå = å-å
From equation (3) we have () yx bm nn bymx n åå =å-å =
Thus the solutions are () 33 22 15,6,, zzz-+ where z is any real number. Since the solutions
35. Let x = the cost per pound of rice, and y = the cost per pound of potatoes.
The system to be solved is
201016.20()
301223.04.() xy1 xy2 += +=
Multiply equation (1) by 1 20
Make the coefficient of the first term in equation (l) equal 1. 1 11 8 5 RR525() 8 221200() xy3 xy2 += +=
1 11 20 RR0.50.81()
301223.04() xy3 xy2 += +=
Eliminate x in equation (2).
Eliminate x in equation (2).
30RRR31.26() xy3 y4 += -+-=-
122
0.50.81()
Multiply equation (4) by 1 3 -
0.50.81() RR0.42() xy3 y5 += -=
1 22 3
Substitute 0.42 for y in equation (3).
0.5(0.42)0.81 0.210.81 0.60 x x x += += =
The cost of 10 pounds of rice and 50 pounds of potatoes is 10(0.60)50(0.42)27, += that is, $27.
36. Let x = the number of new releases Blake downloaded y = the number of older songs Blake downloaded.
The system to be solved is 33()
1.290.9935.97() xy1 xy2 += +=
Eliminate x in eqatuon (2).
212 33() R(1.29)RR0.36.67() xy1 y3 += +−→−=−
Solve equation (3) for y.
6.6 22 0.3 y ==
Substitute 22 for y in equation (1) and solve for x 2233 11 x x += =
Blake bought 11 new releses and 22 older songs.
37. Let x = the number of seats on the main floor, and y = the number of seats in the balcony.
The system to be solved is 854200()
0.25(8)0.40(5)1200.() xy1 xy2 += +=
122 5 525() 8 6 2RRR150() 8 xy3 y4 += -+=
Make the coefficient of the first term in equation (4) equal 1. 8 22 6 5 525() 8 RR200() xy3 y5 += =
Substitute 200 for y in equation (3). 5 (200)525 8 125525 400 x x x += += =
There are 400 main floor seats and 200 balcony seats.
38. Let x = the number of skirts originally in the store, and y = the number of blouses originally in the store.
The system to be solved is 453551,750() 12 453530,600.() 23 xy1 xy2 += æöæö çç÷÷+= çç÷÷ ÷÷ çç÷÷ èøèø
Simplify each equation. Multiply equation (1) by 1 5 and equation (2) by 6 5 . 9710,350() 272836,720() xy3 xy4 += +=
Eliminate x from equation (4).
122 9710,350() 3RRR75670() xy3 y5 += -+=
Make each leading coefficient equal 1. 1 11 9 1 22 7 7 RR1150() 9 RR810() xy6 y7 += =
Substitute 810 for y in equation (6). 7 (810)1150 9 6301150 520 x x x += += =
Half of the skirts are sold, leaving half in the store, so 11 (520)260. 22 x ==
Two-thirds of the blouses are sold, leaving onethird in the store, so 11 (810)270. 33 y ==
There are 260 skirts and 270 blouses left in the store.
39. Let x = the number of model 201 to make each day, and y = the number of model 301 to make each day.
The system to be solved is 2334() 1827335.() xy1 xy2 += +=
Make the coefficient of the first term in equation (1) equal 1.
1 11 2 3 RR17() 2 1827335() xy3 xy2 += +=
Eliminate x in equation (2).
122 3 17() 2 18RRR029() xy3 4 += -+=
Since equation (4) is false, the system is inconsistent. Therefore, this situation is impossible.
40. Let x = the number of shares of Disney stock, and y = the number of shares of Intel stock.
41. (a) Let x = the number of fives y = the number of tens z = the number of twenties
We want to solve the following system. 70() 51020960() xyz1 xyz2 ++= ++=
Eliminate x in equation (2).
122 70 5RRR515610 xyz yz ++= -++=
Let z be the parameter. Solve for y and for x in terms of z 515610 561015 1223 yz yz yz += ==(1223)70 252 252 xzz xz xz +-+= -==-
The solutions are () 252,1223,, zzz where z is any real number. All variables must represent nonnegative integers. 12230 3122 40.67 z z z -³ -³£ 2520 252 26 z z z -³ ³ ³
Therefore, z must be an integer such that 2640. z ££ Thus, there are a total of 15 solutions.
(b) The smallest value of z is 26. 252 2(26)520 xz==-= 1223 1223(26)44 yz ==-=
The solutions with the smallest number of five-dollar bills is no fives, 44 tens, and 26 twenties.
307016,000()
4510525,500() xy1 xy2 += +=
Simplify each equation. Multiply equation (1) by 1 10 and equation (2) by 1 15 371600() 371700() xy3 xy4 += +=
Since 37xy + cannot equal both 1600 and 1700 for one point (x, y), we have an inconsistent system. Therefore, this situation is not possible.
(c) The largest value of z is 40. 252 2(40)5228 xz==-= 1223 1223(40)2 yz ==-=
The solutions with the largest number of fivedollar bills is 28 fives, 2 tens, and 40 twenties.
42. Let x = the number of EZ models, y = the number of compact models, and z = the number of commercial models.
Make a table.
102060440() 10828248() xyz1 xyz2 ++= ++=
Eliminate x from equation (2). 122 102060440() R(1)RR1232192() xyz1 yz3 ++= +-+=
Make the leading coefficients equal 1. 1 11 10 1 22 12 RR2644() 8 RR16() 3 xyz4 yz5 ++= +=
Solve equation (5) for y.
Substitute this expression for y into equation (4) and solve for x.
8(6) 2644 3 16(6) 446 3 132189616 3 362 3 2(18) 3 z xz z xz zz x z x z x éù++= êú êú ëû
The solution of the system is 2(18)8(6) 33,,. zz z æö ÷ ç ÷ ç ÷ ÷ ç èø
The solutions must be nonnegative integers. Therefore, 06. z ££ (Any larger values of z would cause y to be negative, which would make no sense in the problem.)
Values of z Solutions 0 (12, 16, 0) 1 () 3440 33,,1 2 ( ) 3232 33,,2 3 (10, 8, 3) 4 () 2816 33,,4 5 () 26833,,5
6 (8, 0, 6)
Ignore solutions containing values that are not integers. There are three possible solutions:
1. 12 EZ models, 16 compact models, and 0 commercial models;
2. 10 EZ models, 8 compact models, and 3 commercial models; or
3. 8 EZ models, 0 compact models, and 6 commercial models.
43. Let x = the number of buffets produced each week,
y = the number of chairs produced each week.
z = the number of tables produced each week.
Make a table.
Buffet Chair Table Totals
Construction 30 10 10 350
Finishing 10 10 30 150
The system to be solved is 301010350() 101030150.() xyz1 xyz2 ++= ++=
Make the coefficient of the first term in equation (1) equal 1. 1 11 30 1135 RR() 333 101030150() xyz3 xyz2 ++= ++=
Eliminate x from equation (2). 122 1135 () 333 2080100 10RRR() 333 xyz3 yz4 ++= -++=
Solve equation (4) for y. Multiply by 3. 2080100 45 54 yz yz yz += += =-
Substitute 54 z - for y in equation (1) and solve for x. 3010(54)10350 30504010350 3030030 10 xzz xzz xz xz +-+= +-+= =+ =+
The solution is (10,54,). zzz+- All variables must be nonnegative integers. Therefore, 540 54 5 , 4 z z z -³ ³ £
so 0 z = or 1. z = (Any larger value of z would cause y to be negative, which would make no sense in the problem.) If 0, z = then the solution is (10, 5, 0). If 1, z = then the solution is (11, 1, 1).
Therefore, the company should make either 10 buffets, 5 chairs, and no tables or 11 buffets, 1 chair, and 1 table each week.
44. (a) The system to be solved is
43,5001,295,000(1) 27,000440,000(2) xy xy -= -=
Eliminate x in equation (2).
27,000 122 43,500 43,5001,295,000() 1110,550,000 RRR() 2929 xy1 y3 -= -+-=-
Make the coefficient of the first term in equation (3) equal 1. 29 22 11 43,5001,295,000() 10,550,000 RR() 11 xy1 y4 -= -=
Substitute 10,550,000 11 for y in equation (1). 10,550,000 43,5001,295,000 11 24,795,000 43,500 11 570 11 x x x -= = =
The solution is 57010,550,000 1111,. æö ÷
The profit/loss will be equal after 570 11 weeks or about 51.8 weeks. At that point, the profit will be 10,550,000 11 or about $959,091.
(b) If the show lasts longer than 51.8 weeks, Broadway is a more profitable venue. If it lasts less than 51.8 weeks, off Broadway is a more profitable venue.
45. (a) For the first equation, the first sighting in 2000 was on day 7590.338(2000)83, y =-= or during the eighty-third day of the year. Since 2000 was a leap year, the eighty-third day fell on March 23.
For the second equation, the first sighting in 2000 was on day 1637 y =- 0.779(2000) = 79, or during the seventy-ninth day of the year. Since 2000 was a leap year, the seventy-ninth day fell on March 19.
(b) 7590.338() 16370.779() yx1 yx2 ==-
Rewrite equations so that variables are on the left side and constant term is on the right side. 0.338759() 0.7791637() xy3 xy4 += +=
Eliminate y from equation (4). 122 0.338759() 1RRR0.441878() xy3 x5 += -+=
Make leading coefficient for equation (5) equal 1.
The two estimates agree in the year closest to 878 0.441 1990.93, x =» so they agree in 1991.
The estimated number of days into the year when a robin can be expected is
0.338759 0.441
46. (a) We are given the equation 2 yaxbxc =++
Since a car traveling at 0 mph has a stopping distance of 0 feet, then y = 0 when 0. x =
Substituting these values into 2 yaxbxc =++ yields 2 0(0)(0),so 0. abc c =++ =
Therefore, we have 2 .yaxbx =+
After substituting the given values for the stopping distances (y) and speeds (x) in mph, the system to be solved is 2
2 61.7(25)(25)() 106(35)(35).() ab1 ab2 =+ =+
These equations can be written as 6252561.7() 122535106.() ab1 ab2 += +=
Multiply equation (1) by 1 625 ; also eliminate the decimal in 61.7 by multiplying the numerator and denominator of the fraction by 10.
1 11 625 1617 RR() 256250 122535106() ab3 ab2 += +=
Eliminate a in equation (2).
1617 () 256250 3733 1225RRR14() 250 ab3 b4 += -+-=-
122
Multiply equation (4) by 1 141 22 14 1617 () 256250 3733 RR() 3500 ab3 b5 += -=
Substitute 3733 3500 for b in equation (3).
13733617 2535006250 4905 87,500 a a æö ÷ ç += ÷ ç ÷ ÷ ç èø =
Therefore, 4905 0.056057, 87,500 3733 and1.06657. 3500 a b =» =»
(b) Substitute the values from part (a) for a and b and 55 for x in the equation 2 yaxbx =+ Solve for y. 2 0.056057(55)1.06657(55) 228 y y =+ »
The stopping distance of a car traveling 55 mph is approximately 228 ft.
47. Let x = number of field goals, and y = number of foul shots.
Then 64() 2100(). xy1 xy2 += +=
Eliminate x in equation (2).
122 64() 2RRR28() xy1 y3 += -+-=-
Make the coefficients of the first term of each equation equal 1.
22 64() 1RR28() xy1 y4 += -=
Substitute 28 for y in equation (1) to get 36. x = Wilt Chamberlain made 36 field goals and 28 foul shots.
48. Let x = flight time eastward y = difference in time zones
The flight time westward is one hour longer than the time eastward, so flight time westward is 1. x + The system to be solved is
13 (1)2 xy xy += +−= or 1() 13() xy1 xy2 −= += 122 22 1 (1)RRR212 1 1 RR6 2 xy y xy y -= -+= -= =
Substitute 6 y = into (1). 61 7 x x −= =
The flight time is 7 hours, and the difference in time zones is 6 hours.
49. (a) Since 8 and 9 must be two of the four numbers combined using addition, subtraction, multiplication, and/or division to get 24, begin by finding two numbers to use with 8 and 9. One possibility is 8 and 3 since (98)8324. -⋅⋅=
If we can find values of x and y such that either 8 xy+= and 323, xy+= or 3 xy+= and 328, xy+= we will have found a solution. Solving the first system gives 13 x =- and 21. y = This, however, does not satisfy the condition that x and y be singledigit positive integers. Solving the second system gives 2 x = and 1. y = Since both of these values are single-digit positive integers, we have one possible system. Thus, one system is
3
328 xy xy ì += ï ï í ï += ï î
Its solution is (2, 1). These values of x and y give the numbers 8, 9, 8, and 3 on the game card. These numbers can be combined as (98)83 -⋅⋅ to make 24.
Copyright © 2016 Pearson Education, Inc.
2.2 Solution of Linear Systems by the Gauss-Jordan Method
Your Turn 1 4510 7819 xy xy += +=
Write the augmented matrix and transform the matrix. 4510 7819 éù êú êú ëû
1 11 4 RR 55 42 1 7819 éù êú êú êú ëû
122 7RRR-+ 55 42 33 42 1 0 éù êú êú êúêú ëû
4 22 3 RR- 55 42 1 012 éù êú êú êúëû
5 211 4 RRR-+ 105 012 éù êú êúëû
The solution is 5 x = and 2, y =- or (5,2). -
Your Turn 2 232 22327 32510 xyz xyz xyz ++= +-= ++=
Write the augmented matrix and transform the matrix. 122 133 923 1 22 222 211 923 22 233 923 22 1 33 14 1232 22327 32510 1232 2RRR02923 3RRR0444 1232 RR01 0444 2RRR10625 01 4RRR 001442 10625 01 RR 0013 éù êú êúêú êú ëû éù êú -+--êú êú -+--êú ëû éù êú --êú êú êú êú ëû
êú êúêú êú +êú ëû é - ê - ë ù ú êú êú êú êú û 311 9 322 2 6RRR1007 RRR0102 0013 é ù + ê ú ê ú -+ ê ú ê ú - ë û
The solution is 7,2,xy== and 3, z =- or (7,2,3). -
Your Turn 3 22346 32537 4228 xyzw xyzw xyzw -+-= ++-= ++-=
Write the augmented matrix and transform the matrix. 22346 32537 41228
1 11 2 RR 3 2 1123 32537 41228 é ù ê ú ê ú ê úê ú ê ú - ê ú ë û 122 133 3RRR 4RRR -+ -+ 3 2 1 2 1123 0532 05464 é ù ê ú ê ú ê ú - ê ú ê ú ê ú ê ú ë û 1 22 5 RR 3 2 132 1055 1123 01 05464 é ù ê ú ê ú ê ú - ê ú ê ú ê ú ê ú ë û 8713 211 555 132 1055 9 233 2 10 RRR 01 5RRR 0032 é ù - + ê ú ê ú ê ú - ê ú ê ú ê ú - + ê ú ë û 2 33 9 RR- 8713 555 132 1055 24 39 10 01 001 é ù - ê ú ê ú ê ú - ê ú ê ú ê ú - ê ú ë û 8117 311 539 124 322 1039 24 39 RRR100 RRR010 001 é ù -+ê ú ê ú ê ú -+ê ú ê ú ê ú - ê ú ë û
We cannot change the values in column 4 without changing the form of the other three columns. So, let w be the parameter. The last matrix gives these equations. 117171 , or 3993 2442 , or 3993 2442 , or 3993 xwxw
The solution is ( ) 17 14242 939393,,,, wwww+--+ where w is a real number.
2.2 Warmup Exercises
W1. Using z as the parameter, the general solution is (243,102,). zzz+−
For nonnegative solutions we require 2430,1020,andz0. zz +≥−≥≥
Solving each inequality for z, we have 8,5,andz0.zz≥−≤≥
Thus, 05 z ≤≤ and there are 6 solutions in nonegative intergers.
W2. Using z as the parameter, the general solution is (252,143,). zzz−−+
For nonnegative solutions we require 2520,1430,and0. zzz −≥−+≥≥
Solving each inequality for z, we have 2514 23,,and0.zzz ≤≥≥
Thus, 1425 , 32 z ≤≤ and for interger ,512,zz≤≤ so there are 8 solutions in nonnegative intergers.
2.2 Exercises
1. 36 2515 xy xy += +=
The equations are already in proper form. The augmented matrix obtained from the coefficients and the constants is
2. 428 712 xy y -= -=-
2515
The equations are already in proper form. The augmented matrix obtained from the coefficients and the constants is 428 0712 é
-
3. 23 3427 2 xyz xyz xyz ++= -+=++= leads to the augmented matrix 2113 3427. 1112 é ù ê ú ê ú ê ú ê ú ë û 4. 2534 4275 38 xyz xyz xy -+= -+-=-=
The equations are already in proper form. The augmented matrix obtained from the coefficients and the constants is 2534 4275. 3108 é ùê ú ê ú ê
- ë û
5. We are given the augmented matrix 102 . 013 é ù ê ú ê ú ë û
This is equivalent to the system of equations 2 3, x y = = or x = 2, y = 3. 6. 105 013 é ù ê ú ê ú - ë û is equivalent to the system 5 3. x y = =-
7. 1004 0105 0011 éù êú êúêú êú ëû
The system associated with this matrix is 4 5 1, x y z = == or 4, x = 5, y =- 1. z =
8. 1004 0102 0013 éù êú êú êú êú ëû is equivalent to the system 4 2 3. x y z = = = or 4, x = 2, y = 3. z =
9. Rowoperations on a matrix correspond to transformations of a system of equations.
11. 37410 1236 04511 éù êú êú êú êú ëû
Find 12 R(3)R. +-
In row 2, column 1, 3(3)10. +-=
In row 2, column 2, 7(3)21. +-=
In row 2, column 3, 4(3)35. +-=-
In row 2, column 4, 10(3)68. +-=-
Replace R2 with these values. The new matrix is 37410 0158. 04511 éù êú êú êú êú ëû
12. Replace 3 R by 13 1R3R. +The original matrix is 32618 2257. 10520 éù êú êúêú êú ëû
The resulting matrix is 32618 2257. 02942 é ù ê ú ê ú - ê ú ê ú - ë û
13. 1647 0325 0537
ê
ê
é ù ê ú ê ú ê ú ê ú ë û
Find 211 (2)RRR -+
(2)01(2)36(2)24(2)57 0325 0537 1003 0325 0537 é ù -+-+-+-+ ê
ê ú ë û éùêú êú = êú êú ë û
14. Replace R1 by 31 R(3)R. +-
The original matrix is 10421 06530. 001215 é ù ê ú ê ú ê ú ê ú ë û
The resulting matrix is 30048 06530. 001215 é ù ê ú ê ú ê ú ê ú ë û
15. 30018 0509 0048 é ù ê ú ê ú ê ú ê ú ë û 1111 1 3333 11 3 (3)(0)(0)(18) 1006 RR 05090509 00480048 é ù é ù êú ê ú êú ê ú êú = ê ú êú ê ú êú ë û êú ë û
16. Replace R3 by 1 3 6 R. 10030 01017. 006162 é ù ê ú ê ú ê ú ê ú ë û
The resulting matrix is 10030 01017. 00127 é ù ê ú ê ú ê ú ê ú ë û
17. 5 3212 xy xy += +=
Write the augmented matrix and use row operations.
122 22 211 115 3212 115 3RRR 013 115 1RR 013 1RRR102 013 é ù ê ú ê ú ë û é ù ê ú ê ú -+ ë û éù êú êú - ëû éù -+ êú êú ëû
The solution is (2, 3).
18. 25 22 xy xy += +=-
To begin, write the augmented matrix for the given system.
125 212 éù êú êúëû
The third row operation is used to change the 2 in row 2 to 0.
122 125 0312 2RRR éù êú êú -+ ëû
Next, change the 2 in row 1 to 0.
211 2R3RR309 0312 +-éù êú êú ëû
Finally, change the first nonzero number in each row to 1. 1 11 3 1 22 3 RR 103 014 RR éùêú êú ëû
The final matrix is equivalent to the system 3 4, x y == so the solution of the original system is ( 3,- 4).
19. 7 4322 xy xy += +=
Write the augmented matrix and use row operations.
122 22 211 117 4322 117 4RRR 016 117 1RR 016 1RRR101 016 éù êú êú ëû éù êú êú -+ ëû éù êú êú - ëû éù -+ êú êú ëû
The solution is (1, 6).
20. 423 231 xy xy -= -+=
The augmented matrix for the system is 111 88 15 44 122 211 11 22 423 . 231 423 R2RR 045 R2RR8011 045 10 RR 01 RR éùêú êúëû éùêú êú + ëû éù + êú êú ëû éù êú êú êú êú ëû
The solution is () 5 11 84,.
21. 232 461 xy xy -= -=
Write the augmented matrix and use row operations.
122 232 461 232 2RRR 003 éùêú êúëû éùêú êú -+ëû
The system associated with the last matrix is 232 003. xy xy -= +=-
Since the second equation, 03, =- is false, the system is inconsistent and therefore has no solution.
22. 239 467 xy xy += +=
Write the augmented matrix and use row operations. 122 239 467 239 2RRR 0011 éù êú êú ëû éù êú êú -+ëû
The system associated with the last matrix is 239 0011. xy xy += +=-
Since the second equation 011, =- is false, the system is inconsistent and therefore has no solution.
23. 631 1262 xy xy -= -+=-
Write the augmented matrix of the system and use row operations. 1 6 122 1 1 11 2 6 631 1262 631 2RRR 000 1 RR 000 éùêú êú ëû éùêú êú + ëû éù - êú êú êú ëû
This is as far as we can go with the Gauss-Jordan method. To complete the solution, write the equation that corresponds to the first row of the matrix. 11 26xy-=
Solve this equation for x in terms of y. 1131 266 y xy + =+=
The solution is () 31 6 ,,yy + where y is any real number.
24. 1 1 xy xy -= -+=-
The augmented matrix is
The row of zeros indicates dependent equations. (Both equations have the same line as their graph.) The remaining equation is 1. xy-= Solving for x gives 1. xy=+ There are an infinite number
of solutions, each of the form ( 1, y + y), for any real number y.
25. 3 1 4 yx yz zx ==+ =-
First write the system in proper form. 3 1 4 xy yz xz -+=-= +=
Write the augmented matrix and use row operations.
1RR1103 0111 1014 1103 0111 1RRR0111 RRR1014 0111 1RRR0020 R2RR2008 R2RR0202 0020 RR RR R éù êú êúêú êú ëû --éù êú êúêú êú ëû éùêú êúêú êú -+ ëû +-éù êú êúêú êú -+ ëû éù + êú êú + êú êú ëû 3 1004 0101 R0010 éù êú êú êú êú ëû
The solution is (4, 1, 0).
26. 1 2 22 xy xz zy == =--
0111
Put the equations in proper form to obtain the system 1 20 22. xy xz yz += -= +=-
2RRR0212 0122
R2RR2010 0212
R2RR0036
R3RR6006
R3RR06012 0036
RR 1001 RR0102 0012 RRéù êú êúêú êúëû éù êú -+---êú êú êúëû +-éù êú êú êú +-êú ëû +-éù êú +--êú êú êúëû éê ê ê ê - ë
The augmented matrix is 122 211 233 311 322 1 11 6 1 22 6 1 33 3 1101 2010 0122 1101
The solution is ( 1,- 2, 2- ).
27. 225 20 27 xy yz xz -=+= +=-
Write the augmented matrix and use row operations.
The last row indicates an infinite number of solutions. The remaining equations are 3and9.xzyz -=-+=
Solve these for x and y, the solutions are (3,9,) zzz --+ for any real number z.
29. 44424 29 231 xyz xyz xyz +-= -+=-+=
Write the augmented matrix and use row operations.
133 211 233 2205 0210 2017 2205 0210
1RRR0212 RRR2015 0210
1RRR0002 éù êú êú êú êúëû éù êú êú êú -+-êú ëû +-éù êú êú êú -+-êú ëû
This matrix corresponds to the system of equations 25 20 02. xz yz +=+= =-
This false statement 02 =- indicates that the system is inconsistent and therefore has no solution.
28. 3 9 23533 xz yz xyz -=+= -++=
Write the augmented matrix and use row operations. 133 233 1013 0119 23533 1013 0119 2RRR03327 1013 0119 3RRR0000 éù êú êú êú êúëû éù êú êú êú êú + ëû éù êú êú êú êú -+ ëû
122 133 44424 2119 1231 44424
R(2)RR06642 R(4)RR0121620 éùêú êú êú êúëû éùêú +--êú êú +--êú ëû
211 233 322 1 11 12 1 22 12 1 33 4 2R(3)RR120012 06642 2RRR00464 120012 3R2RR0120276 00464 RR 1001 RR01023 00116 RR +--éù êú êúêú -+--êú ëû éùêú êú -+ êú êú ëû - éùêú êú êú êú - ëû
The solution is ( 1,- 23, 16).
30. 272 2521 3549 xyz xyz xyz +-=--+= ++=-
2RRR01123
3RRR01253
2RRR10318 01123
1RRR00370
31R37RR3700296
êú
ëû
12R37RR0370111 00370 RR éù êú êú êú êúëû éù êú êú +--êú êú -+-ëû éù +-êú êú êú êú -+ ëû +-éù êú
Write the augmented matrix and use row operations. 122 133 211 233 311 322 1 11 37 1 1272 2521 3549 1272
- 22 37 1 33 37 1008 RR0103 0010 RR éùêú êú êú êú ëû
The solution is (8, - 3, 0).
31. 350 421 741 xyz xyz xyz +-= -+= ++=
Write the augmented matrix and use row operations. 3510 4121 7411 éùêú êúêú êú ëû
122 133 3510 4R(3)RR023103 7R(3)RR023103 éùêú êú +--êú êú +--ëû
121 233 23R(5)RR6902715 023103 R(1)RR0000 éù +- êú êú êú êú +- ëû 95 1 11 692323 103 1 22 2323 23 RR10 01 RR 0000 éù êú êú êú êú êú êú ëû
The row of zeros indicates dependent equations. Solve the first two equations respectively for x and y in terms of z to obtain 9595 232323 z xz -+ =-+= and 103103 . 232323 z yz=-=
The solution is () 95103 2323,,. zz z -+-
32. 36311 22 5526 xyz xyz xyz -+= +-= -+=
Write the augmented matrix and use row operations.
122 133 36311 2112. 5526 36311 2R3RR015916 5R3RR015937 éùêú êúêú êúëû éùêú -+--êú êú -+--êú ëû
211 233 2R5RR150323 015916 RRR00021 +-éù êú êú êú -+-êú ëû
The last row indicates inconsistent equations. There is no solution to the system.
33. 5426 5311 155323 xyz xyz xyz -+= +-= -+=
Write the augmented matrix and use row operations. 5426 53111 155323 éùêú êúêú êúëû 122 133 5426 1RRR0735 3RRR0735 éùêú -+-êú êú -+-êú ëû 211 233 4R7RR350262 0735 1RRR0000 éù + êú êúêú êú -+ ëû 1 62 2 11 35 3535 35 1 22 77 7 RR 10 01 RR 0000 éù êú êú - êú êú êú êú ëû
The row of zeros indicates dependent equations. Solve the first two equations respectively for x and y in terms of z to obtain 262262 353535 z xz -+ =-+= and 3535 777 z yz + =+=
The solution is ( ) 26235 357,,. zz z -++
34. 3216 64312 5224 xyz xyz xyz +-=-+= -+=
Write the augmented matrix and use row operations.
32116 64312 5224 éù êú êúêú êúëû 122 133 32116
2RRR08544 5R3RR0161192 éù êú -+-êú êú -+-êú ëû 211 233 311 322 1 11 12 1 22 8 R4RR120120 08544
2RRR0014 1RRR120024 5RRR08024 0014 RR 1002 0103 RR 0014 +-éù êú êúêú êú -+ ëû -+-éù êú -+-êú êú êú ëû éùêú êú - - êú êú ëû
Read the solution from the last column of the matrix. The solution is ( 2,- 3,- 4).
35. 239 46218 1319 2444 xyz xyz xyz ++= ++= ---=-
Write the augmented matrix and use row operations. 1139 2444 122 1 133 4 2319 46218 2319 2RRR0000 RRR0000 éù êú êú êú êú êú ëû éù êú êú -+ êú êú + ëû
The rows of zeros indicate dependent equations. Since the equation involves x,y, and z, let y and z be parameters. Solve the equation for x to obtain 93 2 xyz =
The solution is ( ) 93 2 ,,,zyzyz where y and z are any real numbers.
36. 3529 4311 8546 xyz xyz xyz --=-++= -+=
8R3RR0252890
5R11RR330384 01153 25R11RR00183915 R61RR2013006039 5R183RR0201304026 00183915 RR RRé ù êú êú +--êú êú -+ ëû -+-éù êú êú êú êú + ëû é ù -+ê ú ê ú +ê ú ê ú ë û 1 33 183 1003 0102 0015 RR éùêú êúêú êú ëû
Write the augmented matrix and use row operations. 3529 43111 8546 é ù ê ú ê ú ê- ú ê ú - ë û 122 133 211 233 311 322 1 11 2013 1 22 2013 3529 4R3RR01153
Read the solution from the last column of the matrix. The solution is (3, - 2,- 5).
37. 23 2426 26 23 xyw xzw xyz xyzw +-= ++=+-= -++=-
Write the augmented matrix and use row operations. 12013 20426 12106 21113 éùêú êúêú êúêú êú êú ëû 122 133 144 12013 2RRR 044412 1RRR 00113 2RRR 05139 éùêú êú -+ êú êú -+êú êú -+ êú ëû 211 244 20426 044412 00113 0016824 R2RR 5R4RRéù + êú êú êú êú êú êú -+ ëû ( ) 311 322 344 20066 04080 00113 0002424 4RRR 4RRR 16R1RR+éù êú êú + êú êú êú êú +- ëû
411 422 433 80000 0120024 0024048 0002424
R(4)RR R(3)RR R(24)RRé ù +- ê ú ê ú +- ê ú ê ú +- ê ú ê ú ë
1 11 8 1 22 12 1 33 24 1 44 24 RR 10000 RR 01002 00102 RR 00011 RR - éù êú êú êú êú - êú êú êú ëû The solution is 0,2,2,1,xyzw ===-= or (0, 2, 2,- 1).
329 24210
39. 220 211 32227 xyzw xyzw xyzw +-+=-++= -+-= 122 133 1 22 3 211 233 33 111220 211111 321227 111220 2RRR033351 3RRR054887 111220 RR011117 054887 1RRR10013 011117 5RRR00132 10013 1RR éù êú êúêú êú ëû éù êú -+--êú êú -+--êú ëû éù êú ---êú
122 133 144 13219
2RRR 02448 3RRR 044412 1RRR 04133 éù êú êú -+ êú êú + êú êú -+ êú ëû
211 233 244 3R2RR 208106 02448
2RRR 00444 2RRR 009513 éù +êú êú êú êú +êú êú -+ êú ëû
311 322 344 2RRR 20022 1RRR 02004 00444 9R4RR 0001616 éù -+ êú êú -+ êú êúêú êú + êú ëû
411 433 R(8)RR160000 02004 R(4)RR0016032 0001616 +--éù êú êú êú +--êú êú êú êú ëû
1 11 16 1 22 2 1 33 16 1 44 16 RR 10000 RR 01002 00102 RR 00011 RR - éù êú êú - êú êú - - êú êú êú ëû
The solution is (0, 2, 2,- 1).
This is as far as we can go using row operations. To complete the solution, write the equations that correspond to the matrix. 3 419 32 xw yw zw +=+=+=-
Let w be the parameter and express x, y, and z in terms of w. From the equations above, 3, xw=-419,yw=-- and 32.zw=--
The solution is ( 3, w 419, w 32,ww ), where w is any real number.
Interchange rows 1 and 2.
This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is (11.844, 1.153,- 0.609, 14.004).
43. Insert the given values, introduce variables, and the table is as follows. 3 8 ab cd 1 4 efg
From this, we obtain the following system of equations.
The last row indicates that there are an infinite number of solutions. The remaining equations are 243and35 xzwyzw -++=---=-
Solve these for x and y to obtain 43 and35. 2 zw xyzw ++ ==+-
There are an infinite number of solutions, each of the form 43 ,35,,, 2 zw zwzw æö ++ ÷ ç +- ÷ ç ÷ ÷ ç èø or (1.50.52,53,,), zwzwzw++-++ for any real numbers z and w.
The augmented matrix and the final form after row operations are as follows.
41. 10.473.522.586.42218.65
8.624.931.752.83157.03
4.926.832.972.65462.3
2.8619.106.248.73398.4 xyzw xyzw xyzw xyzw ++-= --+= +-+= +--=
Write the augmented matrix of the system.
10.473.522.586.42218.65
8.624.931.752.83157.03
4.926.832.972.65462.3
2.8619.106.248.73398.4 éùêú êú êú êúêú êú êú ëû
This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is 28.9436, x » 36.6326, y » z » 9.6390, and 37.1036, w » or (28.9436, 36.6326, 9.6390, 37.1036).
42. 28.694.516.02.94198.3
16.744.327.38.9254.7
12.538.792.522.4562.7
1100000 0100000 0011000 0010000 00001111 0010100 0001000 10010101 0000100 0100001 0000010 0001001 0000001 01011001 00000000 é éù ê êú ê êú ê êú ê êú ê êú ê êú ê êú ê êú ê êú ê êú êú êú êú êú êú êú êú êú êú ëû ë
êú êú êú êú êú êú êú êú êú êú û
The solution to the system is read from the last column. ,,,, , 5 1111 624123 57 1 and 24224 , abcd efg ==== === So the magic square is: 3 8 1 6 11 24 5 12 1 3 1 4 5 24 1 2 7 24
44. Let x = the number of hours to hire the Garcia firm, and y = the number of hours to hire the Wong firm.
The system to be solved is 1020500() 3010750() 510250.() xy1 xy2 xy3 += += +=
Write the augmented matrix of the system. 1 10 1 10 1 5 11 22 33
The solution is (20, 15). Hire the Garcia firm for 20 hr and the Wong firm for 15 hr.
45. Let x = amount invested in U.S. savings bonds, y = amount invested in mutual funds, and z = amount invested in a money market account.
Since the total amount invested was $10,000, 10,000. xyz++=
Katherine invested twice as much in mutual funds as in savings bonds, so 2.yx =
The total return on her investments was $470, so 0.0250.060.045470. xyz ++=
The system to be solved is 10,000() 20() 0.0250.060.045470(). xyz1 xy2 xyz3 ++= -= ++=
Eliminate x in equations (2) and (4). 122 133 10,000() 2RRR3220,000() 25RRR3520220,000() xyz1 yz5 yz6 ++= -+--=-++=
Eliminate y in equation (6). 233 10,000() 3220,000() 35R3RR1040,000() xyz1 yz5 z7 ++= --=+-=-
Make each leading coefficient equal 1. 1 22 3 1 33 10 10,000() 220,000 RR() 33 RR4000() xyz1 yz8 z9 ++= -+= -=
Substitute 4000 for z in equation (8) to get 4000. y = Finally, substitute 4000 for 2 and 4000 for y in equation (1) to get 2000. x = Ms. Chong invested $2000 in U.S. savings bonds, $4000 in mutual funds, and $4000 in a money market account.
46. Let x = the number of inkjet printers purchased y = the number of LCD monitors purchased z = the number of memory chips purchased
We want to solve the following system.
46 109129894774 2 xyz xyz zy ++= ++= =
Write the augmented matrix and transform the matrix. 11146 109129894774 0210 éù êú êú êú êúëû
Simplify the system by multiplying equation (3) by 1000. 33 10,000() 20() 1000RR256045470,000() xyz1 xy2 xyz4 ++= -= ++=
122 109RRR-+ 11146 02020240 0210 éù êú êú êú êúëû 1 22 20 RR 11146 01112 0210 éù êú êú êú êúëû
211 233 1RRR 2RRR -+ + 10258 01112 00124 éù êú êú êú êú ëû
311 322 2RRR 1RRR + -+ 10010 01012 00124 éù êú êú êú êú ëû 1RR33- 10010 01012 00124 éù êú êú êú êú ëû
The solution is (10, 12, 24). Pyro-Tech, Inc. purchases 10 inkjet printers, 12 LCD monitors, and 24 memory chips.
47. Let x = the number of chairs produced each week, y = the number of cabinets produced each week, and z = the number of buffets produced each week.
Make a table to organize the information.
Chair Cabinet Buffet Totals
Cutting 0.2 0.5 0.3 1950
Assembly 0.3 0.4 0.1 1490
Finishing 0.1 0.6 0.4 2160
The system to be solved is 0.20.50.31950
24,950 11 14 7 1 22 14 16421,600 RR01 07523,700 éù êú êú - êú êú êú ëû 51500 77 1124,950 147 1 2 211 233 10 6RRR 01 7RRR 001250éù -+ êú êú êú êú êú êú + êú ëû 51500 77 1124,950 147 33 10 01 2RR 0012500 -
5 7 11 14 311 322 RRR 1002000 1600 010 RRR 0 012500é ù +
The solution is (2000, 1600, 2500). Therefore, 2000 chairs, 1600 cabinets, and 2500 buffets should be produced.
48. Let x = the number of vans to be purchased, y = the number of small trucks to be purchased, and z = the number of large trucks to be purchased.
0.30.40.11490
0.10.60.42160. xyz xyz xyz ++= ++= ++=
Write the augmented matrix of the system
The system to be solved is 200 35,00030,00050,0007,000,000 2. xyz xyz xy ++= ++= =
0.20.50.31950
0.30.40.11490
0.10.60.42160 éù êú êú êú êú ëû 11 22 33 10RR25319,500 10RR34114,900 10RR16421,600 éù êú êú êú êú ëû
Interchange rows 1 and 3. 16421,600 34114,900 25319,500 éù êú êú êú êú ëû 122 133 16421,600 3RRR0141149,900 2RRR07523,700 éù êú -+---êú êú -+---êú ëû
To simplify the system, divide the second equation by 1000. Write the system in proper form, obtain the augmented matrix, and use row operations to solve. 111200 3530507000 1200 éù êú êú êú êúëû 122 133 111200 35RRR05150 1RRR031200 éù êú -+-êú êú -+---êú ëû
211 233 R5RR50201000 05150 3R5RR00501000 éù + êú êúêú -+--êú ëû
311 322 2R5RR25003000 3R10RR05003000 00501000 éù + êú +--êú êú êú ëû
1 25 1 50 1 50 11 22 33 RR 100120 RR01060 00120 RR éù êú êú êú êú ëû
The solution is (120, 60, 20). U-Drive Rent-ATruck should buy 120 vans, 60 small trucks, and 20 large trucks.
49. Let x = the amount borrowed at 8%, y = the amount borrowed at 9%, and z = the amount borrowed at 10%.
(a) The system to be solved is 25,000
0.080.090.102190 1000 xyz xyz yz ++= ++= =+
Multiply the second equation by 100 and rewrite the equations in standard form. 25,000 8910219,000 1000. xyz xyz yz ++= ++= -=
Write the augmented matrix and use row operations to solve 11125,000 8910219,000 0111000 éù êú êú êú êúëû
122 11125,000 8RRR01219,000 0111000 éù êú êú -+ êú êúëû
211 233 1RRR1016000 01219,000 1RRR00318,000 -+-éù êú êú êú -+--êú ëû
311 322 1R3RR30036,000 2R3RR03021,000 00318,000 éù -+ êú êú + êú êú ëû 1 11 3 1 22 3 1 33 3 RR10012,000 RR0107000 RR0016000 éù êú êú êú êú êú êú ëû
The solution is (12,000, 7000, 6000). The company borrowed $12,000 at 8%, $7000 at 9%, and $6000 at 10%.
(b) If the condition is dropped, the initial augmented matrix and solution is found as before. 11125,000 8910219,000
122 11125,000 8RRR 01219,000 éù êú êú -+ ëû
211 1RRR1016000 01219,000 -+-éù êú êú ëû
This gives the system of equations 6000 219,000 xz yx =+ =-+
Since all values must be nonnegative, 60000 z +³ and 219,0000 z -+³ 6000 z ³- 9500. z £
The second inequality produces the condition that the amount borrowed at 10% must be less than or equal to $9500. If $5000 is borrowed at 10%, 5000, z = and 500600011,000 2(5000)19,0009000. x y =+= =-+=
This means $11,000 is borrowed at 8% and $9000 is borrowed at 9%.
(c) The original conditions resulted in $12,000 borrowed at 8%. So, if the bank sets a maximum of $10,000 at the 8% rate, no solution is possible.
(d) The total interest would be 0.08(10,000)0.09(8000)0.10(7000) 800720700 2220 ++ =++ = or $2220, which is not the $2190 interest as specified as one of the conditions of the problem.
50. (a) Let x = the number of deluxe models y = the number of super-deluxe models z = the number of ultra models
Make a table to organize the information.
We want to solve the following system.
Write the augmented matrix and transform the matrix.
21254 33272 526148 éù êú êú êú êú ëû
1 11 2 RR 1 2 1127 33272 526148 éù êú êú êú êú êú êú ëû
2
2
3
êú êú êú êú ëû
4 311 3 2 322 3 RRR RRR -+ + 10010 0104 00115 éù êú êú êú êú ëû
The solution is (10,4,15). Each week 10 deluxe models, 4 super-deluxe models, and 15 ultra models should be produced. (b) We want to solve the following system.
2254 33272 56148 xyz xyz xyz ++= ++= ++=
Write the augmented matrix and transform the matrix.
122 133 3RRR 5RRR -+ -+ 1 2 3 2 3 2 1127 019 0113 éù êú êú êú êú êú êúêú ëû
2 22 3 RR 1 2 2 3 3 2 1127 016 0113 éù êú êú êú êú êú êúêú ëû 1 211 2 3 233 2 RRR RRR -+ + 4 3 2 3 1030 016 0004 éù êú êú êú êú êú êú êú ëû
The last row indicates that there is no solution to the system.
(c) We want to solve the following system.
2254 33272 56144 xyz xyz xyz ++= ++= ++=
Write the augmented matrix and transform the matrix.
21254 33272 516148 éù êú êú êú êú ëû
1 11 2 RR 1 2 1127 33272 516144 éù êú êú êú êú êú êú ëû
21254 33272 516148 éù êú êú êú êú ëû 1 11 2 RR 1 2 1127 33272 516148 éù êú êú êú êú êú êú ëû
122 133 3RRR 5RRR -+ -+ 1 2 3 2 3 2 1127 019 019 éù êú êú êú êú êú êúêú ëû
2 22 3 RR 1 2 2 3 3 2 1127 016 019 éù êú êú êú êú êú êúêú ëû
The system is dependent. Let z be the parameter and solve the first two equations for x and y, which gives 42 33 30 and 6. xzyz =-=-
Since x, y, and z must be nonnegative integers, we have
Thus, z is an integer such that 922.5, z ££ and z must be a multiple of 3 so that x and y are integers. The permissible values of z are 9, 12, 15, 18, and 21. There are 5 solutions.
51. (a) Let x be the number of trucks used, y be the number of vans, and z be the number of SUVs. We first obtain the equations given here.
23325 24533 3222 xyz xyz xyz ++= ++= ++=
Write the augmented matrix and use row operations.
23325 24533 32122 éù êú êú êú êú ëû 122 133 23325 1RRR0128 3R2RR05731 éù êú êú -+ êú -+---êú ëû 211 233 3RRR2031 0128 5RRR0039 -+-éù êú êú êú êú + ëû
-+
311 322 RRR20010 2R3RR0306 0039 é ù + ê
1 11 2 1 22 3 1 33 3 RR1005 RR0102 RR0013 éù êú êú êú êú êú êú ëû
Read the solution from the last column of the matrix. The solution is 5 trucks, 2 vans, and 3 SUVs.
(b) The system of equations is now 23325 24533. xyz xyz ++= ++=
Write the augmented matrix and use row operations.
23325 24533 éù êú êú ëû
122 23325 RRR 0128 éù êú êú -+ ëû
211 3RRR2031 0128 -+-éù êú êú ëû
Obtain a one in row 1, column 1. 131 222 11 10 RR 0128éù êú êú êú ëû
The last row indicates multiple solutions are possible. The remaining equations are 31 22 xz-= and 28.yz+=
Solving these for x and y, we have 31 22 xz=+ and 28.yz=-+
The form of the solution is ( ) 3 1 22 ,28,.zzz +-+
Since the solutions must be whole numbers, 31 0 22 31 22 1 3 z z z +³ ³³and 280 28 4 z z z -+³ -³£
Thus, there are 4 possible solutions but each must be checked to determine if they produce whole numbers for x and y.
When 1 0,,8,0 2 z æö ÷ ç = ÷ ç ÷ ÷ ç èø which is not realistic. When () 1,2,6,1. z = When 7 2,,4,2 2 z æö ÷ ç = ÷ ç ÷ ÷ ç èø which is not realistic. When 3,(5,2,3). z =
When 13 4,,0,4 2 z æö ÷ ç = ÷ ç ÷ ÷ ç èø which is not realistic.
The company has 2 options. Either use 2 trucks, 6 vans, and 1 SUV or use 5 trucks, 2 vans, and 3 SUVs.
52. Let x = the number of two-person tents, y = the number of four-person tents, and z = the number of six-person tents that were ordered.
(a) The problem is to solve the following system of equations.
246200 4064883200 1291792298950 xyz xyz xyz ++= ++= ++=
Write the augmented matrix and use row operations to solve.
246200 4064883200 1291792298950 éù êú êú êú êú ëû
When this occurs, 2(0)5050. y =-+= And since , xz = the solution with the most four-person tents is 0 two-person tents, 50 four-person tents, and 0 six-person tents.
(c) The number of two-person tents is given by the value of the variable x. Since , xz = the most two-person tents will result when y is as small as possible, or 0. When this occurs, 246200
2()4(0)6()200 8200
25. xyz zz z z ++= ++= = =
The solution with the most two-person tents is 25 two-person tents, 0 four-person tents, and 25 six-person tents.
53. Let 1x = the number of units from first supplier for Roseville,
2x = the number of units from first supplier for Akron,
122 133 246200
20R(1)RR01632800
129R(2)RR01583167900 éù êú êú +- êú êú +- ëû
3x = the number of units from second supplier for Roseville, and 4x = the number of units from second supplier for Akron.
211 233 R(4)RR8080 01632800
79R(8)RR0000 +--éù êú êú êú êú +- ëû
1 11 8 1 22 16 RR 1010 01250 RR 0000 éù -êú êú êú êú êú êú ëû
Since the last row is all zeros, there is more than one solution. Let z be the parameter. The matrix gives 0 250. xz yz -= +=
Solving these equations for x and y, the solution is (z, 250, z -+ z). The numbers in the solution must be nonnegative integers. Therefore, 0 2500 25. y z z ³ -+³ £
Thus, z Î {0, 1, 2, 3, . . . , 25} In other words, depending on the number of six-person tents, there are 26 solutions to this problem.
(b) The number of four-person tents is given by the value of the variable y. Since 250,yz=-+ the most four-person tents will result when z is as small as possible, or 0.
Roseville needs 40 units so 13 40. xx+= Akron needs 75 units so 24 75. xx+=
The manufacturer orders 75 units from the first supplier so 12 75. xx+=
The total cost is $10,750 so 1234 70908012010,750. xxxx+++= The system to be solved is 13 24 12 1234 40
+++= Write augmented matrix and use row operations. 101040 010175 110075 70908012010,750
090101207950
233 244 101040 010175
1RRR 001140 90RRR 0010301200 éù êú êú êú êú -+ êú êú -+ êú ëû 33 344 101040 010175 1RR 001140 10RRR 00020800 éù êú êú êú êú - êú êú + êú ëû 1 20 44 101040 010175 001140 000140 RR éù êú êú êú êú êú êú êú ëû
422 433 101040 1RRR010035 1RRR00100 000140 éù êú êú -+ êú êú -+ êú êú êú ëû
311 1RRR100040 010035 00100 000140 éù -+ êú êú êú êú êú êú êú ëû
The solution of the system is 1 40, x = 2 35, x =
3 0, x = 4 40, x = or (40, 35, 0, 40). The first supplier should send 40 units to Roseville and 35 units to Akron. The second supplier should send 0 units to Roseville and 40 units to Akron.
54. Let 1x = the number of cars sent from I to A,
2x = the number of cars sent from II to A,
3x = the number of cars sent from I to B, and
4x = the number of cars sent from II to B.
A B
I x1 x3 II x2 x4
Plant I has 28 cars, so
13 28. xx+=
Plant II has 8 cars, so
24 8. xx+=
Dealer A needs 20 cars, so
12 20 xx+=
Dealer B needs 16 cars, so
34 16. xx+=
The total transportation cost is $10,640, so 1234 22040030018010,640. xxxx +++=
The system to be solved is 13 24 12 34 1234 28 8 20 16 22040030018010,640. xx xx xx xx xxxx += += += += +++=
Write the augmented matrix and use row operations. 101028 01018 110020 001116 22040030018010,640
133 155 101028 01018 1RRR 01108 001116 220RRR 0400801804480 -+ -+
233 255 101028 01018 1RRR 001116 001116
400RRR 00802201280 éù êú êú êú êú -+ êú êú êú êú -+êú ëû 131 344 355 RRR 100112 01018 001116 RRR 00000 80RRR 0003000 é ù +ê ú ê ú ê ú ê ú ê ú ê ú + ê ú ê ú + - ê ú ë û
There is a 0 now in row 4, column 4, where we would like to get a 1. To proceed, interchange the fourth and fifth rows. 100112 01018 001116 0003000 00000 é ùê ú ê ú ê ú ê ú ê ú ê ú - ê ú ê ú ê ú ë û
Each of the original variables has a value, so the last row of all zeros may be ignored. The solution of the system is 1 12, x = 2 8, x = 3 16, x =
4 0. x = Therefore, 12 cars should be sent from I to A, 8 cars from II to A, 16 cars from I to B, and no cars from II to B.
55. Let w = the number of Italian style vegetable packages
x = the number of French style vegetable packages
y = the number of Oriental style vegetable packages
z = the number of Combo style vegetable packages
The system to be solved is
zucchini
broccoli
carrots
cauliflower
0.30.216,200 wy+=
0.30.50.30.642,300 wxyz +++=
0.40.20.221,000 wxy++=
0.30.30.429,500 xyz++=
Expressing the amount of each vegetable in thousands and multiplying each equation by 10 and forming the augmented matrix we get 3020162 3536423 4220210 0334295 éù
Therefore, the company should produce 18,000 packages of Italian style, 15,000 packges of French style, 54,000 packages of Oriental style, and 22,000 packages of Combo style.
56. Let w = the number of Triple Berry packages x = the number of Summer Blend packages y = Berry Banana packages
z = Strawberry Banana Chunk packages
The system to be solved is strawberries 64410302,400 wxyz+++= raspberries 543153,700 wxy++= blueberries 53108,100 wy+= bananas 866255,000 xyz++=
The augmented matrix of this system is 64410302,400 5430153,700
5030108,100 0866255,000
64410302,400 (1)RRR040045,600 5030108,100 0866255,000
211 244 6040256,800 (1)RRR 040045,600 5030108,100
60410256,800 040045,600 125
00108,100
0066163,800
010011,400
000112,100
311 344 (1)RRR100012,500 010011,400 000112,100 001015,200 (1)RRR
100012,500 010011,400 001015,200 000112,100 RR
The company should prepare 12,500 packages of Triple Berry, 11,400 packages of Summer Blend, 15,200 packages of Berry Banana, and 12,100 packages of Strawberry Banana Chunk.
57. Let x = the number of grams of group A, y = the number of grams of group B, and z = the number of grams of group C.
(a) The system to be solved is 400(1) 1 (2) 3 2.(3) xyz xy xzy ++= = +=
Rewrite equations (2) and (3) in proper form and multiply both sides of equation (2) by 3.
Write the augmented matrix.
Interchange rows 2 and 3.
The solution is () 4004002000 939,,. Include 400 9 g of group A, 400 3 g of group B, and 2000 9 g of group C.
(b) If the requirement that the diet include onethird as much of A as of B is dropped, refer to the first two rows of the fifth augmented matrix in part (a).
This gives
Therefore, for any positive number z of grams of group C, there should be z grams less than 800 3 g of group A and 400 3 g of group B.
(c) Since there was a unique solution for the original problem, by adding an additional condition. the only possible solution would be the one from part (a). However, by substituting those values of A, B, and C for x, y, and z in the equation for the additional condition, 0.020.020.038.00, xyz++= the values do not work. Therefore, a solution is not possible.
58. Let 1x = the number of cases of Brand A, 2x = the number of cases of Brand B, 3x = the number of cases of Brand C, and 4x = the number of cases of Brand D. 1234 1234 1234 2550751001200 30303060600 30202030400 xxxx xxxx xxxx +++= +++= +++=
The augmented matrix of the system is 2550751001200 30303060600 30202030400
Interchange rows 1 and 2. 111220 5101520240 322340
322 RRR10010
We cannot change the values in column 4 further without changing the form of the other three columns. Therefore, let x4 be arbitrary. This matrix gives the equations 1414 2424 3434 0or, 412or124, 8or8. xxxx xxxx xxxx -== +==-==+
The solution is (x4, 12 - 4x4, 8 + x4, x4). Since all solutions must be nonnegative, 4 4 1240 3. x x -³ £
If 4 0, x = then 1 0, x = 2 12, x = and 3 8. x =
If 4 1, x = then 1 1, x = 2 8, x = and 3 9. x =
If 4 2, x = then 1 2, x = 2 4, x = and 3 10. x =
If 4 3, x = then 1 3, x = 2 0, x = and 3 11. x =
Therefore, there are four possible solutions. The breeder should mix
1. 0 cases of A, 12 cases of B, 8 cases of C, and 0 cases of D;
2. 1 case of A, 8 cases of B, 9 cases of C, and 1 case of D;
3. 2 cases of A, 4 cases of B, 10 cases of C, and 2 cases of D; or
4. 3 cases of A, 0 cases of B, 11 cases of C, and 3 cases of D.
59. Let x = the number of species A, y = the number of species B, and z = the number of species C.
Use a chart to organize the information.
60. Let x = the number of the first species, y = the number of the second species, and z = the number of the third species.
The system to be solved is 1.322.10.86490
2.90.951.52897
1.750.62.01653. xyz xyz xyz ++= ++= ++=
Use graphing calculator or computer methods to solve this system. The solution, which may vary slightly, is to stock about 244 fish of species A, 39 fish of species B, and 101 fish of species C.
1.31.18.116,000
1.32.42.928,000
2.33.75.144,000 xyz xyz xyz ++= ++= ++=
Write the augmented matrix of the system.
1.31.18.116,000
1.32.42.928,000 2.33.75.144,000
This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is 2340 of the first species, 10,128 of the second species, and 224 of the third species. (All of these are rounded to the nearest whole number.)
61. Let x = the number of acres for honeydews, y = the number of acres for yellow onions, and z = the number of acres for lettuce.
(a) 220 12015018029,100 180808032,600 4.974.454.65480 xyz xyz xyz xyz ++= ++= ++= ++=
Write the augmented matrix for this system. 111220 12015018029,100 180808032,600 4.974.454.65480
Using graphing calculator or computer methods, we obtain 100150 01050 . 00120 000581
There is no solution to the system. Therefore, it is not possible to utilize all resources completely.
(b) If 1061 hr of labor are available, the augmented matrix becomes, 111220 12015018029,100 . 180808032,600 4.974.454.651061
Again, using graphing calculator or computer methods we obtain
100150 01050 00120 0000 éù êú êú êú êú êú êú êú ëû
The solution is (150, 50, 20). Therefore, allot 150 acres for honeydews, 50 acres for onions, and 20 acres for lettuce.
62. (a) In 1980, 0 t = and 183.9. R = 2 183.9(0)(0) 183.9 abc c =++ =
In 1990, 10 t = and 203.3. R = 2 203.3(10)(10) 203.310010 abc abc =++ =++
In 2000, 20 t = and 196.5. R = 2 196.5(20)(20) 196.540020 abc abc =++ =++
The linear system to be solved is 40020196.5 10010203.3 183.9 abc abc c ++= ++= =
Write the augmented matrix and use row operations to solve.
400201196.5 100101203.3 001183.9 éù êú êú êú êú ëû
122 400201196.5 1R4RR0203616.7 001183.9 éù êú êú -+ êú êú ëû
211 1RRR40002420.2 0203616.7 001183.9 -+--éù êú êú êú êú êú ëû
311 322 2RRR4000052.4 3RRR020065 001183.9 +-éù êú êú -+ êú êú ëû 1 11 400 1 22 20 RR 1000.131 0103.25 RR 001183.9 éùêú êú êú êú êú ëû
The solution is 0.131, a - = 3.25, b = and 183.9. c =
(b) In 2010, 30. t = 2 2 0.1313.25183.9 0.131(30)3.25(30)183.9 163.5 Rtt=-++ =-++ »
The actual value is 186.2 deaths per 100,000. (c) Let t equal 0, 10, 20, and 30 to obtain four equations involving the coefficients a, b, c, and d. The linear system to be solved is 27,00090030186.2 80040020196.5 100010010203.3 183.9 abcd abcd abcd d +++= +++= +++= =
The augmented matrix is 27,000900301186.2 8000400201196.5 1000100101203.3 0001183.9
Use a graphing calculator or computer methods to obtain the solution 0.003783 a = 0.2445 b =- 4.007, c = and 183.9. d =
63. (a) Bulls:
The number of white ones was one half plus one third the number of black greater than the brown.
11 23 5 6 656 XYT XYT XYT æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ -=
The number of the black, one quarter plus one fifth the number of the spotted greater than the brown. 11 45 9 20 20920 20920 YZT YZT YZT YZT æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ =+ -=
The number of the spotted, one sixth and one seventh the number of the white greater than the brown.
11 67 13 42 421342 421342 ZXT ZXT ZXT ZXT æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ =+ -=
So the system of equations for the bulls is 656 20920 421342. XYT YZT ZXT -= -= -=
Cows:
The number of white ones was one third plus one quarter of the total black cattle.
11 () 34 7 () 12 1277 1277 xYy xYy xYy xyY æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ =+ -=
The number of the black, one quarter plus one fifth the total of the spotted cattle.
11 () 45 9 () 20 2099 2099 yZz yZz yZz yzZ æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ =+ -=
The number of the spotted, one fifth plus one sixth the total of the brown cattle.
11 () 56 11 () 30 301111 301111 zTt zTt zTt ztT æö ÷ ç =++ ÷ ç ÷ ÷ ç èø =+ =+ -=
The number of the brown, one sixth plus one seventh the total of the white cattle. 11 () 67 13 () 42 421313 421313 tXx tXx tXx txX æö ÷ ç =++ ÷ ç ÷ ÷ ç èø
So the system of equations for the cows is
1277 2099 301111 134213 xyY yzZ ztT xtX -= -= -= -+=
(b) For T = 4,149,387, the 33 ´ system to be solved is 6524,896,322 20982,987,740 1342174,274,254 XY YZ XZ -= -= -+=
Write the augmented matrix of the system. 65024,896,322 020982,987,740 13042174,274,254 é ù -
This exercise should be solved by graphing calculator or computer methods. The solution is 10,366,482 X = white bulls, 7,460,514 Y = black bulls, and 7,358,060 Z = spotted bulls.
For 10,366,482, X = 7,460,514, Y = and 7,358,060, Z = the 44 ´ system to be solved is 12752,223,598 20966,222,540 301145,643,257 1342134,764,266 xy yz zt xt -= -= -= -+=
Write the augmented matrix of the system. 1270052,223,598 0209066,222,540 00301145,643,257 130042134,764,266
This exercise should be solved by graphing calculator or computer methods. The solution is 7,206,360 x = white cows, 4,893,246 y = black cows, 3,515,820 z = spotted cows, and 5,439,213 t = brown cows.
64. (a) The other two equations are 23 34 700 600. xx xx += +=
1RRR0101100 0110700 0011600
10011000 0101100
1RRR0011600 0011600
10011000 0101100 00116
(b) The augmented matrix is 122 233 344 10011000 11001100 . 0110700 0011600 10011000
1RRR 0000
Let x4 be arbitrary. Solve the first three equations for x1, x2, and x3 14 24 34 1000 100 600 xx xx xx ==+ =-
The solution is 44 (1000,100,600 xx -+- 44,).xx
(c) For x4, we see that 4 0 x ³ and 4 600 x £ since 4 600 x - must be nonnegative. Therefore, 4 0600. x ££
(d) x1: If 4 0, x = then 1 1000. x =
If 4 600, x = then 1 1000600400. x =-=
Therefore, 1 4001000. x ££
x2: If 4 0, x = then 2 100. x =
If 4 600, x = then 2 100600700. x =+=
Therefore, 2 100700. x ££
x3: If 4 0, x = then 3 600. x =
If 4 600, x = then 3 6006000. x =-=
Therefore, 3 0600. x ££
(e) If you know the number of cars entering or leaving three of the intersections, then the number entering or leaving the fourth is automatically determined because the number leaving must equal the number entering.
65. (a) The system to be solved is 0200,0000.50.3 0350,0000.50.7. rb rb =-=--
First, write the system in proper form.
0.50.3200,000
0.50.7350,000 rb rb += +=
Write the augmented matrix and use row operations. 11 22 122 3 211 4 1 11 5 1 22 4 0.50.3200,000 0.50.7350,000 10RR 532,000,000 10RR 573,500,000 532,000,000 041,500,000 1RRR 50875,000 RRR 041,500,000 RR 10175,000 01375,000 RR éù
ëû é ê ë ù ú êú êú û
The solution is (175,000, 375,000). When the rate of increase for each is zero, there are 175,000 soldiers in the Red Army and 375,000 soldiers in the Blue Army.
66. Let x = number of foul shots, y = number of field goals, and z = number of three pointers. Since Bryant made a total of 46 baskets, 46. xyz++=
Since the number of field goals is equal to three times the number of three pointers, 3,yz = or 30.yz-=
And since the total number of points was 81, 2381.xyz++=
The system to be solved is 46() 30() 2381(). xyz1 yz2 xyz3 ++= -= ++=
Eliminate x in equation (3).
133 46() 30() 1RRR235() xyz1 yz2 yz4 ++= -= -++=
Eliminate y in equation (4).
46() 30() (1)RRR535() xyz1 yz2 z5 ++= -= -+=
233
Make each leading coefficient equal 1. 1 33 5 46() 30() RR7() xyz1 yz2 z6 ++= -= =
Substitute 7 for z in equation (2) to get y = 21. Finally, substitute 7 for z and 21 for y in equation (1) to get x = 18. Bryant made 18 foul shots, 21 field goals, and 7 three pointers.
67. Let x = the number of singles, y = the number of doubles, z = the number of triples, and w = the number of home runs hit by Ichiro Suzuki.
The system to be solved is
Write the equations in proper form, obtain the augmented matrix, and use row operations to solve.
262 1111 003 11 0030 1 04500 1 éù êú êú -êú êúêú êú êúëû
262 1111 003 11 0030 1 1RRR 046262 11 éù êú êú -êú êúêú êú -+ êú ëû 23
144
262 1111 0030 1 RR 003 11 046262 11 éù êú êúêú êú « -êú êú êú ëû
211 244 1RRR 1140262 0030 1 003 11 RRR 0046262 4 éù -+ êú êúêú êú -êú êú + êúëû
311 344 1RRR 005265 1 0030 1 003 11 46RRR 00050400 éù -+ êú êúêú êú -êú êú + êú ëû
411 422 433 R10RR 100002250 3R50RR 050001200 1R50RR 00500250 00050400 éù + êú êú -+ êú êú -+ êú êú êú ëû 1 11 10 1 22 50 1 33 50 1 44 50 RR 000225 1 RR 000124 RR 0005 1 RR 0008 1 éù êú êú êú êú êú êú êú êú - êú ëû
Ichiro Suzuki hit 225 singles, 24 doubles, 5 triples, and 8 home runs during the 2004 season.
68. (a) 2 2 2 5.4(8)(8) 5.4648
6.3(13)(13) 6.316913
5.6(18)(18) 5.632418 abc abc abc abc abc abc =++ =++ =++ =++ =++ =++
The linear system to be solved is 6485.4 169136.3 324185.6. abc abc abc ++= ++= ++=
Use a graphing calculator or computer methods to solve this system. The solution is 0.032, a =- 0.852, b = and 0.632. c = Thus, the equation is 2 0.0320.8520.632.yxx =-++ (b)
The answer obtained using Gauss-Jordan elimination is the same as the answer obtained using the quadratic regression feature on a graphing calculator.
69. Let x = the number of balls, y = the number of dolls, and z = the number of cars.
(a) The system to be solved is 100 234295 1216181542. xyz xyz xyz += ++= ++= Write the augmented matrix of the system. 122 133 211 233 1 33 2 311 322 111100 234295 1216181542 111100
2RRR01295
12RRR046342
1RRR1015 01295 4RRR00238 1015 01295 00119 RR RRR10 2RRR01 00 éù êú êú êú êú êú ëû éù êú êú -+ êú êú -+ êú ëû -+-éù êú êú êú -+--êú êú ëû éùêú êú êú êú - êú ëû + -+ 024 057 119 éù êú êú êú êú êú ëû
The solution is (24, 57, 19). There were 24 balls, 57 dolls, and 19 cars.
(b) The augmented matrix becomes 122 133 111100 234295. 1115191542 111100 2RRR01295 11RRR048442 éù êú êú êú êú êú ëû éù êú êú -+ êú êú -+ êú ëû 211 233 1RRR1015 01295 4RRR00062 -+-éù êú êú êú êú -+ êú ëû
Since row 3 yields a false statement, 062, = there is no solution.
(c) The augmented matrix becomes 111100 234295. 1115191480 éù êú êú êú êú êú ëû
122 133 211 233 111100 2RRR01295 11RRR048380 1RRR1015 01295 4RRR0000 éù êú êú -+ êú êú -+ êú ëû -+-éù êú êú êú êú -+ êú ëû
Since the last row is all zeros, there are infinitely many solutions Let z be the parameter. The matrix gives 5 295. xz yz -= +=
Solving these equations for x and y, the solution is (5,952,). zzz+- The numbers in the solution must be nonnegative integers. Therefore, 9520 295 47.5. z z z -³ -³£
Thus, {0,1,2,3,...,47}. z Î There are 48 possible solutions.
(d) For the smallest number of cars, 0, z = the solution is (5, 95, 0). This means 5 balls, 95 dolls, and no cars.
(e) For the largest number of cars, 47, z = the solution is (52, 1, 47). This means 52 balls, 1 doll, and 47 cars.
70. Let x = the number of calories in each gram of fat y = the number of calories in each gram of carbohydrates z = the number of calories in each gram of protein We want to solve the following system. 10362240 14373280 202311295 xyz xyz xyz ++= ++= ++=
Write the augmented matrix and transform the matrix. 10362240 14373280 202311295
14373280 202311295
13.60.224 013.40.256 0497185
1 22 13.4
13.60.224 RR010.0149254.179104
0497185 éù êú --êú êú êú ëû
3.6RRR100.253738.95522
49RRR006.26867519.7761 éù -+ êú êúêú êú + ëû
211 233
010.0149254.179104
100.253738.95522
010.0149254.179104
1 33 6.268675 RR
0013.15475 éù êú êúêú êú ëû
311 322 0.25373RRR 0.014925RRR -+ +
1008.15476 0104.22619
0013.15475 éù êú êú êú êú ëû
The solution is (8.15,4.23,3.15). There are 8.15 calories in a gram of fat, 4.23 calories in a gram of carbohydrates, and 3.15 calories in a gram of protein.
71. (a) 111221 111222 112122 122122 1 1 1 1 xxx xxx xxx xxx ++= ++= ++= ++=
Write the augmented matrix of the system.
122 133 11101 11011 10111 01111 11101 00110 1RRR 01010 1RRR 01111 éù êú êú êú êú êú êú êú ëû éù êú êúêú êú -+êú êú -+ êú ëû
Since 11-= modulo 2, replace 1 - with 1. 11101 00110 01010 01111 éù êú êú êú êú êú êú êú ëû
Interchange rows 2 and 3. 11101 01010 00110 01111 éù êú êú êú êú êú êú êú ëû
211 244 1RRR 10111 01010 00110 RRR 00101 éù -+êú êú êú êú êú êú -+ êú ëû
Again, replace 1- with 1. 0 1111 000 11 000 11 00011 éù êú êú êú êú êú êú êú ëû
311 344 1RRR 000 11 000 11 000 11 1RRR 000 11 éù -+ êú êú êú êú êú êú -+ êúëû
Replace 1- with 1.
422 433 000 11 000 11 000 11 000 11 000 11 1RRR 00011 1RRR 00011 000 11 éù êú êú êú êú êú êú êú ëû éù êú êú -+êú êú -+êú êú êú ëû
Finally, replace 1- with 1. 000 11 00011 00011
000 11 é ù ê ú ê ú ê ú ê ú ê ú ê ú ê ú ë û
The solution (1, 1, 1, 1) corresponds to 11 1, x = 12 1, x = 21 1, x = and 22 1. x = Since 1 indicates that a button is pushed, the strategy required to turn all the lights out is to push every button one time.
(b) 111221 111222 112122 122122 0 1 1 0 xxx xxx xxx xxx ++= ++= ++= ++= Write the augmented matrix of the system.
00 111 0 1111 0 1111 00 111 é ù ê ú ê ú ê ú ê ú ê ú ê ú ê ú ë û
122 133 00 111 00 111
1RRR 00111
1RRR 00 111 éù êú êúêú êú -+êú êú -+ êú ëû
Replace 1- with 1. 00 111 00 111 00111 00 111 éù êú êú êú êú êú êú êú ëû
Interchange rows 2 and 3.
72. (a) Let x = amount of paddy in one top grade bundle y = amount of paddy in one medium grade bundle
z = amount of paddy in one low grade bundle
The system to be solved is 3239 2334 2326 xyz xyz xyz ++= ++= ++=
00 111 00 111
1RRR 0 1111 00111
211 244 00 111 00111
00 111
1RRR 00011 éù êú êú êú êú êú êú êú ëû éù -+ êú êú êú êú êú êú -+ êúëû
Replace 1- with 1.
(b) The augmented matrix of the system is 32139 23134 12326
31 RR12326 23134 32139
122 133 12326 (2)RRR 01518 04839 (3)RRR
22 12326 (1)RR01518 04839
1RRR 0000 1 00111 00 111
311 344 0 1111 00111 00 111 00011
1RRR 0000 1 éù êú êú êú êú êú êú êú ëû éù -+ êú êú êú êú êú êú -+ êúëû
Replace 1 - with 1.
0000 1 00111
00 111
0000 1
211 233 10710 (2)RRR 01518 001233 4RRR
33 10710 01518 11 1 001 RR 4 12
311 322 7RRR 1009.25 0104.25 (5)RRR 0012.75
0000 1
1RRR 00011
0000 1 éù êú êú êú êú êú êú êú ëû éù êú êú -+ êú êú -+ êú êú êú ëû
422 433
1RRR 00011
The solution (0, 1, 1, 0) corresponds to 11 0, x = 12 1, x = 21 1, x = and 22 0. x = Since 1 indicates that a button is pushed and 0 indicates that it is not, the strategy required to turn all the lights out is to push the button in the first row, second column, and push the button in the second row first column.
The bundle of top grade paddy yields 9.25 dou, one bundle of medium grade paddy yields 4.25 dou, and one bundle of low grade paddy yields 2.75 dou.
73. (a) Let x = the cost of a sheep y = the cost of a dog z = the coast of a hen w = the cost of a rabbit
The system to be solved is 54321496 42631175 75958 35861 xyzw xyzw xyzw xyzw +++= +++= +++= +++=
(b) Using a graphing calculator or computer methods, we find the following solution: A sheep costs 177 coins, a dog costs 121 coins, a hen costs 23 coins, and a rabbit costs 29 coins.
2.3 Addition and Subtraction of Matrices
Your Turn 1
(a) It is not possible to add a 24 ´ matrix and a 23 ´ matrix.
(b) 345124429 1232481211 éùéùéùêúêúêú += êúêúêú ëûëûëû
Your Turn 2 345124261 123248365 éùéùéùêúêúêú -=
2.3 Exercises
1. 1135 5737 éùéù êúêú = êúêú ëûëû
This statement is false, since not all corresponding elements are equal.
2. [] 1 123 2 3 éù êú êú = êú êú ëû
This statement is false. For two matrices to be equal, they must be the same size, and each pair of corresponding elements must be equal. These two matrices are different sizes.
3. 2 8 x y éùéùêúêú = êúêú ëûëû if 2 x =- and 8. y =
This statement is true. The matrices are the same size and corresponding elements are equal.
4. 358 2 0 114 éù êú êúëû is a 42 ´ matrix.
This statement is false. Since the matrix has 2 rows and 4 columns, it is a 24 ´ matrix.
5. 9 14 37 2 0 11 é ù - ê ú ê ú ê ú ê ú - ê ú ë û is a square matrix.
This statement is true. The matrix has 3 rows and 3 columns.
6. 241 241 375 375 000 é ùêú é ù - êú ê ú = êú ê ú êú ë û êú ë û
This statement is false since the matrices are different sizes.
7. 8 4 3 2 é ù - ê ú ê ú ë û is a 22 ´ square matrix.
8. 37 2 0 14 é ù - ê ú ê ú ë û
This matrix has 2 rows and 3 columns, so it is a 23 ´ matrix.
9. 6800 9 412 357 1 é ù - ê ú ê ú ê ú ê ú - ê ú ë û is a 34 ´ matrix.
10. 863 24 é ù - ê ú ë û
The matrix has 1 row and 5 columns, so it is a 15 ´ matrix. It is a row matrix since it has only 1 row.
11. 7 5 é ùê ú ê ú ë û is a 21 ´ column matrix.
12. [ ] 9-
This matrix has 1 row and 1 column, so it is a 11 ´ square matrix. It is also a row matrix since it has only 1 row, and a column matrix because it has only 1 column.
13. Undefined
14. Since A is a 52 ´ matrix, and since A and K can be added, we know that K is also a 52 ´ matrix. Also, since ,AKA += all entries of K must be 0.
15. 343 81 x yz é ùéù ê úêú = ê úêú - ë ûëû
Corresponding elements must be equal for the matrices to be equal. Therefore, 4,8,xy==and 1. z =
16. 55 8 y éùéù êúêú = êúêú ëûëû
Two matrices can be equal only if they are the same size and corresponding elements are equal. These matrices will be equal if 8. y =
17. 4262 575 st r éùéù -+ êúêú = êúêú ëûëû
Corresponding elements must be equal 4 6227. 100 str st -=+== ==
Thus, 10,0, and 7. str ===
18. 9735 080 mn r éùéù -+ êúêú = êúêú ëûëû
The matrices are the same size, so they will be equal if corresponding elements are equal. 93758 122 mnr mn =-=+= ==
Thus, 12,2,and8.mnr===
19. 23472615256 78362816 abcb dfd éùéùéù +êúêúêú += êúêúêú ëûëûëû
Add the two matrices on the left side to obtain 234726 78362 abcb dfd éùéù +êúêú + êúêú ëûëû (2)(7)3246 (3)7(6)8(2) abbc ddf éù ++-++ êú = êú +-+-+ëû
5546 2766 abc df éù -+ êú = êú ëû
Corresponding elements of this matrix and the matrix on the right side of the original equation must be equal.
515525466 2050 abc abc -==+= === 28761 41 df df -=--= ==
Thus, 20, a = 5, b = 0, c = 4, d = and 1. f =
20. 2315325 403256 azmazm kk éùéù ++ êúêú + êúêú ëûëû 101480 1059 éùêú = êú ëû
425110101480 6591059 azm k é ++-ùéù ê úêú = ê úêú ë ûëû
42105114 23 az az +=+===610 1080 5 8 3 k m mk = = = = 5 3
Thus,2,3,8,and. azmk ==-==
21. 245780101 6312028911 é ùéù ê úêú + ê úêú ë ûëû 28405(10)71 6(2)3812(9)011 é ù +++--+ ê ú = ê ú +--++-+ ë û 10456 45311 é ù ê ú = ê ú ë û
22. 15231253 23852835 37193(1)79 38 102 216 é ùéùéù ++ ê úêúêú ê úêúêú -+=+-+ ê úêúêú ê úêúêú -+-+ ë ûëûëû éù êú êú = êú êú ëû
é ù ê ú éù - ê ú êú + ê ú êú ëû ê ú - ë û
23. 30 132 64 471 52
These matrices cannot be added since the first matrix has size 23, ´ while the second has size 32. ´ Only matrices that are the same size can be added.
24. 803152 1195398 é ùéù ê úêúê úêú ë ûëû 810(5)32 131995(8) é ù ê ú = ê ú ë û 755 2103 é ùê ú = ê ú - ë û
25. The matrices have the same size, so the subtraction can be done. Let A and B represent the given matrices.
AB-= éù êú êú =------êú êú ëû éùêú êú = êú êú ëû
32. 4856 6325 2546 4242 kyky zxzx kaka mnmn é -+ùéù ê úêú ê -+úêú ê úêú - ê ++úêú ê úêú ê -+-úêú ê úêú ë ûëû
218312609 724(3)1(3)54 18200(2)1017 1569 5721 7227
26. 21180 53532 72675 90210 éùéùêúêú êúêúêúêú + êúêú êúêú êúêúêúêú ëûëû This operation is not possible because the matrices are different sizes. Only matrices of the same size can be added.
27. 234332 247814 éùéùéù êúêúêú +êúêúêúëûëûëû 24333234 27148448 éùéù +-+êúêú == êúêú -+-+ëûëû
28. 431111 121014 éùéùéù êúêúêú -+ êúêúêú ëûëûëû 41131143 11120416 éùéù -+-+ êúêú == êúêú -+-+ ëûëû
29. 2148127 0135753 éùéùéùêúêúêú -+ êúêúêúëûëûëû 2412187102 0(5)51373109 éùéù -+--+êúêú == êúêú --+-+ ëûëû
30. 580158 312261 éùéùéù êúêúêú ++ êúêúêú ëûëûëû 50(5)81(8)01 3(2)61(2)110 éùéù ++-++êúêú == êúêú -+-++-+ ëûëû
31. 423862 63253564 xyxyxyx xyxyyxxy éùéù -+-+-+ êúêú + êúêú ---+ ëûëû (42)(86)(3)2 (63)(35)(25)(64) 128 8 xyxyxyx xyyxxyxy xyxy xxy éù -++-+-++ êú = êú -+--++ ëû éù -+-+ êú = êúëû
48(56)14 63(25)48 25(46)2 42(42)84 kykyky zxzxzx kakaka mnmnmn é --+--ùéù ê úêú ê --+-úêú ê ==úêú ê +-+--úêú ê úêú ê -+---+úêú ê úêú ë ûëû
33. 00 00 00 00 xy OX zw xyxy zwzw é ùéù êúêú -=-êúêú êúêú ëûëû
é ùéù ê úêú == ê úêú ê úêú ë ûëû
34. Verify that XTTX +=+ XTxrys ztwu éù + + êú += êú ++ ëû TXrxsy tzuw éù + + êú += êú ++ ëû
Because of the commutative property for addition of real numbers, .xrrx +=+ This also applies to the other corresponding elements, so we conclude that . TXXT +=+
35. Show that ()(). XTPXTP ++=++
On the left side, the sum TP + is obtained first, and then ().XTP ++
This gives the matrix ()() ()() xrmysn ztpwuq é ù ++++ ê ú ê ú ++++ ë û
For the right side, first the sum XT + is obtained, and then (). XTP ++
This gives the matrix ()() ()() xrmysn ztpwuq é ù ++++ ê ú ê ú ++++ ë û
Comparing corresponding elements, we see that they are equal by the associative property of addition of real numbers. Thus, ()() .XTPXTP ++=++
36. Verify that (). XXO -=
37. Show that .POP += 0000 0000 mnmn PO pqpq mn P pq éùéùéù ++ +=+=êúêúêú êúêúêú ++ ëûëûëû éù êú == êú ëû
Thus, .POP +=
38. All of these properties are valid for matrices that are not square, as long as all necessary sums exist. The sizes of all matrices in each equation must be the same.
39. (a) CarsTrucks New1515 Used3625 éù êú êú ëû
(b) 171215153227 241836256043
(c) 1515171223 36252418127
IIIIII
40. (a)
Bread8810560 Milk487240 PB16210 Cold cuts11214750 éù êú êú êú êú êú êú êú
(b)
880.25(88)105(105)600.1(60)
480.25(48)72(72)400.1(40)
160.25(16)21(21)00.1(0)
1120.25(112)147(147)500.1(50) 11014066 609644 20280 14019655
14.214.8 0.63.0 7.41.7 23.740.3
35.233.9 2.13.9 15.013.8 42.0145.9 58.332.5 67.859.8 224.0247.3
6.84.3 0.92.1 0.210.4 5.465.3 6.322.1 11.819.8 36.261.7 CM
42. (a) 178.9230.8 16.2100.0 111.3135.9 64.9146.5 29.458.5 A
(b) 300.0332.1 122.0440.4 226.2280.5 65.1138.5 47.4114.6 B
(c) 121.4101.3 105.8340.4 114.9144.6 0.28.0 18.056.1
All trade amounts increased from 2000 to 2013 except for U.S. imports from Japan.
43. (a) There are four food groups and three meals. To represent the data by a 34 ´ matrix, we must use the rows to correspond to the meals, breakfast, lunch, and dinner, and the columns to correspond to the four food groups. Thus, we obtain the matrix
(b) There are four food groups. These will correspond to the four rows. There are three components in each food group: fat,
carbohydrates, and protein. These will correspond to the three columns. The matrix is 507 0101 0152 10128 éù êú êú êú êú êú êú êú ëû
(c) The matrix is 8 4. 5 éù êú êú êú êú ëû
44. (a) Length5.66.46.97.66.1 Weight144138149152146 éù êú êú ëû
(b) Length10.211.411.412.710.8 Weight196196225250230 éù êú êú ëû
(c) Subtract the matrix in part (a) from the one in part (b).
10.211.411.412.710.85.66.46.97.66.1 196196225250230144138149152146éùéù êúêú êúêú ëûëû
4.65.04.55.14.7Changein length 5258769884Change in weight éù êú = êú ëû
(d) Add the new matrix to the one from part (b).
10.211.411.412.710.81.81.52.31.82.0 1961962252502302522293320 éùéù êúêú + êúêú ëûëû
12.012.913.714.512.8Final length 221218254283250Final weight éù êú = êú ëû
45.
Obtained Pain Relief YesNo Painfree223 Placebo817 éù êú êú ëû
(a) Of the 25 patients who took the placebo, 8 got relief.
(b) Of the 25 patients who took Painfree, 3 got no relief.
(c)
2232141962328515 81761910153222773 éùéùéùéùéù êúêúêúêúêú +++= êúêúêúêúêú ëûëûëûëûëû
(d) Yes, it appears that Painfree is effective. Of the 100 patients who took the medication, 85% got relief.
46. (a) The matrix for motorcycle helmet usage in 2012 is
Compliant Noncompliant Northeast606 Midwest499 South6116 West824 A éù êú êú êú = êú êú êú êú ëû
The matrix for motorcycle helmet usage in 2013 is 5210 425 6511 923 B éù êú êú êú = êú êú êú êú ëû
(b) The matrix showing the change in helmet usage from 2012 to 2013 is 521060684 42549974 6511611645 923824101
(c) The percentage of compliant helmets incressed in the South and West but decreased elsewhere. The percentage of noncompliant helmets decreased everywhere except the Northeast.
47. (a) The matrix for the life expectancy of African Americans is MF 197060.068.3 198063.872.5 199064.573.6 200068.275.1 201071.878.0 éù êú êú êú êú êú êú êú êú êú ëû
(b) The matrix for the life expectancy of White Americans is MF 197068.075.6 198070.778.1 199072.779.4 200074.779.9 201076.581.3 éù êú êú êú êú êú êú êú êú êú ëû
(c) The matrix showing the difference between the life expectancy between the two groups is 60.068.368.075.68.07.3 63.872.570.778.16.95.6 64.573.672.779.48.25.8 68.275.174.779.96.54.8 71.878.076.581.34.73.3 é ùéùéù ê úêúêú ê úêúêú ê úêúêú ê úêúêú -= ê úêúêú ê úêúêú ê úêúêú ê úêúêú ê úêúêú ë ûëûëû
48. (a) The matrix for the educational attainment of males is
4 Years of 4 Years of High School College or or More More
197055.014.1
198069.220.9
199077.724.4
200084.227.8
201086.030.3 éù êú êú êú êú êú êú êú êú êú ëû
(b) The matrix for the educational attainment of females is
4 Years of 4 Years of High School College or or More More
198045.37.9 198547.98.5 199050.89.2 199553.49.3 200057.010.6 200558.512.0 201062.913.9
197055.48.2
198068.113.6
199077.518.4
200084.023.6
201087.629.6 éù êú êú êú êú êú êú êú êú êú ëû
(c) The matrix showing how much more (or less) education males have attained than females is
(c) The matrix showing the difference in the educational attainment between African and Hispanic Americans is 51.27.945.37.95.90.0 59.811.147.98.511.92 66.211.350.89.215.4 73.813.253.49.320.4 78.516.557.010.621.5 81.317.658.512.022.8 84.219.862.913.921.3
50. (a) MJCaCl
M1111
J0100
55.014.155.48.20.45.9
69.220.968.113.61.17.3
77.724.477.518.40.26.0
84.227.884.023.60.24.2
86.630.387.629.61.00.7 éùéùéùêúêúêú êúêúêú êúêúêú êúêúêú -= êúêúêú êúêúêú êúêúêú êúêúêúêúêúêú ëûëûëû
49. (a) The matrix for the educational attainment of African Americans is 4 Years of 4 Years of High School College or or More More
Ca0111
Cl0111
(b) Rows 1 and 2 will stay the same. Since the cats now like Musk, the zeros in rows 3 and 4 change to ones.
MJCaCl
M1111
J0100
Ca1111
Cl1111
198051.27.9
198559.811.1 199066.211.3 199573.813.2 200078.516.5 200581.317.6 201084.219.8 éù êú êú êú êú êú êú êú êú
(b) The matrix for the educational attainment of Hispanic Americans is 4 Years of 4 Years of High School College or or More More
2.4 Multiplication of Matrices
Your Turn 1 3412 1224
3(1)4(2)3(2)4(4) 1(1)2(2)1(2)2(4) 522 310 AB é ùéùêúêú = êúêú ëûëû é
Your Turn 2 35134 24253 AB éùéù êúêú = êúêú ëûëû
AB does not exist because a 23 ´ matrix cannot be multiplied by a 22 ´ matrix. 34351 53242
3(3)4(2)3(5)4(4)3(1)4(2) 5(3)3(2)5(5)3(4)5(1)3(2) 115 213711 BA éùéù êúêú = êúêú ëûëû éù êú = êú ëû éùêú = êú ëû
2.4 Exercises
In Exercises 1-6, let 2462 and. 0340 AB éùéù ==êúêú êúêú ëûëû
1. 2448 22 0306 A éùéù ==êúêú êúêú ëûëû
2. 62186 33 40120 B éùéù -=-=êúêú êúêúëûëû
3. 241224 66 03018 A éùéù -=-=êúêú êúêúëûëû
4. 623010 55 40200 B éùéù ==êúêú êúêú ëûëû
5. 2462 4545 0340 8163010 012200 226 2012 AB éùéù -+=-+êúêú êúêú ëûëû éùéù =+êúêú êúêúëûëû éù êú = êúëû
6. 6224 7373 4003 4214612 28009 362 289 BA éùéù -=-êúêú êúêú ëûëû éùéù =-êúêú êúêú ëûëû éùêú = êúëû
7. MatrixsizeMatrixsize 22 AB ´´ 22
The number of columns of A is the same as the number of rows of B, so the product AB exists. The size of the matrix AB is 22. ´
MatrixsizeMatrixsize
22 BA ´´ 22
Since the number of columns of B is the same as the number of rows of A, the product BA also exists and has size 22. ´
8. A is 33, ´ and B is 33. ´
The number of columns of A is the same as the number of rows of B, so the product AB exists; its size is 33. ´ The number of columns of B is the same as the number of rows of A, so the product BA also exists; its size is 33. ´
9. MatrixsizeMatrixsize
34 AB ´´ 44
Since matrix A has 4 columns and matrix B has 4 rows, the product AB exists and has size 34. ´
MatrixsizeMatrixsize
44 BA ´´ 43
Since B has 4 columns and A has 3 rows, the product BA does not exist.
10. MatrixsizeMatrixsize
46 AB ´´ 33
Since the number of columns of A, 3, is the same as the number of rows of B, 3, the product AB exists. Its size is 46. ´
MatrixsizeMatrixsize
33 BA ´´ 64
The product BA does not exist since the number of columns of B, 6, is not the same as the number of rows of A, 4.
11. MatrixsizeMatrixsize
44 AB ´´ 23
The number of columns of A is not the same as the number of rows of B, so the product AB does not exist.
MatrixsizeMatrixsize
32 BA ´´ 44
The number of columns of B is the same as the number of rows of A, so the product BA exists and has size 32. ´
12. A is 32 ´ and B is 13. ´
The product AB does not exist, since the number of columns of A is not the same as the number of rows of B.
The product BA exists; its size is 12. ´
13. To find the product matrix AB, the number of columns of A must be the same as the number of rows of B.
14. The product matrix AB has the same number of rows as A and the same number of columns as B.
15. Call the first matrix A and the second matrix B
The product matrix AB will have size 21. ´
Step1: Multiply the elements of the first row of A by the corresponding elements of the column of B and add.
2(3)(1)(2)8 58 éùéù êúêú +--= êúêú ëûëû 3 21 2-
Therefore, 8 is the first row entry of the product matrix AB
Step2: Multiply the elements of the second row of A by the corresponding elements of the column of B and add.
21 5(3)8(2)1 éùéùêúêú +-=- êúêúëûëû 3 58 2
The second row entry of the product is 1.-
Step3: Write the product using the two entries found above.
38 21 58 21 AB éùéùéù==êúêúêú êúêúêú ëûëûëû
16. 61652 5 14 27602 70 42 éùéùéùéù -⋅+⋅êúêúêúêú == êúêúêúêú ⋅+⋅ ëûëûëûëû
17. 5 7 21 10 30 4 2 éù êú éù - êú êú êú êú -ëû êú ëû
25(1)1072 (3)5010(4)2 14 23 éù ⋅+-⋅+⋅ êú = êú -⋅+⋅+-⋅ ëû éù êú = êúëû
18. 5 2 0 14 76 212 0 1 éù êú éù êú êú êú êú êúëû êú ëû
5122542(1)5022 7162746(1)7062 1102140(1)1002 918 4 19 2212 0 14 é ù ⋅+⋅⋅+-⋅+⋅ ê ú ê ú =⋅+⋅⋅+-⋅+⋅ ê ú ê ú ⋅+⋅⋅+-⋅+⋅ ë û éù êú êú = êú êú êú ë û
19. 0 2114 3650 2 é ùéù ê úêú ê úêú - ë ûëû
2(1)(1)520(1)(2)24(1)0 3(1)65306(2) 3460 ⋅-+-⋅⋅+-⋅-⋅+-⋅ = ⋅-+⋅⋅+⋅- ⋅+⋅
78 2 27 1212= -
20. 60 416102(4)06 5 1221122505 10310101(1)2308
é ù é êú-⋅+⋅+-ùéùéù ê úêúêú êú ê úêúêú =⋅+⋅+⋅= êú ê úêúêú êú ê êú-⋅+-+⋅úêúêú ë ûëûëû ë û
é ù êú éù - êú êúêú êú êú ëû êú ëû
21. 0 2 221 14 30 1 0 2
é ù ⋅+-+-⋅+⋅+- ê ú = ê ú ⋅+-+ ⋅+⋅+⋅ ë û
éùêú = êú ë û
202(1)(1)02224(1)2 300(1)1(0) 320412 10 2 08
é ù êú éù - êú êúêú êú êú ëûêú ëû
é ù -⋅+⋅-+⋅- ê ú = ê ú ⋅+⋅-+⋅- ë û
22. 3 30 1 1 608 2 (3)31(1)0(2) 630(1)8(2) 10 2
éùêú = êú ë û
é ùéùêúêú êúêú ëûëû
é ù -+⋅ ⋅+⋅
ê ú = ê ú -+⋅ ⋅+⋅ ë û éù êú = êú
23. 5 121 370 4 1(1)27 1520 3(1)47 3540 135 2515
ë û
24. 2180 750 1 21802081 (7)150(7)051 8 2 75 éùéù êúêú êúêúëûëû
éù ⋅+⋅⋅+⋅ êú = êú -⋅+⋅-⋅+⋅ ëû
éù êú = êúëû
25. 1 2(1)(3)273 37 2 2 115263 56 1 3 13 29 éù êú éùéù -+-+⋅ - - êú êúêú = êú êúêú ⋅+⋅+⋅ ëûëû êú ëû éù êú = êú ëû
26. [] 2 9101 12 2(1) 20 21 (9)1(9)0(9)(1) 12(1) 120 121 0 22 909 0 1212 éù êú êú êú êú êú ëû éù ⋅ êú êú =-⋅-⋅-⋅êú êú ⋅ êú ⋅ ëû
éùêú êú =-êú êúêú ëû
27. 5 214 33 12 3615 12 11 21 4408 7 33 4 æö é ùéù - ÷ ç ê úêú ÷ ç é ùéùéù ÷ ç - ê úêú ÷ ç ê úêúêú ÷ --=- ç ê úêú ÷ ê úêúêú ç ÷ ê úêú - ÷ ç ë ûëûëû ÷ ç - ê úêú ÷ ç èø ë ûëû éù êú êú =-êú êú êú ëû
28. 21 3 12 36 1 21 0 4 7 21 1 3633 5 0 44 é ù êú æö é ùéù ÷ - êú ç ÷ ê úêú ç ÷ êú ç ÷ ê úêú ÷ ç èø êú - ë ûëû êú ëû é ùéù ê úêú éù ê úêú êú =--=ê úêú êú ê úêú ëû ê úêú ë ûëû
29. 370 224 5 11121 3 2211 07 11 8 22 114 æö é ùéùéù ÷ - ç ÷ ê úêúêú ç + ÷ ç ÷ ê úêúêú ÷ ç èø ë ûëûëû éùéùêúêú = êúêúëûëû éùêú = êú - ë û
éùéùéùéùéùéùéù êúêúêúêúêúêúêú +=+= êúêúêúêúêúêúêúëûëûëûëûëûëûëû
30. 370616108 22422222 5385 11121111114
31. 25 30 A
A
2 (2)(2)(5)(3)(2)(5)(5)(0)1910 (3)(2)(0)(2)(3)(5)(0)(0)615
2 (0.5)(0.5)(0.2)(0.7)(0.5)(0.2)(0.2)(0.1)0.390.08 (0.7)(0.5)(0.1)(0.7)(0.7)(0.2)(0.1)(0.1)0.280.15
120 312
35. (a) 16 242122 336719 1 AB éùéùéù ==êúêúêú êúêúêú ëûëûëû
(b) 55 2124 363030 1 BA éùéùéù==êúêúêú êúêúêú ëûëûëû
(c) No, AB and BA are not equal here.
(d) No, AB does not always equal BA.
36. Verify that ()(). PXTPXT = () ()()()() ()()()() ymxnzmynw mnxss PXTrr pqpxqzpyqwzwtutu mxnzrmynwtmxnzsmynwu pxqzrpyqwtpxqzspyqwu mxrnzrmytnwtmxsnzsmyu
æö éùéùéùéùéù + + ÷ ç ÷ êúêúêúêúêú ç == ÷ ç ÷ êúêúêúêúêú ÷ ç ++ èø ëûëûëûëûëû éù ++++++ êú = êú ++++++ ëû +++++ = nwu pxrqzrpytqwtpxsqzspyuqwu éù + êú êú ++++++ ëû () ()()()() ()()()() x yrytxsyumnxsmnPXTr pqpqzrwtzswu zwtu mxrytnzrwtmxsyunzswu pxrytqzrwtpxsyuqzswu mxrmytnzrnwtmxsmyunzs
Thus, ()(). PXTPXT =
æö éùéùéùéùéù ++ ÷ ç ÷ êúêúêúêúêú ç == ÷ ç ÷ êúêúêúêúêú ÷ ç ++ èø ëûëûëûëûëû éù ++++++ êú = êú ++++++ ëû +++++ = nwu pxrpytqzrqwtpxspyuqzsqwu mxrnzrmytnwtmxsnzsmyunwu pxrqzrpytqwtpxsqzspyuqwu éù + êú êú ++++++ ëû éù ++++++ êú = êú ++++++ ëû
37. Verify that (). PXTPXPT +=+
Find ()PXT + and PXPT + separately and compare their values to see if they are the same () ()() ()() ()()()() yxrysmnxsmnPXTr pqpqztwu zwtu mxrnztmysnwumymsnwnu mxmrnznt pxrqztpysqwupxprqzqtpy æöæö éùéùéùéùéù + + çç÷÷ ÷÷ êúêúêúêúêú +=+=çç÷÷ çç÷÷ êúêúêúêúêú çç÷÷ ++ èøèø ëûëûëûëûëû éù++++++ +++ +++ êú == êú+++++++++ ëû psqwqu é ù ê ú ê ú +++ ë û ()() ()() ()()()() mxnzmrntmsnuymynwmnxmnsPXPTr pqpqpxqzpyqwprqtpsqu zwtu mxnzmrntmynwmsnumynw mxnzmrnt pxqzprqtpyqwpsqu éùéùéùéùéùéù +++ + +=+=+êúêúêúêúêúêú êúêúêúêúêúêú ++++ ëûëûëûëûëûëû éù +++ +++ ++ +++ êú == êú ++++++ ëû msnu pxqzprqtpyqwpsqu mxmrnzntmymsnwnu pxprqzqtpypsqwqu é ù + ê ú ê ú ++++++ ë û éù +++ +++ êú = êú ++++++ ëû
Observe that the two results are identical. Thus, (). PXTPXPT +=+
38. Prove that () kXTkXkT +=+ for any real number k.
()kXT xyskr zwtu + æö éùéù ÷ ç ÷ êúêú ç =+ ÷ ç ÷ êúêú ÷ ç èø ëûëû kxrys ztwu
æö éù + + ÷ ç ÷ êú ç = ÷ ç ÷ êú ÷ ç ++ èø ëû
()() ()() kxrkys kztkwu éù ++ êú = êú ++ ëû
Distributive property for real numbers kxkrkyks kzktkwku éù + + êú = êú ++ ëû
kxkrks ky kzkwktku éùéù =+êúêú êúêú ëûëû
xs yr kkkXkT zwtu éùéù êúêú =+=+ êúêú ëûëû
39. Verify that ()khPkPhP +=+ for any real numbers k and h
()() ()() ()() mn khPkh pq khmkhn khpkhq kmhmknhn kphpkqhq kmknhmhn kpkqhphq mnmnkh pqpq kPhP
éù êú +=+ êú ëû éù ++ êú = êú ++ ëû éù ++ êú = êú ++ ëû éùéù =+êúêú êúêú ëûëû éùéù =+êúêú êúêú ëûëû =+
Thus, ()khPkPhP +=+ for any real numbers k and h.
40. (a) 10 01 mnmn IPP pqpq éùéùéù êúêúêú === êúêúêú ëûëûëû
Thus, . IPP = 10 01 mnmn PIP pqpq éùéùéù êúêúêú === êúêúêú ëûëûëû
Thus, .PIP = 10 01 Ixyxy XX zwzw éùéùéù êúêúêú === êúêúêú ëûëûëû
Thus, IXX = (b) ITT =
The matrix I is called an identity matrix because it acts like the multiplicative identity for real numbers, which is 1. If x is a real number, 11. xxx⋅=⋅=
If X is a 22 ´ matrix, IXXIX ==
(c) I is called an identity matrix, because if a matrix is multiplied by I (assuming that these matrices are 22), ´ the resulting matrix is the matrix you originally started with. Hence, I maintains the identity of any 22 ´ matrix under multiplication.
41. 1 123 2 123 3 23 3 21 , 45 5 14 x xxx x xxx x é ù êú éù éù ++ êú êú êú = êú êú êú -+ëû ëû êú ë û and 123 123 23 5 45 8 xxx xxx é ù éù ++ êú êú = êú êú -+ ëû ë û
This is equivalent to 123 123 235 458 xxx xxx ++= -+=
since corresponding elements of equal matrices must be equal. Reversing this, observe that the given system of linear equations can be written as the matrix equation 1 2 3 5 3 21 8 5 14 x x x éù êú é ùéù êú ê úêú = êú ê úêú - ë ûëû êú ëû
42. 1 2 4 12 ,, 12 35 x AXB x éù é ùéùêú ê ===úêú êú ê úêú - ë ûëû ëû
If , AXB = then 1 2 4 12 12 35 x x éù é ùéùêú ê úêú = êú ê úêú - ë ûëû ëû
Using the definition of matrix multiplication, 12 12 12 4 35 12 xx xx é ù éù +êú êú = êú êú -+ ëû ë û
By the definition of equality of matrices, corresponding elements must be equal, so 12 12 24() 3512.() xx1 xx2 +=-+=
This is a linear system of two equations in two variables, x1 and x2
Solve this system by the elimination (addition) method. Multiply equation (1) by 3 and add the result to equation (2). 12 12 2 2 3612 3512 110 0 xx xx x x +=-+= = =
Substitute 0 for x2 in equation (1) to find x1. 1 1 2(0)4 4 x x +==-
The solution of the system is (–4, 0).
Substitute 4- for x1 and 0 for x2 in the matrix equation to check this result.
1(4)2(0)4 124 3(4)5(0)12 350 éùéùéùéù -+-êúêúêúêú == êúêúêúêú --+ -
43. (a) Use a graphing calculator or a computer to find the product matrix. The answer is 610615828 222 1201396475115 146184129 2144 10694116110 24 AC éù êú êú êú = êúêú êú êú ëû
(b) CA does not exist.
(c) AC and CA are clearly not equal, since CA does not even exist.
44. This exercise should be solved by graphing calculator or computer methods. The answers are as follows:
(a)
CD éù êú êúêú êú = êú êú êú êú êú ëû (b) 110963022637 941271348733 ; 52126193153 22 117565514757 5469583731 DC éù êú êú êú êú =êú êúêú êú êú ëû
756033 4411 2016916418105 ; 1138223921855 11983782106 1622017514374
(c) No, .CDDC ¹
45. Use a graphing calculator or computer to find the matrix products and sums. The answers are as follows.
(a) 159131 7172106 189121222 9418103 1610285 CD é ù ê ú ê ú - ê ú ê ú += - ê ú ê
(b) 99077 2 6312762 42 () 4137618056 2919885 44 13720162103 CDB é ù - - ê ú ê ú - -
ù - -
ú ê ú ê
ú = - ê ú
(c) 5645 11 15611976 122 3158611891 171711651 1181912577
ê ú ê ú ê ú ë û (d) 5488932 565160 114 98106235 278234 12 193726 1
CBDB é ù - - ê ú ê ú - - ê ú ê ú += - ê ú ê ú - - ê ú ê ú ê ú ë û
DB é ù - ê ú ê ú - ê ú ê ú = - ê ú ê ú - - ê ú ê ú ê ú ë û (e) 99077 2 6312762 42 4137618056 2919885 44 13720162103
(f) Yes, () CDB + and CBDB + are equal, as can be seen by observing that the answers to parts (b) and (e) are identical.
46. Exercise 45 illustrates the distributive property.
47. (a) 3 2 10356 4 11 738 22 3 4 415010 33 0355 4 12 éù êú éù êú êú êú êú êú êú êú êú êú êú êú êú êú êú ëû êú ëû AB Dept.1 5770
Dept.2 54 41
Dept.3 2740
Dept.4 3940 é ù ê ú ê ú ê ú = ê ú ê ú ê ú ê ú ë û
(b) The total cost to buy from supplier A is 57412739$164, +++= and the total cost to buy from supplier B is 7054++ 4040$204. += The company should make the purchase from supplier A, since $164 is a lower total cost than $204.
48. (a) CCMMAD
S0.50.40.3
C0.20.30.3 éù êú êú ëû
(b) SC SD43 MC25 M17 éù êú êú êú êú ëû
(c) 43 0.50.40.3 25 0.20.30.3 17 éù êú éù êú êú êú êú ëû êú ëû
CCMMAD
The rows give the average price per pair of footwear sold by each store, and the columns give the state.
(d) The average price of footwear at a Fred’s outlet in Arizona is $108.
SD2.62.52.1
MC22.32.1
M1.92.52.4 éù êú êú = êú êú ëû
(d) Look at the entry in row 3, column 2 of the last matrix. The cost is $2.50.
(e)
(a) 20 10202180 50120260 25 QR éù êú êú éù êú êú = êú êú ëû êú êú êú ëû
2.62.52.1 1001810
2.32.1 22001710
1.92.52.4 5001890
éù éùéù êú êúêú êú êúêú = êú êúêú êú êúêú êú ëûëû ëû
The total sugar and chocolate cost is $1810 in San Diego, $1710 in Mexico City, and $1890 in Managua, so the order can be produced for the lowest cost in Mexico City.
49. (a) ShSaB Sal's8040120 Fred's6030150 P éù êú = êú ëû
10(20)2(180)0(60)2(25)
50(20)1(180)20(60)2(25) 610 2430 é ù +++ ê ú = ê ú +++ ë û éù êú = êú ë û
The rows represent the cost of materials for each type of house.
(c) 1 1 5 2 8040120 11 45 6030150 1 3 4 5 PF éù êú éù êú êú êú êú = êú êú êú êú êú ëû êú êú ëû ( ) ( ) ( ) ( ) ( ) ( ) () () () () () () 3 11111 80401208040120 244555 3 11111 60301506030150 244555 8096 75108 ++++ = ++++ é ù ê ú ê
Sh Sa B F éù êú êú = êú êú êú êú ëû
(b) 1 1 5 2 11 45 13 4 5 CAAR
(b) From part (a), 610 . 2430 QR éù êú = êú ëû 03072,900 610 ()102054,700 2430 202060,800 PQR é ù é ù êú ê ú éù êú ê ú êú == êú ê ú êú êú ëû ê ú êú ë û ë û 20 030 10202180 ()1020 50120260 2020 25 PQR é ù æö éù ê ú ÷ ç êú÷ ç ê ú éù ÷ ç êú ÷ ê ú ç êú ÷ = ç êú ÷ ê ú êú ç ÷ êú ÷ ç ëû ê ú ÷ ç êú ÷ ç ê ú èø ëû ê ú ë û
15003060060 180 11004040060 60 12006040080 25 72,900 54,700 60,800 éù êú éù êú êú êú êú = êú êú êú êú ëûêú êú ëû éù êú êú = êú êú ëû
Therefore, ()(). PQRPQR =
51. 1712151519121713 1 2418362533203121 3
52. (a) To increase by 25%, multiply by 1.25 or 5 4 To increase by 1 3 , multiply by 1 3 1 or 4 3 . To increase by 10% multiply by 1.10 or 11 10 The matrix is 5 4 4 3 11 10 éù êú êú êú
(b) 5 4 4 3 11 10 8810560316 487240200 1621048 11214750391 éùéù éù êúêú êú êúêú êú êúêú êú = êúêú êú êúêú êú êúêú êú êúêú ëû ëûëû
53. (a) 507 2121205227 0101 3221256235 0152 4321307243 10128 XY éù êú éùéù êú êúêú êú êúêú == êú êúêú êú êúêú êú ëûëû êú ëû
The rows give the amounts of fat, carbohydrates, and protein, respectively, in each of the daily meals.
(b) 50775 8 010145 4 015270 5 10128168 YZ éùéù êúêú éù êúêú êú êúêú êú ==êúêú êú êúêú êú êúêú ëû êúêú ëûëû
The rows give the number of calories in one exchange of each of the food groups.
(c) Use the matrices found for XY and YZ from parts (a) and (b).
2052278503 ()2562354623 3072435743
75 2121503 45 ()3221623 70 4321743 168 XYZ
The rows give the number of calories in each meal.
55.
60.068.368.075.666.774.4 63.872.570.778.169.677.2 15 64.573.672.779.471.378.4 66 68.275.174.779.973.679.1
56. (a) Let n represent the present year. Then 900, nj = 500, ns = 2600. na =
For the year, 1, n + we have 000.33 1 0.1800 1 00.710.94 1 000.33900 0.1800500 00.710.942600 858 162. 2799 nn nn nn jj ss aa é ù éù éù êú + êú êú êú êú êú = êú êú + êú êú êú êú êú êú ëû êú + ëû ëû
é ùéù ê úêú ê úêú = ê úêú ê úêú ë ûëû éù êú êú = êú êú ë û
For the year 1, n + there is a total of 858 + 16227993819 += female owls.
é ùéù éù êúêú êú êúêú êú êúêú êú êúêú êú êúêú ëû ëûëû
é ùéùéù ê úêúêú ê úêúêú ê úêúêú ê úêúêú ë ûëûëû
For the next year, 2, n + we have 21 21 21 000.33 0.1800 00.710.94 000.33858924 0.1800162154. 00.710.9427992746 nn nn nn jj ss aa ++ ++ ++ = =»
For the year 2, n + there is a total of approximately 92415427463824 ++= female owls.
For the next year, 3, n + we have
32 32 32 000.33 0.1800 00.710.94
000.33924906 0.1800154166. 00.710.9427462691 nn nn nn jj ss aa ++ ++ ++ = =» éùéù éù êúêú êú êúêú êú êúêú êú êúêú êú êúêú ëû ëûëû
éùéùéù êúêúêú êúêúêú êúêúêú êúêúêú ëûëûëû
For the year 3, n + there is a total of approximately 90616626913763 ++= female owls.
For the next year, 4, n + we have 43 43 43 000.33 0.1800 00.710.94
000.33906888 0.1800166163. 00.710.9426912647 nn nn nn jj ss aa ++ ++ ++ éùéù éù êúêú êú êúêú êú = êúêú êú êúêú êú êúêú ëû êúêú ëûëû
êúêúêú êúêúêú =» êúêúêú êúêúêú ëûëûëû
For the year 4, n + there is a total of approximately 88816326473698 ++= female owls.
For the year 5, n + we have 54 54 54 000.33 0.1800 00.710.94
000.33888 0.1800163 00.710.942647 874 160. 2604 nn nn nn jj ss aa ++ ++ ++ éùéù éù êúêú êú êúêú êú = êúêú êú êúêú êú êúêú ëû êúêú ëûëû éùéù êúêú êúêú = êúêú êúêú ëûëû éù êú êú » êú êú ëû
For the year 5, n + there is a total of approximately 87416026043638 ++= female owls.
(b) Each year, the population is about 98 percent of the population of the previous year. In the long run, the northern spotted owl will become extinct.
(c) Change 0.18 in the original matrix equation to 0.40.
For the next year, 1, n + we have 1 1 1 000.33 0.4000 00.710.94 nn nn nn jj ss aa
ëû
900 000.33 0.4000500 00.710.942600 858 360. 2799 é ù é ù ê ú êú ê ú êú = ê ú êú ê ú êú ëû ë û éù êú êú = êú êú ë û
For the year 1, n + there would be a total of 85836027994017 ++= female owls. For the next year, 2, n + we have
21 21 21 000.33 0.4000 00.710.94 000.33858924 0.4000360343. 00.710.9427992887 nn nn nn jj ss aa ++ ++ ++ é ùéù é ù êúêú êú êúêú êú = êúêú êú êúêú êú êúêú ëû ëûëû é ùéùéù ê úêúêú ê =»úêúêú ê úêúêú ê úêúêú ë ûëûëû
For the year 2, n + there would be a total of approximately 92434328874154 ++= female owls. For the next year, 3, n + we have 32 32 32 000.33 0.4000 00.710.94 000.33924953 0.4000343370. 00.710.9428872957 nn nn nn jj ss aa ++ ++ ++ é ùéù é ù êúêú êú êúêú êú = êúêú êú êúêú êú êúêú ëû ëûëû é ùéùéù ê úêúêú ê ==úêúêú ê úêúêú ê úêúêú ë ûëûëû
For the year 3, n + there would be a total of approximately 95337029574280 ++= female owls.
For the next year, 4, n + we have 43 43 43 000.33 0.4000 00.710.94 nn nn nn jj ss aa ++ ++ ++ é ùéù éù ê úêú êú ê úêú êú = ê úêú êú ê úêú êú ê úêú ëû ë ûëû 000.33953 0.4000370 00.710.942957
976 381. 3042 é ùéù ê úêú ê úêú = ê úêú ê úêú ë ûëû éù êú êú » êú êú ë û
For the year 4, n + there would be a total of approximately 97638130424399 ++= female owls.
For the next year, 5, n + we have 4 5 54 5
éù éù êú êú êú êú êú êú = êú êú êú êú êú êú êú êú ëû ëû êú ëû éùéù êúêú êúêú = êúêú êúêú ëûëû éù êú êú » êú êú ëû
ss aa + + ++ + +
00.710.94 4 000.33976 0.4000381 00.710.943042 1004 390. 3130 nn nn nn
For the year 5, n + there would be a total of approximately 100439031304524 ++= female owls.
Assuming that better habitat management could increase the survival rate of juvenile female spotted owls from 18 percent to 40 percent, the overall population would increase each year and would, therefore, not become extinct.
57. (a) The matrices are
0.03460.0118
0.01740.0073
0.01890.0059
êú êú êú êú = êú êú êú êú êú ëû éù êú êú êú êú = êú êú êú êú êú ëû
0.01350.0083
0.00990.0103
3612038286227460
4732494362252484
6272978443278499
8033435524314511
10133824591344522
(b) The total number of births and deaths each year is found by multiplying matrix B by matrix A
36120382862274600.03460.0118
47324943622524840.01740.0073
62729784432784990.01890.0059
80334355243145110.01350.0083 101338245913445220.00990.0103 BA = éùéù êúêú êúêú êúêú êúêú êúêú êúêú êúêú êúêú ëûëû
Births Deaths
197060.9827.45
198074.8033.00
199090.5839.20
2000106.7545.51 2010122.5751.59 éù êú êú êú êú = êú êú êú êú êú ëû
2.5 Matrix Inverses
Your Turn 1
Use row operations to transform the augmented matrix so that the identity matrix is the first three columns. 231100 121010 332001 A
12 RR121010 231100 332001
1312 22 7777 121010 RR010 095031
3513
937 888 RRR100 RRR010
888 513
888 937 888
Your Turn 2
Solve the linear system
5423 436 xy xy += -=
Let 54 43 A éù êú = êúëû be the coefficient matrix,
23 , 6 B éù êú = êú ëû and . x X y éù êú = êú ëû
First find 1 A5410 4301 éù êú êúëû
141 11 555 RR10 4301 éù êú êú êúëû
41 55 314 122 55 10 4RRR 01 éù êú êú êú -+ êú ëû
41 55 5 45 22 31 3131 10 RR 01 éù êú êú êú -êú ëû
34 4 211 3131 5 45 3131 10 RRR 01 éù -+ êú êú êúêú ëû
34 3131 1 45 3131 Aéù êú êú = êúêú ëû
34 3131 1 45 3131 233 62 XABéù éùéù êú êúêú êú ===êúêú êú ëûëûêú ëû
The solution to the system is (3,2).
Your Turn 3
(a) The word “behold” gives two 31 ´ matrices: 215 5and12 84 éùéù êúêú êúêú êúêú êúêú ëûëû
Use the coding matrix 134 213 421 A éù êú êú = êú êú ëû to find the product of A with each column matrix.
1342491341567 213533 and 2131254 421826421488 == é ùéùéùéùéùéù ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ë ûëûëûëûëûëû
The coded message is 49, 33, 26, 67, 54, 88. (b) Use the inverse of the coding matrix to decode the message 96, 87, 74, 141, 117, 114. 1 0.20.20.296141
0.40.60.2, 87, and 117 00.40.274114 Aé ùéùéùê úêúêú ê úêúêú =ê úêúêú ê úêúêú - ë ûëûëû
0.20.20.29613
0.40.60.2871 00.40.27420
0.20.20.214118
0.40.60.21179 00.40.211424
The message is the word “matrix.”
2.5 Warmup Exercises
W1.
W3.
233 22 33 1024 01617 2RRR 001133 1024 (1)RR 01617 1 0013 RR
311 22 (2)RRR 1002 0101 (6)RR 0013
The solution is (2,1,3).
0.20.54(0.2)(4)(0.5)(6)3.8 0.41.16(0.4)(4)(1.1)(6)5.0
W4. 1101(1)(1)(1)(2)(0)(3)3 0352(0)(1)(3)(2)(5)(3)9 2423(2)(1)(4)(2)(2)(3)16
2.5 Exercises
---+
1. 21316522 5352151556 10 01 31216533 5253101056 10 01 I I
2. 1474 2721 éùéù êúêú êúêú ëûëû 7844 141487 éù -+êú = êú -+ëû 10 01 I éù êú == êú ëû 7414 2127 é ùéù ê úêú ê úêú ë ûëû 782828 2287 10 01 I é ù -+- ê ú = ê ú -+- ë û éù êú == êú ë û
Since the products obtained by multiplying the matrices in either order are both the 22 ´ identity matrix, the given matrices are inverses of each other.
3. 26121020 2424612 I é ùéùéù êúêúêú=¹ êúêúêú ë ûëûëû
No, the matrices are not inverses of each other since their product matrix is not I.
4. 125210 353101 I é ùéùéù êúêúêú=¹ êúêúêú ë ûëûëû
Since this product is not the 22 ´ identity matrix, the given matrices are not inverses of each other.
éù êú == êú ëû éùéùéù êúêúêú = êúêúêú --+-+ ëûëûëû éù êú == êú ëû
Since the products obtained by multiplying the matrices in either order are both the 22 ´ identity matrix, the given matrices are inverses of each other.
5. 201111 112010 010122 é ùéùê úêú ê úêú ê úêú ê úêú ë ûëû 201202202 102114104 000010000 100 123 010 I é ù +-+--++ ê ú ê ú =+-+--++ ê ú ê ú ++++++ ë û éù êú êú =--¹ êú êú ë û
No, the matrices are not inverses of each other since their product matrix is not I.
6. 010101100 002100020 110010001 I é ùéùéù êúêúêú êúêúêú -=¹ êúêúêú êúêúêú ë ûëûëû
Since this product is not the 33 ´ identity matrix, the given matrices are not inverses of each other.
7. 133733100 143110010 134101001 733133100 110143010 101134001 I I éùéùéù êúêúêú êúêúêú -== êúêúêú êúêúêúëûëûëû éùéùéù êúêúêú êúêúêú -== êúêúêú êúêúêúëûëûëû
Yes, these matrices are inverses of each other.
8. 112 333 100100 123001 010 éù éù êú êú êú êú êú êú êú êú êú ëû êú ëû 100000000 101001022 000000010 éù ++++++ êú êú
100 010 001 I éù êú êú == êú êú ëû 112 333 100100 001123 010 éù éù êú êú êú êú êú êú êú êú êú ëû êú ëû 1122 3333 100000000 000001000 00 010 éù ++++++ êú êú êú =++++++ êú êú -+-+ ++ êú ëû
100 010 001 I éù êú êú == êú êú ëû
Since the products obtained by multiplying the matrices in either order are both the 33 ´ identity matrix, the given matrices are inverses of each other.
9. No, a matrix with a row of all zeros does not have an inverse; the row of all zeros makes it impossible to get all the 1’s in the main diagonal of the identity matrix.
10. Since the inverse of A is 111 ,()AA would be the inverse of the inverse of A, which is A.
11. Let 11 . 20 A éùêú = êú ëû
Form the augmented matrix [ ] |.AI
[] 1110 | 2001 AI éùêú = êú ëû
Perform row operations on [ ] | AI to get a matrix of the form [ ] |.IB
122 121 11 11 22 11 22 22 1110 2001 1110 2RRR 0221 2RRR2001 0221 RR100 | RR011 IB éùêú êú ëû éùêú êú -+ëû éù + êú êúëû éù êú êú = êú -êú ëû
The matrix B in the last transformation is the desired multiplicative inverse. 1 2 1 1 2 0 1 Aéù êú êú = êúêú ëû
This answer may be checked by showing that –1 AAI = and –1 AAI =
12. Find the inverse of 11 , 23 A éù êú = êú ëû if it exists.
Write the augmented matrix [ ] |.AI [] 1110 | 2301 AI éù êú = êú ëû
Perform row operations on [ ] | AI to get a matrix of the form [ ] |.IB 122 211 1110
2RRR 0121
1RRR1031 0121 éù êú êú -+ëû -+-éù êú êúëû
The last augmented matrix is of the form [ ] |,IB so the desired inverse is 1 31 . 21 Aéùêú = êúëû
13. Let 31 52 A éùêú = êúëû
[] 3110 | 5201 AI éùêú = êúëû
122 121 1 11 3 3110 5R3RR 0153 RRR3063 0153 1021 RR | 0153 IB éùêú êú + ëû éù + êú êú ëû éù êú = êú ëû
The desired inverse is 1 21 . 53 Aéù êú = êú ëû
14. Find the inverse of 38 , 13 A éù êú = êú ëû if it exists.
Write the augmented matrix [ ] |.AI
[] 3810 | 1301 AI éù êú = êú ëû
Perform row operations on [ ] | AI to get a matrix of the form [ ] |.IB
122 211 1 11 3 3810 R3RR 0113 8RRR30924 0113 1038 RR 0113 éù êú êú + ëû +-éù êú êúëû
éù - êú êú ëû
The last augmented matrix is of the form [ ] |,IB
so the desired inverse is –1 38 . 13 A éù êú = êú ëû
Note that matrix A is its own inverse: . AAI =
15. Let 13 . 26 A éùêú = êúëû
[] 1310 | 2601 AI éùêú = êúëû
122 1310 2RRR 0021 éùêú êú + ëû
Because the last row has all zeros to the left of the vertical bar, there is no way to complete the desired transformation. A has no inverse.
16. Find the inverse of 510 , 36 A éù êú = êú ëû if it exists.
[] 122 51010 | 3601 51010 3R5RR 0035 AI éù êú = êú ëû éù êú êú + ëû
At this point, the matrix should be changed so that the first row, second column element will be 0. Since this cannot be done using row operations, the inverse of the given matrix does not exist.
17. Let 100 010. 101 A é ù ê ú ê ú =ê ú ê ú ë û
[] 100100 |010010 101001 AI é ù ê ú ê ú =ê ú ê ú ë û 133 100100 010010 1RRR001101 é ù ê ú ê ú - ê ú ê ú -+ë û 22 100100 1RR010010 001101 é ù ê ú ê ú -ê ú ê ú - ë û 1 100 010 101 Aé ù ê ú ê ú =ê ú ê ú - ë û
18. Find the inverse of 130 021, 102 A éù êú =-êú êú êú ëû if it exists.
[] 130100 |021010 102001 AI é ù ê ú ê ú =ê ú ê ú ë û 133 130100 021010 1RRR032101 é ù ê ú ê ú - ê ú ê ú -+-ë û 211 233 3R2RR203230 021010 3R2RR001232
é ù -+ê ú ê ú - ê ú ê ú +ë û
311 322 3RRR2008126 RRR020242 001232 éù -+-êú +-êú êú êúëû
1 11 2 1 22 2 RR 100463 RR010121 001232 éù êú -êú êú êúëû 1 463 121 232 Aéù êú =-êú êú êúëû
19. Let 111 450. 013 A éù êú êú = êú êúëû [] 111100 |450010 013001 AI éù êú êú = êú êúëû 122 111100 4RRR014410 013001 éù êú +-êú êú êúëû
211 233 RRR105510 014410 1RRR001411 +--éù êú êúêú -+--êú ëû
311 322 5RRR1001545 4RRR0101234 001411 +---éù êú +--êú êú êú ëû 11 1RR1001545 0101234 001411 éù -êú êú êú êú ëû 1 1545 1234 411 Aéùêú =--êú êú êú ëû
20. Find the inverse of 210 031, 413 A éù êú êú = êú êú ëû if it exists. [] 210100 |031010 413001 AI éù êú êú = êú êú ëû
133 210100 031010 2RRR033201 éù êú êú êú -+---êú ëû
211 233 1R3RR601310 031010 RRR002211 -+--éù êú êú êú +--êú ëû
311 322 1R2RR1200831 R2RR010231 001211 éù -+-êú +-êú êú êú ëû
211 1 11 3412 12 1111 22 6326 111 33 222 100 RR RR010 RR0011 éù êú êú êú -êú êú êú --êú ëû 211 3412 111 1 326 11 22 1 Aé ù ê ú ê ú ê ú =ê ú ê ú ê ú ê ú
21. Let 123 321. 101 A éù êú =---êú êú êúëû
[] 123100 |321010 101001 AI éù êú =---êú êú êúëû
122 133 123100 3RRR048310 RRR024101 éù êú êú + êú êú + ëû
211 233 R(2R)R202110 048310 R(2R)R000112 +--éù êú êú êú êú +-ëû
Because the last row has all zeros to the left of the vertical bar, there is no way to complete the desired transformation. A has no inverse.
22. Find the inverse of 204 101, 002 A éù êú =-êú êú êúëû if it exists.
[] 204100 |101010 302001 AI éù êú =-êú êú êúëû
122 133 204100 R(2)RR006120 3R(2)RR0016302 éù êú +--êú êú êú +-ëû
Since the second column is all zeros, it will not be possible to get a 1 in the second row, second column position. Therefore, the inverse of the given matrix does not exist.
23. Find the inverse of 132 273, 385 A éùêú =-êú êú êúëû if it exists. [] 132100 |273010 385001 AI éùêú =-êú êú êúëû 122 133 132100 2RRR011210 3RRR011301 éùêú -+-êú êú -+--êú ëû
211 233 3RRR105730 011210 RRR002511 -+--éù êú êúêú +-êú ëû
311 322 5R2RR2001115 1R2RR020111 002511 +--éù êú êú -+êú êúëû 11115 11 2222 1111 22 2222 1511 33 2222 RR100 RR010 RR001 éù --êú êú êú êú êú êú -êú ëû 1115 222 111 1 222 511 222 Aéù êú êú êú =-êú êú êúêú ëû
24. Find the inverse of 414 211, 245 A éùêú =-êú êú êú ëû if it exists. [] 414100 |211010 245001 AI éùêú =-êú êú êú ëû 122 133 414100 R2RR012120 R2RR076102 éùêú -+-êú êú +-êú ëû
211 233 1RRR406220 012120 7RRR00206142 é ù -+-ê ú ê ú - ê
ê
-+--
ê ú +ë û 311 322 3R10RR40002226 1R10RR0100462 00206122 é ù +
ú ê ú - ë û 11113 11 40202020 1231 22 10555 1371 33 20101010 RR100 RR010 RR001 é ù ê
ê ú =-ê ú ê ú ê ú - ê ú ë û 25. Let 1230 0111 2224 0231 A é ùê ú ê ú - ê ú = ê ú ê
ê ú - ê ú ë û [] 12301000 01110100 | 22240010 02310001 AI é ùê ú ê ú - ê ú = ê ú ê ú ê ú - ê ú ë û 133 12301000 01110100 02442010 2RRR 02310001 é ùê ú ê ú - ê ú ê ú - ê ú ê ú + - ê ú ë û 211 233 244 2RRR 10121200 01110100 2RRR 00262210
2RRR 00110201 é ù + ê ú ê ú - ê ú ê ú + ê ú ê ú -+ ê ú ë û
311 322 344 R(2)RR 20020210
R2RR 02082410 00262210
R2RR 00042212 é ù +- ê ú ê ú + ê ú ê ú ê ú ê ú + - ê ú ë û
411 422 433 R(2)RR40002212
2RRR02002814
3R2RR004021016 00042212
é ù +-ê ú ê ú -+--ê ú ê ú -+--ê ú ê ú - ê ú ë û
Carry out these row operations to get the next matrix: 122 2RRR,-+ 133 3RRR,-+ 144 –1RRR. + 11021000 03152100 00283010 01121001 éù êú êú êú êú êú êú êú ëû
Carry out the row operations 211 R3RR, + 244 R3RR + 30111100 03152100 00283010 004115103 éù êú êú êú êú êú êú êú ëû
311 R(2)RR, +- 322 R(2)RR, +-
344 2RRR-+
600105210 06021210 00283010 00051123 éù êú êúêú êú êú êúêú ëû
411 2RRR, + 422 2R5RR,-+
- ê
û 1 11 6 RR,- 1 22 30 RR, 1 33 10 RR, 1 44 5 RR 11 22 1231 105105 741112 105105 1123 5555 100001 0100 0010 0001 é
ë
- ê
=
11 22 1231 105105 1 741112 105105 1123 5555 01 Aé ù - ê ú
- ê ú ë û
27. 2515 49 xy xy += +=
First, write the system in matrix form. 2515 149 x y é ùéùéù ê úêúêú = ê úêúêú ë ûëûëû
Let 2515 ,,and. 149 x AXB y é ùéùéù ê ===úêúêú ê úêúêú ë ûëûëû
The system in matrix form is . AXB = We wish to find –1–1 .XAAXAB == Use row operations to find A–1
[] 2510 | 1401 AI éù êú = êú ëû
122 2510 1R2RR 0312 éù êú êú -+ëû
211 5R3RR60810 0312 -+-éù êú êúëû
145 11 633 112 22 333 RR10 RR01 éù -êú êú êú -êú ëû
433 8R5RR + 60003036 0300031296 00100781124 00051123
45 33 1 12 33 45 1 12 3 Aéù - éù êúêú êú == êú êúëûêú ëû
Next find the product 1 AB45 33 1 12 33 15 9 4515 1 129 3 155 1 31 3 XABéù - éù êú êú êú == êú êú ëûêú ëû éùéùêúêú = êúêúëûëû éùéù ==êúêú êúêú ëûëû
Thus, the solution is (5, 1).
28. 215 220 xy xy -+= --=
This system may be written as the matrix equation 1215 . 2120 x y éùéùéùêúêúêú = êúêúêú ëûëûëû
In Exercise 12, it was calculated that 12 1 55 21 55 12 , 21 - éù éù êúêú êú = êú êú ëûêú ëû
and we now know that –1 XAB = is the solution to AXB = Thus, 12 55 21 55 1511 , 202 x y éù éùéùéù - êú êúêúêú êú == êúêúêú êú ëûëûëûêú ëû and (11, - 2) is the solution of the system.
29. 2 5 5313 xy xy += +=
Let 215 ,,. 5313 x AXB y éùéùéù ===êúêúêú êúêúêú ëûëûëû
Use row operations to obtain 1 31 52 Aéùêú = êúëû 1 3152 52131 XABéùéùéù===êúêúêú êúêúêúëûëûëû
The solution is (2, 1).
30. 2 8 3424 xy xy --= +=
This system may be written as the matrix equation 128 3424 x y é ùéùéù ê úêúêú = ê úêúêú ë ûëûëû
In Exercise 14, it was calculated that 1 31 22 21 12 , 34 - éù éù êú êú = êú êú êú ëû ëû and since –1 ,XAB = we get 31 22 21 840 2424 x y é ù é ùéùéù êú ê úêúêú == êú ê úêúêú - êú ë ûëûëû ë û
Therefore, the solution is (40, 24).-
31. 323 750 xy xy -= -=
First, write the system in matrix form. 323 750 x y é ùéùéùê úêúêú = ê úêúêú - ë ûëûëû
Let 32 , 75 A é ùê ú = ê ú - ë û , x X y éù êú = êú ëû and 3 0 B é ù ê ú = ê ú ë û
The system is in matrix form AXB = We wish to find 1 XA= 1 . AXAB= Use row operations to find 1 A[] 122 211 1 11 3 22 3210 | 7501 3210 7R3RR 0173 2RRR 30156 0173 RR 1052 1RR 0173 AI éùêú = êúëû éùêú êú -+ ëû éù -+êú êú ëû éùêú êú -ëû 1 52 73 Aéùêú = êúëû
Next find the product 1 AB1 52315 73021 XABé ùéùéùê ===úêúêú ê úêúêú - ë ûëûëû
Thus, the solution is (15, 21).
32. 361 591 xy xy -= -+=-
This system may be written as the matrix equation 361 . 591 x y éùéùéùêúêúêú = êúêúêú ëûëûëû
Calculate the inverse of 36 59 A éùêú = êúëû 1 5 3 32 , 1 Aéù êú = êú êú ëû so 52 33 321 1 . 1 1 x y éùéùéùéù êúêúêúêú êúêú==êúêúëûëûêúêú ëûëû
The solution is () 2 3 1,.
33. 812
32436 xy xy --= +=-
Let 18 , 324 A éù êú = êú ëû , x X y éù êú = êú ëû 12 36 B
Using row operations on [ ] | AI leads to the matrix 1810 , 0031 éùêú êú ëû
but the zeros in the second row indicate that matrix A does not have an inverse. We cannot complete the solution by this method.
Since the second equation is a multiple of the first, the equations are dependent. Solve the first equation of the system for x. 812 812 812 xy xy xy --= -=+ =--
The solution is (812,), yy where y is any real number.
[] 2710 | 41401 AI éù êú = êú ëû
122 2710 2RRR 0021 éù êú êú -+ ëû
The zeros in the second row indicate that matrix A does not have an inverse. Since the second equation is a multiple of the first, the equations are dependent. Solve the first equation for x. 2714 2147 147 2 xy xy y x += =-=
The solution is ( ) 147 2 ,,yy - where y is any real number.
35. 1 452 33 xyz xy yz ---= +=-= has coefficient matrix 111 450. 013 A é ù ê ú ê ú = ê ú ê ú - ë û
In Exercise 19, it was found that 1 1 111 450 013 1545 1234. 411 Aé ù ê ú ê ú = ê ú ê ú ë û é ùê ú ê ú =-ê ú ê ú ë û
Since 1 ,XAB= 154518 123426. 41131 x y z é ùéùéùéù ê úêúêúêú ê úêúêúêú =---= ê úêúêúêú ê úêúêúêú ë ûëûëûëû
34. 2714
41428 xy xy += +=
First, write the system in matrix form. 2714 41428 x y éùéùéù êúêúêú = êúêúêú ëûëûëû
Let 27 414 A éù êú = êú ëû Use row operations to find 1 A-
The solution is (8, - 6, 1).
36. 21 38 438 xy yz xyz += += --=
has coefficient matrix
210 031. 413 A éù êú êú = êú êú ëû
In Exercise 20, it was calculated that 211 1 3412 111 1 326 11 22 210 031. 413 1 Aéù êú éù êú êú êú êú ==-êú êú êú êú êú ëû êú ëû
Since 1 ,XAB= 211 3412 111 326 11 22 12 85. 87 1 x y z éù êú éùéùéù - êú êúêúêú êú êúêúêú =-= êú êúêúêú êú êúêúêú - êú ëûëûëû êú ëû
Thus, the solution is (2, - 5, 7).-
37. 324 2738 3854 xyz xyz xyz +-= +-= +-=has coefficient matrix 132 273. 385 A éùêú =-êú êú êúëû
In Exercise 23, it was calculated that 1115 1 222 111 1 222 511 222 132 273 385 1115 1 111. 2 511 Aéù êú éù - êú êú êú êú =-=-êú êú êú êú - êú ëûêú ëû éù êú =-êú êú êúëû
Since 1 XAB= 11154 1 1118 2 5114 7236 1 168 2 168 x y z éùéùéù êúêúêú êúêúêú =êúêúêú êúêúêú ëûëûëû éùéù êúêú ==êúêú êúêú êúêú ëûëû
Thus, the solution is (36, - 8, 8).-
38. 4417 212 24517 xyz xyz xyz +-= +-= --+=
has coefficient matrix 414 211. 245 A é ùê ú ê ú =ê ú ê ú ë û
In Exercise 24, it was calculated that 1113 202020 231 1 555 371 101010 414 211 245 1113 1 8124. 20 6142 Aéù êú éù - êú êú êú êú =-=--êú êú êú êú êú ëûêú ëû éù êú =--êú êú êú - ë û
Since 1 ,XAB= 111317 1 812412 20 614217 x y z é ùéùéù ê úêúêú ê úêúêú =-- ê úêúêú ê úêúêú - ë ûëûëû
20010 1 603. 20 1005 é ùéù ê úêú ê =-=-úêú ê úêú ê úêú ë ûëû
Thus, the solution is (10, 3,- 5).
39. 225 487 21 xy yz xz -= += += has coefficient matrix 220 048. 102 A é ùê ú ê ú = ê ú ê ú ë û
However, using row operations on [ ] | AI shows that A does not have an inverse, so another method must be used.
Try the Gauss-Jordan method. The augmented matrix is 2205 0487. 1021 é ùê ú ê ú ê ú ê ú ë û
After several row operations, we obtain the matrix 17 4 7 4 102 012. 00013 éù êú êú êú êú êú êú êú ëû
The bottom row of this matrix shows that the system has no solution, since 013 = is a false statement.
40. 21 5 27 xz yz xy +=-= +=
The system can be written as the matrix equation 1021 0115 1207 x y z éùéùéùêúêúêú êúêúêú -= êúêúêú êúêúêú ëûëûëû
First calculate 1 A- using row operations.
The zeros in the third row indicate that matrix A does not have an inverse. Since the equations are not multiples of each other, the system has no solution.
Since 1 ,XAB= 1111 2242 1 2 1513 2242 1111 2242
x y z
é ùêú é ùéùéù - êú ê úêúêú êú ê úêúêú êú ê úêúêú == êú ê úêúêú êú - ê úêúêú êú ê úêúêú êú - ê úêúêú ë ûëûëû êú - ê ú ë û
The solution is (7,34,19,7). 42. 23 23 33225 23 xyw xyzw xyzw xyz ++= -+-= ++-= ++=
This system may be written as the matrix equation 11023 21113 . 33225 12103 x y z w é ùéùéù ê úêúêú ê úêúêú ê úêúêú = ê úêúêú - ê úêúêú ê úêúêú ê úêúêú ë ûëûëû
The inverse of this coefficient matrix was calculated in Exercise 26. Use that result to obtain 11 22 1231 105105 741112 105105 1123 5555 01 31 30 52 31 x y z w é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú êú ê úêúêú == êú ê úêúêú êú ê úêúêú êú ê úêúêú êú ê úêúêú ë ûëûëû êú - ê ú ë û
The solution is (1, 0, 2, 1).
In Exercises 43– 48, let . ab A cd éù êú = êú ëû
43. 10 01 abab IAA cdcd é ùéùéù êúêúêú === êúêúêú ë ûëûëû
Thus, . IAA =
In Exercise 25, it was found that
44. 10 01 1001 1001 ab AI cd abab cdcd ab A cd éùéù êúêú = êúêú ëûëû éù ⋅+⋅⋅+⋅ êú = êú ⋅+⋅⋅+⋅ ëû éù êú == êú ëû
45. 0000 0000 ab AOO cd éùéùéù êúêúêú ⋅=== êúêúêú ëûëûëû Thus, .AOO ⋅=
46. 10 01 ab cd éù êú êú ëû 122 10 0 RRR ab caadbcca éù êú êú -+ ëû 211 11 22 1 1 0 RRR 0 RR10 RR 01 badab adbcadbcadbc db aadbcadbc ca adbcadbcadbc a adbcca--éù -+ êú êú êú ëû éù êú êú êú êú ëû
Thus, 1 , db adbcadbc ca adbcadbc A-éù êú êú = êú êú ëû if 0. adbc-¹
47. In Exercise 46, it was found that 1 1 . db A adbccaéùêú = êú - - ëû 1 1 1 0 1 0 10 01 dbab AA adbccacd dbab adbccacd adbc adbcadbc Iæöéùéù - ÷ ç êúêú ÷ ç = ÷ ç êúêú ÷ ÷ ç -èøëûëû æö éùéù - ÷ ç êúêú ÷ ç = ÷ ç êúêú ÷ ÷ ç - - èø ëûëû éùêú = êú - - ëû éù êú == êú ëû
Thus, 1 AAI=
48. Show that 1 . AAI= From Exercise 46, 1 db adbcadbc ca adbcadbc A-éù êú êú = êú êú ëû 1 10 01 db adbcadbc ca adbcadbc adbcabab adbcadbc cdcdbcad adbcadbc ab AA cd I---+ --+ é ù éù êú êú êú = êú êú ëû êú ëû éù éù êú êú êú === êú êú ëû êú ë û
49. 11 1 () () ABO AABAO AABO IBO BO= =⋅ = ⋅= =
Thus, if ABO = and 1 A- exists, then .BO =
50. Assume that matrix A has two inverses B and C. Then () and.() ABBAI1 ACCAI2 == ==
Multiply equation (1) by C. ()() CABCBACI ==
Since matrix multiplication is associative, ()(). CABCAB =
Since I is an identity matrix, .CIC =
Combining these results, we have () (). CABC CABC = =
From equation (2), , CAI = giving IBC BC = =
Thus, if it exists, the inverse of a matrix is unique.
51. This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is 1 0.04770.02300.02920.08950.0402 0.09210.01500.03210.02090.0276 0.06780.03150.04040.03260.0373 0.01710.02480.00690.00030.0246 0.02080.07400.00960.10180.0646 . Cé
(Entries are rounded to 4 places.)
52. Use a graphing calculator or a computer to perform this calculation. With entries rounded to 6 places, the answer is
0.0101460.0118830.0027720.0207240.012273
0.0063530.0142330.0018610.0291460.019225
0.0006380.0067820.0048230.0226580.019344
0.0052610.0037810.0061920.0048370.006910
0.0122520.0011770.006
1260.0067440.002792
53. This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is
58. Use a graphing calculator or a computer to obtain 0.489558 1.00104 2.11853 1.20793 0.961346 X é ù ê ú ê ú ê ú ê
59. (a) The matrix is 72 48. 60 B éù êú êú = êú êú ëû
0.03940.08800.00330.05300.1499
0.14920.02890.01870.10330.1668
0.13300.05430.03560.17680.1055
0.14070.01750.04530.13440.0655
0.01020.06530.09930.00850.0388 1 D= éù êú êú êú - êú êú êú êú êú ëû (Entries are rounded to 4 places.)
54. Use a graphing calculator or a computer to determine that, no, 11CD and 1 () CD - are not equal.
55. This exercise should be solved by graphing calculator or computer methods. The solution may vary slightly. The answer is, yes, 111 ().DCCD =
56. Use a graphing calculator to obtain 241 272727 231 1 8 999999 5379 14 297297297 Aéù -
Use a graphing calculator again to solve the matrix equation 1 XAB= 1.18519 0.95960 1.30976 X éù êú =-êú êú êúëû
57. This exercise should be solved by graphing calculator or computer methods. The solution, which may vary slightly, is 1.51482 0.053479 0.637242 0.462629 éù êú êú êú êúêú êú êú ëû
(b) The matrix equation is 1 2 3 24272 21248. 21360 x x x éù é ùéù êú ê úêú êú ê úêú = êú ê úêú êú ê úêú ë ûëû ëû
(c) To solve the system, begin by using row operations to find 1 A[] 122 133 211 233 311 242100 |212010 213001 242100
R(1)RR030110 R(1)RR031101 4R3RR606140 030110 R(1)RR001011 6RRR6001106 030110 001011 AI é ù êú êú = êú êú ëû éù êú +--êú êú êú +--ëû -+-éù êú êúêú +--êú ëû éù -+-êú êúêú êúëû 15 1 63 11 6 11 1 22 33 3 1001 RR 0100 RR 001011 é ù ê ú ê ú ê ú - ê ú ê ú ê ú - ê ú ë û
The inverse matrix is 15 63 11 1 33 1 0. 011 Aé ù ê ú ê ú ê ú =ê ú ê ú ê ú - ê ú ë û
Since 1 ,XAB=
1
There are 8 daily orders for type I, 8 for type II, and 12 for type III.
60. Let x = the number of transistors, y = the number of resistors, and z = the number of computer chips.
Solve the following system:
332amount of copper available; 2amount of zinc available; 22amount of glass available. xyz xyz xyz ++= ++= ++=
First, find the inverse of the coefficient matrix
(a) 810 units of copper, 410 units of zinc, and 490 units of glass 41 33 21 33 1 810100 0410110 49090 111 x y z é ù êú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú - ë ûëûëû êú ë û
100 transistors, 110 resistors, and 90 computer chips can be made.
(b) 765 units of copper, 385 units of zinc, and 470 units of glass
41 33 21 33 1 76595 0385100 47090 111 x y z é ù êú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú - ë ûëûëû êú ë û
95 transistors, 100 resistors, and 90 computer chips can be made.
(c) 1010 units of copper, 500 units of zinc, and 610 units of glass 41 33 21 33 1 1010140 0500130 610100 111 x y z é ù êú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú - ë ûëûëû êú ë û 140 transistors, 130 resistors, and 100 computer chips can be made.
61. Let x = the amount invested in AAA bonds, y = the amount invested in A bonds, and z = amount invested in B bonds.
(a) The total investment is 25,000. xyz++= The annual return is 0.060.0650.08 xyz ++= 1650. Since twice as much is invested in AAA bonds as in B bonds, 2.xz = The system to be solved is 25,000 0.060.0650.081650 20 xyz xyz xz ++= ++= -= Let 11125,000 0.060.0650.08,1650, 1020 AB é ùéù ê úêú ê ==úêú ê úêú ê úêú - ë ûëû and . x Xy z é ù ê ú ê ú = ê ú ê ú ë û
Use a graphing calculator to obtain 1 264003 406004. 132001 Aé ùê ú ê ú =-ê ú ê ú - ë û
Use a graphing calculator again to solve the matrix equation 1 XAB=
the weight of pretzels as dried fruit is to be used, 2.xy =
10,000 10,000 5000 x y z éùéùéùêúêúêú êúêúêú =-êúêúêú êúêúêúëûëûëû
26400325,000 4060041650 1320010
éù êú êú = êú êú ëû
$10,000 should be invested at 6% in AAA bonds, $10,000 at 6.5% in A bonds, and $5000 at 8% in B bonds.
(b) The matrix of constants is changed to 30,000 1985. 0 B éù êú êú = êú êú ëû
(a) For a total of 140 pounds at $6 per pound, a total value of 140$6$840, ´= the system to be solved is 140 459840 20 xyz xyz xy ++= ++= -= Let 111 459, 120 A
-
140 840, 0 B éù êú êú = êú êú ëû and . x Xy z é ù
26400330,000 4060041985 1320010
14,000 9,000 7000 x y z éùéùéùêúêúêú êúêúêú =-êúêúêú êúêúêúëûëûëû éù êú êú = êú êú ëû
$14,000 should be invested at 6% in AAA bonds, $9000 at 6.5% in A bonds, and $7000 at 8% in B bonds.
(c) The matrix of constants is changed to 40,000 2660. 0 B éù êú êú = êú êú ëû
Use row operations to obtain the inverse of the coefficient matrix. 9 12 777 1 95 1 141414 133 1 141414
Since 1 ,XAB= 1824140 1 915480 14 13310 84060 1 42030. 14 70050 x y z é ùéùéùê úêúêú ê úêúêú =-- ê úêúêú ê úêúêú - ë ûëûëû éùéù êúêú ==êúêú êúêú êúêú ëûëû
Use 60 pounds of pretzels, 30 pounds of dried fruits, and 50 pounds of nuts.
24,000 4000 12,000 x y z éùéùéùêúêúêú êúêúêú =-êúêúêú êúêúêúëûëûëû
26400340,000 4060042660 1320010
éù êú êú = êú êú ëû
$24,000 should be invested at 6% in AAA bonds, $4000 at 6.5% in A bonds, and $12,000 at 8% in B bonds.
62. Let x = the number of pounds of pretzels, y = the number of pounds of dried fruit, and z = the number of pounds of nuts.
The total amount of trail mix is .xyz ++ The cost of the trail mix is 459. xyz ++ Since twice
(b) For a total of 100 pounds at $7.60 per pound, a total value of 100$7.60$760, ´= the system to be solved is
100 459760 20 xyz xyz xy ++= ++= -=
The matrix of constants is changed to
100 760. 0 B é ù ê ú ê ú = ê ú ê ú ë û
Since 1 ,XAB=
1824100 1 915760 14 13310
28020 1 14010. 14 98070 x y z éùéùéùêúêúêú êúêúêú =-êúêúêú êúêúêúëûëûëû éùéù êúêú ==êúêú êúêú êúêú ëûëû
Use 20 pounds of pretzels, 10 pounds of dried fruits, and 70 pounds of nuts.
(c) For a total of 125 lb at $6.20 per pound, a total value of 125$6.20$775, ´= .the system to be solved is
The matrix of constants is changed to
Since –1 .XAB = 1824125 1 915775 14 13310
70050 1 35025. 14 70050 x y z éùéùéùêúêúêú êúêúêú =-êúêúêú êúêúêúëûëûëû
éùéù êúêú ==êúêú êúêú êúêú ëûëû
Use 50 pounds of pretzels, 25 pounds of dried fruits, and 50 pounds of nuts.
63. Let x = the number of Super Vim tablets, y = the number of Multitab tablets, and z = the number of Mighty Mix tablets.
The total number of vitamins is xyz ++
The total amount of niacin is 152025. xyz ++
The total amount of Vitamin E is 121535. xyz ++
(a) The system to be solved is 225 1520254750 1215355225. xyz xyz xyz ++= ++= ++=
Let 111225 152025,,4750. 2015355225 x AXyB z éùéùéù êúêúêú êúêúêú === êúêúêú êúêúêú ëûëûëû
Thus, AXB = and
111225 1520254750. 1215355225 x y z é ùéùéù ê úêúêú ê úêúêú = ê úêúêú ê úêúêú ë ûëûëû
Use row operations to obtain the inverse of the coefficient matrix. 6541 171717 45232 1 178517 331 178517 Aé ù - ê ú ê ú ê ú =ê--ú ê ú ê ú ê ú ê ú ë û
Since –1 ,XAB = 6541 171717 45232 178517 331 178517 22550 475075. 5225100 x y z é ù êúêú é ùéùéù êú ê úêúêú êú ê úêúêú êú =--= ê úêúêú êú ê úêúêú êú ë ûëûëû êú êú êú ë û
There are 50 Super Vim tablets, 75 Multitab tablets, and 100 Mighty Mix tablets.
(b) The matrix of constants is changed to 185 3625. 3750 B éù êú êú = êú êú ëû 6541 171717 45232 178517 331 178517 18575 362550 375060 x y z é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú =--= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êú ê ú ë û
There are 75 Super Vim tablets, 50 Multitab tablets, and 60 Mighty Mix tablets.
(c) The matrix of constants is changed to 230 4450. 4210 B éù êú êú = êú êú ëû 6541 171717 45232 178517 331 178517 23080 4450100 421050 x y z é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú =--= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êú ê ú ë û
There are 80 Super Vim tablets, 100 Multitab tablets, and 50 Mighty Mix tablets.
64. (a) First, divide the letters and spaces of the sentence into groups of 3, writing each group as a column vector.
éùéùéùéù êúêúêúêú êúêúêúêú êúêúêúêú êúêúêúêú ëûëûëûëû
0.20.20.2 0.40.60.2, 00.40.2 A é ùê ú ê ú =ê ú ê ú - ë û and each of the
(space)(space) ,,,, (space) ,,(space),, (space)(space) Ai lifr lsa ilenw noda ar υ
éùéùéùéùéù êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú ëûëûëûëûëû
Next, convert each letter into a number, assigning 1 to A, 2 to B, and so on, with the number 27 used to represent each space between words.
1272799
12,9,6,18,14, 121912727 1251423 15,27,4,1 2212718
éùéùéùéùéù êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú ëûëûëûëûëû éùéùéùéù êúêúêúêú êúêúêúêú êúêúêúêú êúêúêúêú ëûëûëûëû
Now find the product of the coding matrix
presented in Example 7, 134 213, 421 A éù êú êú = êú êú ëû and each column vector above. This produces a new set of vectors, which represents the coded message.
8513049171159 50,120,63,117,113, 401451219991 1459013498 105,40,113,101 1007591112
éùéùéùéùéù êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú ëûëûëûëûëû éùéùéùéù êúêúêúêú êúêúêúêú êúêúêúêú êúêúêúêú ëûëûëûëû
The message will be transmitted as 85, 50, 40, 130, 120, 145, 49, 63, 121, 171, 117, 99, 159, 113, 91, 145, 105, 100, 90, 40, 75, 134, 113, 91, 98, 101, 112.
(b) First, divide the coded message into groups of three numbers and form each group into a column vector.
Next, find the product of the decoding matrix presented in Example 7, the inverse of matrix A in part (a) above, –1
column vectors above. This produces a new set of vectors, which represents the decoded message.
9152721 27,22,25,27 1251527 é ùéùéùéù ê úêúêúêú ê úêúêúêú ê úêúêúêú ê úêúêúêú ë ûëûëûëû
Lastly, convert each number into a letter, assigning A to 1, B to 2, and so on, with the number 27 used to represent each space between words.
Decoded message: I love you
65. (a) First, divide the letters and spaces of the message into groups of three, writing each group as a column vector.
úêúêúêúêúêú
úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ë ûëûëûëûëûëû
(space)(space) ,,,,, (space)(space)(space)
toe
Next, convert each letter into a number, assigning 1 to A, 2 to B, and so on, with the number 27 used to represent each space between words.
20215142727 15,5,18,15,20,2 27272720155
é ùéùéùéùéùéù ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ë ûëûëûëûëûëû
Now, find the product of the coding matrix B and each column vector. This produces a new set of vectors, which represents the coded message.
26218626420822492 161,103,168,134,152,50 12229553
é ùéùéùéùéùéù ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ê úêúêúêúêúêú ë ûëûëûëûëûëû
This message will be transmitted as 262, 161,- 12,- 186, 103,- 22,- 264, 168, 9,- 208, 134,- 5,- 224, 152,- 5, 92, 50,- 3.-
138101162210
81,67,124,150 102109173165
éùéùéùéù êúêúêúêú êúêúêúêú êúêúêúêú êúêúêúêú ëûëûëûëû
(b) Use row operations or a graphing calculator to find the inverse of the coding matrix B. 1 1.752.53 0.250.50 0.250.51 Béù êú =--êú êú êú ëû
(c) First, divide the coded message into groups of three numbers and form each group into a column vector.
11629414896264 60,197,92,64,182 152942
éùéùéùéùéù êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú ëûëûëûëûëû
Next, find the product of the decoding matric 1 B- and each of the column vectors. This produces a new set of vectors, which represents the decoded message.
8162201 1,25,9,8,25 162718427
éùéùéùéùéù êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú êúêúêúêúêú ëûëûëûëûëû
Last, convert each number into a letter, assigning A to 1, B to 2, and so on, with the number 27 used to represent a space between words. The decoded message is HAPPY BIRTHDAY.
66. (a) [] 505045100 | 01520010
Interchange rows 1 and 3.
505045100 111001 01520010 50RRR0051050 RRR15001165 4RRR015041200 0051050 RR100 RR010 RR001010 éù êú êú êú êú ëû éù êú êú êú
R(15)RR15050115 01520010 0051050 +---éù êú êú êú êú ëû 1113
(c) Denoting the original positions of the band members as x1, x2, and so on, the original shape was
We are given that band member x9, originally positioned at (50, 0), moved to (40, 10); band member x6, originally at (50, 15), moved to (55, 10); band member x2, originally at (45, 20), moved to (60, 15).
To find the new position of x1, originally at (40, 20), multiply A by the vector 40 20. 1
4001404060 2010602020 100111
The new position of band member x1 is (60, 20).
To find the new position of x3, originally at (50, 20), multiply A by the vector 50 20. 1 é ù ê ú ê ú ê ú ê ú ë û 5001405060 2010602010 100111 A é ùéùéùéù ê úêúêúêú ê úêúêúêú =-= ê úêúêúêú ê úêúêúêú ë ûëûëûëû
The new position of band member x3 is (60, 10).
To find the new position of x4, originally at (55, 20), multiply A by the vector 55 20. 1 é ù ê ú ê ú ê ú ê ú ë û
5501405560
The new position of band member x4 is (60, 5).
To find the new position of x5, originally at (60, 20), multiply A by the vector
6001406060 201060200 100111 A
The new position of band member x5 is (60, 0).
To find the new position of x7, originally at (50, 10), multiply A by the vector
êúêúêúêú êúêúêúêú =-= êúêúêúêú êúêúêúêú ëûëûëûëû
The new position of band member x7 is (50, 10).
To find the new position of x8, originally at (50, 5), multiply A by the vector
êúêúêúêú êúêúêúêú =-= êúêúêúêú êúêúêúêú ëûëûëûëû
The new position of band member x8 is (45, 10). The new position of the band is
2.6 Input-Output Models
Your Turn 1 1 ()
1.3950.4960.589322
0.8371.3640.620447
0.5580.4651.302133
749.239 961.682 560.697 XIAD X=é ùéù
The productions of 749 units of agriculture, 962 units of manufacturing, and 561 units of transportation are required to satisfy the demands of 322, 447, and 133 units, respectively.
Your Turn 2
Thus, the new shape is a sideways T whose vertical and horizontal intersection is at mark (60, 10).
Clearing fractions gives the following
The solution is () 42 99333,,, xxx where 3x is any real number. For 3 9, x = the solution is (4,2,9). So, the production of the three commodities should be in the ratio 4:2:9.
2.6 Warmup Exercises
W1. 12 122 11 211 3210 1101 1101 RR 3210 1101
3RRR 0113 1101 (1)RR 0113 R+RR1012 0113
The inverse is 12 13
W2.
2RRR 014110 (1)RRR 001111 (12)RRR100151412 010554 (4)RRR 001111
The inverse is 151412 554. 111
2.6 Exercises 1. 0.80.22 , 0.20.73 AD é ùéù ê ==úêú ê úêú ë ûëû
To find the production matrix, first calculate IA100.80.20.20.2 010.20.70.20.3 IA é ùéùéùê -=-=úêúêú ê úêúêú - ë ûëûëû
Using row operations, find the inverse of IA11 22 122 211 1 11 2 1 0.20.210 [|] 0.20.301 10RR 22100 10RR 23010 22100 RRR 011010 2RRR 203020 011010 RR 101510 011010 1510 () 1010 IAI IAéùêú -= êúëû éù êú êú ëû éùêú êú + ëû éù + êú êú ëû éù êú êú ëû éù êú -= êú ëû
Since 1 (), XIAD=- the product matrix is 1510260 1010350 X é ùéùéù ê ==úêúêú ê úêúêú ë ûëûëû
2. 0.20.043 , 0.60.0510 AD é ùéù ê ==úêú ê úêú ë ûëû
To find the production matrix X, first calculate IA -
100.20.04
010.60.05
0.80.04
0.60.95 IA é ùéù ê -=-úêú ê úêú ë ûëû éùêú = êú - ë û
Next, use row operations to find the inverse of .IA1 1.290.054 () 0.8151.087 IAé ù ê ú -= ê ú ë û
Since 1 (), XIAD=1.290.05434.4 . 0.8151.0871013.3 X é ùéùéù ê ==úêúêú ê úêúêú ë ûëûëû
3. 0.10.035 , 0.070.610 AD éùéù ==êúêú êúêú ëûëû
First, calculate IA0.90.03 0.070.4 IA éùêú -= êúëû
Use row operations to find the inverse of ,IAwhich is 1 1.1180.084 (). 0.1962.515 IAéù êú -» êú ëû
Since 1 (), XIAD=- the production matrix is 1.1180.08456.43 . 0.1962.5151026.12 X éùéùéù ==êúêúêú êúêúêú ëûëûëû
4. 0.020.03100 , 0.060.08200 AD éùéù ==êúêú êúêú ëûëû
To find the production matrix, first calculate IA100.020.03 010.060.08 0.980.03 0.060.92 IA éùéù -=-êúêú êúêú ëûëû éùêú = êúëû
Using row operations, find the inverse of IA -
0.980.0310 | 0.060.9201
100RR 9831000 100RR 6920100 9831000
3R49RR 044993004900
3R4499RR RR 101.0224 RR
IAI éùêú -= êúëû éù êú êú ëû éùêú êú + ëû éù + êú êú ëû 490.033341 010.0666811.089131 éù êú êú ëû
11 22 122 211 11 22 440,9020450,80014,700 044993004900 1 440,902 1 4499
1.0224490.033341 1 () 0.0666811.089131 IA éù - êú -= êú ëû
Since 1 (), XIAD=- the production matrix is 1.0224490.033341100 0.0666811.089131200 108.91 (rounded). 224.49 X éùéù êúêú = êúêú ëûëû éù êú = êú ëû
5. 0.800.11 0.10.50.2,6 00.20.73 AD é ùéù ê úêú ê ==úêú ê úêú ê úêú ë ûëû
To find the production matrix, first calculate IA1000.800.1 0100.10.50.2 00100.20.7
0.200.1 0.10.50.2 00.20.3 IA é ùéù ê úêú ê -=-úêú ê úêú ê úêú ë ûëû éùêú =--êú êú êú ë û
Using row operations, find the inverse of .IA[] 11 22 33 122 0.200.1100 |0.10.50.2010 00.20.3001 10RR2011000 10RR1520100 10RR00.230010 2011000 R2RR010510200 00.230010 IAI é ùê
233 33 311 322 11 22 2011000 010510200 1 RRR 0022410 5 2011000 010510200 1 RR 001125 2 RRR2001125 5RRR0100153025 001125 1115 RR1001 222 135 RR0103 1022 001125
1 115 1 22 35
Since 1 (), XIAD the production matrix is 115 1 22 119 35 3627. 22 328 125
6. 0.10.5010
123 123 123 0.70.10.80 0.50.40.10 0.20.30.90 xxx xxx xxx --= -+-= --+=
Rewrite the equations without decimals. 123 123 123 780() 540() 2390() xxx1 xxx2 xxx3 --= -+-= --+=
IA XIADéùêú
1.1580.9470.421 ()0.0841.7050.758
0.1470.4841.33
1.1580.9470.42110 ()0.0841.7050.7584
0.1470.4841.332
êú êú ëû éù êú êú -= êú êú ëû éùéù êúêú êúêú =-= êúêú êúêú ëûëû = 6.06 éù êú êú êú êú ëû
7. ABC
A0.30.10.8
B0.50.60.1
C0.20.30.1
Use row operations to solve this system of equations. Begin by eliminating x1 in equations (2) and (3) 123 12123 13323 780() 5R7RR23470() 2R7RR23470() xxx1 xx4 xx5 --= +-= +-+=
Eliminate x2 in equations (1) and (5). 12113 23 233 23RRR1612310() 23470() RRR00() xx6 xx4 7 +-= -= +=
The true statement in equation (7) indicates that the equations are dependent. Solve equation (6) for x1 and equation (4) for x2, each in terms of x3 133 23 23133 16123 47 23 xxx xx == =
The solution of the system is 333 3347 2323,,. xxx æö ÷ ç ÷ ç ÷ ÷ ç èø
0.70.10.8
0.50.40.1 0.20.30.9 A IA éù êú êú = êú êú ëû éù êú -=--êú êú êú ëû
Set ()IAXO -= to obtain the following. 1 2 3 123 123 123
If 3 23, x = then 1 33 x = and 2 47, x = so the production of the three commodities should be in the ratio 33:47:23.
8. ABC
A0.30.20.3
0.70.10.80
0.50.40.10
0.20.30.90
B0.10.50.4
C0.60.30.3 A éù êú êú = êú êú ë û
0.70.10.80
0.50.40.10
0.20.30.90 x x x xxx xxx xxx éù éùéù êú êúêú êú êúêú --= êú êúêú êú êúêú ëûëû ëû
éù éù êúêú êú êú -+-= êú êú êúêú --+ ëû ëû
Rewrite this matrix equation as a system of equations.
Calculate ,IA - and then set ()IAXO -= to find X
0.70.20.3
0.10.50.4
0.60.30.7
0.70.20.3
Solving this system with x3 as the parameter will give the solution
11. In Example 4, it was found that 1 1.38820.1248 (). 0.51471.1699 IAéù êú -» êú ëû
Since 1 (), XIAD=- the production matrix is 1.38820.12489251440.085 . 0.51471.169912501938.473 X é ùéùéù ê ==úêúêú ê úêúêú ë ûëûëû
Thus, about 1440. metric tons of wheat and 1938 metric tons of oil should be produced.
If 312 33, then23 and 31, xxx=== so the production of A, B, and C should be in the ratio 23:31:33.
9. Use a graphing calculator or a computer to find the production matrix 1 (). XIAD=- The answer is
7697 4205 . 6345 4106 X éù êú êú êú = êú êú êú êú ëû
Values have been rounded.
10. This exercise should be solved using a graphing calculator or a computer, to find 1 (1) XAD =− The answer, rounded to the nearest whole number, is
7022 4845 . 5116 4647 X éù êú êú êú = êú êú êú êú ëû
12. 1 3 1 5 0 500 , 1000 0 AD é ù êú éù êú êú == êú êú ëû êú ë û 1 1 1 3 1 5 155 1414 315 1414 () 1 500 1000 1 500 1000 892.9 1178.6 XIAD=éùêú éù êú êú = êú êú ëûêú ëû éù êú éù êú êú = êú êú ëû êú ëû éù êú = êú ë û
Produce about 893 metric tons of wheat and about 1179 metric tons of oil.
13. In Example 3, it was found that 1 1.39530.49610.5891 ()0.83721.36430.6202. 0.55810.46511.3023
Since 1 (), XIAD=- the production matrix is 1.39530.49610.58916071505.66 0.83721.36430.62026071712.77. 0.55810.46511.30236071411.58 X é ùéùéù ê úêúêú ê ==úêúêú ê úêúêú ê úêúêú ë ûëûëû
Thus, about 1506 units of agriculture, 1713 units of manufacturing, and 1412 units of transportation should be produced.
14. 1 2 1 3 1 4 00 1000 00,1000 1000 00 AD é ù êú éù êú êú êú êú == êú êú êú êú êú ëû êú ê ú ë û
0.3478261.0434780.173913
0.0869570.2608701.043418 1695.6 1565.2 1391.3
Produce about 1696 units of agriculture, about 1565 units of manufacturing, and about 1391 units of transportation.
15. From the given data, we get the input-output matrix
16. 11 24 11 44 11 24 0 500 0,500 500 0 AD éù êú éù êú êú êú êú == êú êú êú êú êú ëû êú êú ëû 11 24 11 44 11 24 1 1 1 IA éù êú êú êú -=--êú êú êú êú êú ëû 1 1 1.5380.92310.6154 ()0.61541.4360.5128 0.92310.82051.436 1538.3 ()1282.1 1589.8 IA XIADéù êú êú -= êú êú ëû éù êú êú =-= êú êú ëû
Use row operations to find the inverse of ,IAwhich is 1 1.5380.9230.615 ()0.6151.4360.513. 0.9230.8211.436 IAéù êú êú -» êú êú ëû
Since 1 (), XIAD=- the production matrix is 1.5380.9230.61510003077 0.6151.4360.51310002564. 0.9230.8211.43610003179 X éùéùéù êúêúêú êúêúêú =» êúêúêú êúêúêú ëûëûëû
Thus, the production should be about 3077 units of agriculture, 2564 units of manufacturing, and 3179 units of transportation.
Produce about 1538 units of agriculture, about 1282 units of manufacturing, and about 1590 units of transportation.
17. From the given data, we get the input-output matrix 11 46 1 2 31 46 1 2 0 1 A IA éù êú êú = êú êú ëû éùêú êú -= êú ê-ú ëû
Use row operations to find the inverse of ,IAwhich is 31 4 2 1 9 3 8 4 (). IAéù êú -= êú êú êú ëû
(a) The production matrix is
1 (). 1 XIADé ù éù éù ê ú êú êú =-== ê ú êú êú ê ú êú ëû êú ê ú ëû ë û
Thus, 7 4 bushels of yams and 15 8 2 » pigs should be produced.
(b) The production matrix is 31 24 1 39 48 100167.5 (). 70153.75 XIADéù êú éùéù êú =-==êúêú êú êúêú ëûëû êú ëû
Thus, 167.5 bushels of yams and 153.75154 » pigs should be produced.
18. For this economy, 0.20.40.2 0.40.20.1. 00.10.2 A é ù ê ú ê ú = ê ú ê ú ë û
Use a graphing calculator or a computer to find –1 ()IAD - where
1000 1000. 1000 D éù êú êú = êú êú ëû
The solution, which may vary slightly, is 3179.35 3043.48. 1630.43 éù êú êú êú êú ëû
Produce about 3179 units of oil, 3043 units of corn, and 1630 units of coffee.
19. Use a graphing calculator or a computer to find the production matrix –1 (–). XIAD = The answer is
848 516. 2970 éù êú êú êú êú ëû
Values have been rounded.
Produce 848 units of agriculture, 516 units of manufacturing, and 2970 units of households.
20. Use a graphing calculator or a computer to find –1 (–). IAD The solution, which may vary slightly, is
18.2 73.2. 66.7 éù êú êú êú êú ëû
Values have been rounded.
Produce $18.2 billion of agriculture, $73.2 billion of manufacturing, and $66.7 billion of households.
21. Use a graphing calculator or a computer to find the production matrix –1 (–). XIAD = The answer is
195,492 25,933. 13,580 éù êú êú êú êú ëû
Values have been rounded. Change from thousands of pounds to millions of pounds.
Produce about 195 million Israeli pounds of agriculture, 26 million Israeli pounds of manufacturing, and 13.6 million Israeli pounds of energy.
22. (a) Use a graphing calculator or a computer to find –1 (–). IAD The solution, which may vary slightly, is
183,464 304,005. 42,037 éù êú êú êú êú ëû
Values have been rounded. In 100,000 RMB, produce about 183,000 for agriculture, 304,000 for industry/construction, and 42,000 for transportation/commerce.
(b) The entries in matrix –1 (–) IA are called multipliers, and they give the desired economic values. 1 1.240.340.04 ()0.301.850.14 0.030.081.02 IAéù êú
(Each entry has been rounded to two decimal places.) Since we are interested in the result when there is a 1 RMB increase in demand for agricultural exports, we are interested in the first column of this matrix. Interpreting the multipliers shown, we find that an increase of 1 RMB in demand for agricultural exports will result in a 1.24 RMB increase in production of agricultural commodities, a 0.30 RMB increase in production of industrial/ construction commodities, and a 0.03 RMB increase in production of transportation/ commercial commodities.
23. Use a graphing calculator or a computer to find the production matrix –1 (–). XIAD = The answer is 532 481 . 805
1185
Values have been rounded.
Produce about 532 units of natural resources, 481 manufacturing units, 805 trade and service units, and 1185 personal consumption units. Units are millions of dollars.
24. (a) Use a graphing calculator or a computer to find –1 (–). IBC The solution, which may vary slightly, is 3 60 27. 42 1002 éù êú êú êú êú êú êú êú êú êú ëû
Values have been rounded.
A $50 million increase in manufacturing demand will result in a $3 million production increase in natural resources, a $60 million production increase in manufacturing, a $27 million production increase in trade and services, a $42 million production increase in personal consumption, and 1002 new jobs.
(b) The matrix –1 (–) IB is
1.10.1000
0.21.20.20.10
0.70.51.90.70
The bottom row of this matrix indicates the total employment requirement, per million dollars, of a sector. For example, the total employment requirement, per million dollars, of natural resource output is 40.9 employees.
25. (a) Use a graphing calculator or a computer to find the matrix 1 (). IA- The answer is 1.670.560.56 0.191.170.06. 3.153.274.38 éù
Values have been rounded.
(b) These multipliers imply that if the demand for one community’s output increases by $1 then the output in the other community will increase by the amount in the row and column of this matrix. For example, if the demand for Hermitage’s output increases by $1, then output from Sharon will increase $0.56, from Farrell by $0.06, and from Hermitage by $4.38.
26. Find the value of ,IA - then set (). IAXO -= 11 42 1 31 2 42 31 31 12 42 42 1 1 31 3 2 4212 4 2
If 2 3, x = then 1 2. x = Therefore, produce 2 units of yams for every 3 units of pigs.
27. Calculate ,IA - and then set ()IAXO -= to find X 31 43 1 12 2 43 11 43 1 11 2 43 11 4312 11 4312 10 () 01 0 0 x IAX x x x xx xx æö éù ÷ ç êú ÷ é ù éù ç ÷ ç êú ÷ ê ú êú -=ç ÷ ç êú ÷ ê ú êú ç ÷ ëû ë û êú ÷ ç ÷ ç èø ëû éùêú éù êú êú = êú êú ëû ê-ú ëû éùêú éù êú êú = êú êú ëû ê-+ú ë û Thus, 11 4312 11 4312 4 12 3 0 xx xx xx -= = = If 213,4.xx== Therefore, produce 4 units of steel for every 3 units of coal.
28. Find the value of ,IA - then set (). IAXO -=
()010 001 0 x IAXx x x x x æö éù ÷ ç êú ÷ ç ÷ é ù éù ç êú ÷ ç ÷ ê ú êú ç êú ÷ ç ÷
system to be solved is
Use x3 as the parameter. Therefore, 3 13 8 xx = and 1 23 2 , xx = and the solution is () 31 82333,,. xxx
If 3 8, x = then 1 3 x = and 2 4. x =
Produce 3 units of agriculture to every 4 units of manufacturing and 8 units of transportation.
29. For this economy, 13 55 214 555 211 555 0 A éù êú êú êú =êú êú êú êú êú ëû
Find the value of ,IA - then set (). IAXO -=
Use x3 as the parameter. Therefore, 6 13 5 xx = and 8 23 5 , xx = and the solution is () 68 55333 , ,.xxx If 3 5, x = then 1 6 x = and 2 8. x = Produce 6 units of mining for every 8 units of manufacturing and 5 units of communication.
Chapter 2 Review Exercises
1. True
2. True
3. False; a system with three equations and four unknowns has an infinite number of solutions.
4. False; only row operations can be used.
5. True
6. False; matrix A is a 22 ´ matrix and matrix B is a 32 ´ matrix. Only matrices having the same dimension can be added.
7. False; only matrices having the same dimension can be added.
8. True
9. False; in general, matrix multiplication is not commutative.
10. False; only square matrices can have inverses.
11. True; for example, 000 AA ⋅=⋅=
12. False; any nn ´ zero matrix does not have an inverse, and the matrix 24 12 éùêú êúëû is an example of another square matrix that doesn’t have an inverse.
13. False; if and ABCA = has an inverse, then 1 .BAC=
14. True
15. False; implies ABCBAC == only if B is the identity matrix or B has an inverse.
16. True
17. For a system of m linear equations in n unknowns and ,mn = there could be one, none, or an infinite number of solutions. If ,mn < there are an infinite number of solutions. If ,mn > there could be one, none, or an infinite number of solutions.
19. 2314() 325() xy1 xy2 -= +=-
Eliminate x in equation (2).
122 2314() 3R2RR1352() xy1 y3 -= -+=-
Make each leading coefficient equal 1. 13 11 22 1 22 13 RR7() RR4() xy4 y5 -= =-
Substitute –4 for y in equation (4) to get 1. x = The solution is ( ) 1,–4.
20. 3 24 4 42 xy xy += -=
First, multiply both equations by 4 to clear fractions. 212 216 xy xy += -=
Now proceed by the echelon method. 122 11 11 22 1 22 5 212 R(2)RR520 RR6 RR4 xy y xy y += +-=+= =-
Back-substitution gives 1 2 (4)6 8. x x +-= =
The solution is (8,4). -
21. 235() 55314() xyz1 xyz2 -+=++=
Eliminate x in equation (2).
122 235 5R(2)RR2553 xyz yz -+=+---=-
Let z be the parameter. Solve for y and for x in terms of z.
2553 2553 53 25 yz yz z y --=-=-+= 53 235 25 1593 25 2525 2834 2 2525 3428 2 2525 3428 50 z xz z xz xz xz z x æö÷ ç -+=÷ ç ÷ ç èø -++=-
The solutions are 342853 5025,,, zz z æö ÷ ç ÷ ç ÷ ç èø where z is any real number.
22. 2345(1) 3456(2) xyz xyz -+= ++=
Eliminate x in equation (2).
122 2345 3R(2)RR2523 xyz yz -+= +-+=
Let z be the parameter. Solve for y and for x in terms of z 23 32 yz yz += =23(32)45
The solutions are (2,32,), zzz where z is any real number.
23. 246 3512 xy xy +=--=
Write the augmented matrix and use row operations.
246 3R2RR 026 2RRR2018 026
RR 109 013 RR éùêú êú êú ëû éùêú êú + êú ëû -+-éù êú êú êú ëû éù êú êú êú ëû
122 211 1 11 2 1 22 2 246 3512
The solution is (9,3). -
24. 410 53119 xy xy -= +=
Write the system in augmented matrix form and use row operations to solve.
1410 53119 éùêú êú ëû 122 1410 5RRR 02369 éùêú êú -+ êú ëû
RR 1022 013 RR éù + êú êú êú ëû éù êú êú êú ëû
211 1 11 23 1 22 23 4R23RR230506 02369
The solution is (22, 3).
25. 313 4217 3221 xyz xyz xyz -+= ++= ++=
Write the augmented matrix and use row operations.
4RRR051035
R5RR50530 051035
1RRR0033
133 211 233 311 322 11313 41217 3221 11313
2(1)3(3)9 2. x x +-+= =
The solution is (2,1,3). -
27. 36912 234 27 xyz xyz xyz -+= -+-=++=
3RRR05738
5R(3R)R1500105 10R3RR0150135 0033 éùêú êú êú êú êú êú ëû éùêú êú -+--êú êú êú -+--êú ëû éù + êú êú êú êú êú -+-êú ëû +---éù êú êú +-êú ê ê êëû 1 11 15 1 22 15 1 33 3 RR1007 RR0109 RR0011 ú
The solution is (7,9,1).
26. 239() 24() 3212() xyz1 xy2 xz3 ++= -= +=
Eliminate x in equations (2) and (3).
122 133 239() 1RRR435() 3RRR6715() xyz1 yz4 yz5 ++= -+--=-+--=-
Eliminate y in equation (5).
233 239() 435() 3R(2)RR515() xyz1 yz4 z6 ++= --=+-=
Make each leading coefficient equal 1. 1 22 4 1 33 5 239() 35 RR() 44 RR3() xyz1 yz7 z8++= += =
Back-substitution gives 35 (3) 44 59 1 44 y y += =-=and
Write the augmented matrix and use row operations. 122 133 36912 1234 1127 36912 R3RR0000 1R3RR0939 éùêú êú êú êú êú ëû éùêú êú + êú -+-êú êú ëû
The zero in row 2, column 2 is an obstacle. To proceed, interchange the second and third rows. 36912 0939 0000 é ù - ê ú ê ú - ê ú ê
ë û 121 1 7 11 3 9 1 1 22 3 9 902154 3R2RR 0939 0000 106 RR 011 RR 0000 éù + êú êú êúêú êú êú ëû éù êú êú êú - êú êú êú ëû
The row of zeros indicates dependent equations. Solve the first two equations respectively for x and y in terms of z to obtain 6and171 33 xzyz =-=+
The solution of the system is 71 6,1,, 33 zzz æö ÷ ç -+ ÷ ç ÷ ÷ ç èø where z is any real number.
28. 25 328 210 xz xy xz -= += -+=
Write the system in augmented matrix form and apply row operations.
122 133 1025 3208 10210 1025 3RRR0267 RRR00015 éùêú êú êú êúêú ëû éùêú -+-êú êú êú + êú ëû
The last row says that 015, = which is false, so the system is inconsistent and there is no solution. In Exercises 29–32, corresponding elements must be equal.
29. 23 59 ab qc éùéù êúêú = êúêú ëûëû
Size: 22;2,3,5,9; abcq ´==== square matrix
30. 21 646 57 xa y zp éùéùêúêú êúêú = êúêú êúêú ëûëû
The size of these matrices is 32. ´ For matrices to be equal, corresponding elements must be equal, so 2,1,4,5,andaxyp ==-== 7. z =
31. [ ] [ ] 24312 12193 mz kr=+--
Size: 14;6,3,3,9; mkzr ´===-=- row matrix
32. 5367223 4233214 141128 abbk cdd pqm éùéù +-+êúêú êúêú = +-êúêú êúêú ëûëû
These are 33 ´ square matrices. Since corresponding elements must be equal, 57,so a +=- 12; a =-
32,so bb=+ 1; b = 623,so k =9 2 ; k =
43,so c = 3 4 ; c = 221,so dd +=- 3; d = 34,so-= 3 4 ; =- 1; m =412,so p = 3;and p = 18,so q -= 9 q =
33. 41050 2313 6947 910 30 1016 AC é ùéù ê úêú ê +=--+-úêú ê úêú ê úêú ë ûëû éù êú =-êú êú êú ë û
34. 2014 2424 1537 40416 2101228 016 1018 GF é ùéù ê -=-úêú ê úêú ë ûëû é ùéù ê =-úêú ê úêú ë ûëû éùêú = êú ë û
35. 50410 32313223 4769 150820 3946 12211218 2320 73 2439 CA é ùéù ê úêú ê +=-+--úêú ê úêú ê úêú ë ûëû é ùéù ê úêú ê =-+--úêú ê úêú ê úêú ë ûëû
éù êú =-êú êú êú ë û
36. Since B is a 33 ´ matrix, and C is a 32 ´ matrix, the calculation of BC - is not possible.
37. 41050 25223513 6947 820250 46515 12182035 1720 121 817 AC é ùéù ê úêú ê -=----úêú ê úêú ê úêú ë ûëû é ùéù ê úêú ê =----úêú ê úêú ê úêú ë ûëû
éùêú êú =êú êú ë û
38. A has size 32 ´ and G has 22, ´ so AG will have size 32. ´
é ùéù ê úêú éùê úêú êú =--=ê úêú êú ëû ê úêú - ë ûëû
410250 20 23115 15 69345 AG
39. A is 32 ´ and C is 32, ´ so finding the product AC is not possible.
32 AC ´´ 23 (The inner two numbers must match.)
40. D has size 31 ´ and E has size 13, ´ so DE will have size 33. ´ [] 661824 1134134 0000 DE éùéùêúêú êúêú =-=êúêú êúêú ëûëû
41. 6 [134]1 0 [1.63.1(4)0][9] ED éù êú êú =- êú êú ëû =++-=
42. B has size 33 ´ and D has size 31, ´ so BD will have size 31. ´ 232615 240116 01201 BD éùéùéùêúêúêú ==êúêúêú êúêúêú êúêúêú ëûëûëû
43. [] 50 13413 47 EC éù êú =--êú êú êú ëû [ ] [153(1)(4)410 33(4)7] 1419 =⋅+-+-⋅⋅ +⋅+-⋅ =--
44. [] 14 37 1410 | 3701 F FI éùêú = êú ëû éùêú = êú ëû 122 211 174 11 191919 131 22 191919 1410 3RRR 01931 4R(19R)R 19074 01931 RR10 RR01 éùêú êú + ëû éù +-êú êú ëû éù -êú êú êú êú ëû
74 1919 1 31 1919 Fé ù - ê ú ê ú = ê ú ê ú ë û
45. Find the inverse of 232 240, 012 B éùêú êú = êú êú ëû if it exists.
Write the augmented matrix to obtain [] 232100 |240010. 012001 BI é ùê ú ê ú = ê ú ê ú ë û 122 211 233 232100 1RRR012110 012001 3RRR208430 012110 1RRR000111 éùêú -+-êú êú êú ëû -+--éù êú êúêú -+-êú ëû
No inverse exists, since the third row is all zeros to the left of the vertical bar.
46. A and C are 32 ´ matrices, so their sum AC + is a 32 ´ matrix. Only square matrices have inverses. Therefore, 1 () AC+ does not exist.
47. Find the inverse of 13 , 27 A éù êú = êú ëû if it exists.
Write the augmented matrix [ ] |.AI [] 1310 2701 AI éù êú = êú ëû
Perform row operations on [ ] | AI to get a matrix of the form [ ] |.IB 122 211 1310 2RRR 0121 1073 3RRR 0121 éù êú êú -+ëû éù - -+ êú êúëû
The last augmented matrix is of the form [ ] |,IB so the desired inverse is 1 73 . 21 Aé ùê ú = ê ú - ë û
48. Let 42 03 A éùêú = êú ëû [] 4210 | 0301 AI éùêú = êú ëû 211 11 1 11 46 12 11 22 3 3 11 46 1 1 3 2R(3R)R12032 0301 10 RR RR 010 0 A+--éù êú êú ëû éù - êú êú êú êú ëû éùêú êú = êú êú ëû
49. Find the inverse of 36 , 48 A éùêú = êúëû if it exists.
Write the augmented matrix [ ] |.AI [] 3610 | 4801 AI éùêú = êúëû
Perform row operations on [ ] | AI to get a matrix of the form [ ] |.IB
122 3610 4R3RR 0043 éùêú êú + ëû
Since the entries left of the vertical bar in the second row are zeros, no inverse exists.
50. Let [] 122 64 32 6410 | 3201 6410 R(2)RR 0012 A AI éù êú = êú ëû éù êú = êú ëû éù êú êú +-ëû
The zeros in the second row indicate that the original matrix has no inverse.
51. Find the inverse of 210 101, 120 A éùêú êú = êú êúëû if it exists.
The augmented matrix is [] 210100 |101010. 120001 AI éùêú êú = êú êúëû 122 133 211 233 311 322 21 11 3 22 33 210100 R(2)RR012120 R(2)RR030102 1RRR202020 012120 3RRR006462 R3RR600402 R(3)RR030102 006462 1 RR 1000 6 1 RR 3 1 RR 6 éùêú +----êú êú êú +-ëû éù -+ êú êú êú êú +--ëû éù +êú êú +-êú êú ëû - 3 12 33 21 33 0100 0011 éù êú êú êúêú êú êú êúêú ëû 21 33 12 1 33 21 33 0 0 1 Aéùêú êú êú =ê-ú êú êú êúêú ëû 52. Let 204 110. 012 A é ù ê ú ê ú =ê ú ê ú - ë û [] 204100 |110010 012001 AI é
ê ú ê ú =ê ú ê ú - ë û 212 204100 024120 2RRR 012001 éù êú êú - -+ êú êúëû 323 204100 024120 2RRR008122 éù êú êúêú êú -+-ëû 311 322 1R2RR400122 1R2RR040122 008122 éù -+ êú -+-êú êú êú ëû
RR001 . Aéù êú êú êú êú êúêú êú êú êú ëû éù êú êú êú =-êú êú êú êú êú ëû
53. Find the inverse of 136 409, 51530 A éù êú êú = êú êú ëû if it exists
The last row is all zeros to the left of the bar, so no inverse exists.
54. Find the inverse of 234 157, 468 A éùêú êú = êú êú ëû if it exists.
122 133 234100 |157010 468001 234100 1R2RR01310120 2RRR000201 AI éùêú êú = êú êú ëû éùêú -+-êú êú êú + ëû
The zeros in the third row to the left of the vertical bar indicate that the original matrix has no inverse.
55. 518 , 1224 AB éùéù==êúêú êúêú ëûëû
The matrix equation to be solved is , AXB = or 518 . 1224 x y éùéùéùêúêúêú = êúêúêú ëûëûëû
Calculate the inverse of the coefficient matrix A to obtain 11 1 48 15 48 51 . 12éù êú éù êú êú = êú êú ëû ê--ú ëû
Now –1 ,XAB = so 11 48 15 48 81 . 2413 x y é ù êú é ùéùéùêú ê úêúêú == êú ê úêúêú - ë ûëûëû êú ë û
56. 125 , 2410 AB é ùéù ê ==úêú ê úêú ë ûëû
Row operations may be used to see that matrix A has no inverse. The matrix equation AXB = may be written as the system of equations
25() 2410.() xy1 xy2 += +=
Use the elimination method to solve this system. Begin by eliminating x in equation (2).
122 25() 2RRR00() xy1 3 += -+=
The true statement in equation (3) indicates that the equations are dependent. Solve equation (1) for x in terms of y.
25xy=-+
The solution is (25,), yy-+ where y is any real number.
57. 1028 110,4 3046 AB é ùéù ê úêú ê =-=úêú ê úêú ê úêú - ë ûëû
By the usual method, we find that the inverse of the coefficient matrix is 1 31 22 201 211. 0 Aé
-
Since –1 ,XAB = 31 22 201822 211418. 615 0 X é ù êú éùéù êú êúêú êú =-=-êúêú êú êúêú êú êúêúëûëû êúë û
58. 24072 120,24 00348 AB éùéù êúêú =-=-êúêú êúêú êúêú ëûëû
Use row operations to find the inverse of A, which is 11 42 11 1 84 1 3 0 0. 00 Aéù êú êú êú =-êú êú êú êú êú ëû
Since –1 ,XAB = 11 42 11 84 1 3 0 726 02415. 4816 00 X éù êú éùéù êú êúêú êú =--=êúêú êú êúêú êú êúêú êú ëûëû êú êú ëû
59. 24 231 xy xy += -=
The coefficient matrix is 12 . 23 A éù êú = êúëû
Calculate the inverse of A. 32 77 1 21 77 Aéù êú êú = êú ê-ú ëû
Use –1 XAB = to solve. 32 77 21 77 42 11 x y éù êú éùéùéù êú êúêúêú == êú êúêúêú ëûëûëû ê-ú ëû
The solution is (2, 1).
60. 51080 32120 xy xy += -=
Let 51080 ,,. 32120 x AXB y éùéùéù ===êúêúêú êúêúêúëûëûëû
Use row operations to find the inverse of A, which is 11 204 1 31 408 Aéù êú êú = êú ê-ú ëû
Since –1 ,XAB = 11 204 31 408 8034 1209 x y é ù êú é ùéùéù êú ê úêúêú == êú ê úêúêú - ë ûëûëû ê-ú ë û
The solution is () 34,9. -
61. 1 22 32 xyz xy yz ++= +=+=
The coefficient matrix is 111 210. 031 A éù êú êú = êú êú ëû
Find that the inverse of A is 121 555 212 1 555 631 555 . Aé ù - ê ú ê ú ê ú =ê-ú ê ú ê ú ê ú ê ú ë û
Now –1 ,XAB = so 121 555 212 555 631 555 11 20. 22 x y z é ùêú é ùéùéù - êú ê úêúêú êú ê úêúêú =--= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êú ê ú ë û
The solution is () 1, 0, 2. -
62. 421 239 3527 xyz xyz xyz -+=-+-=+-=
Let 1421 213,,9. 3527 x AXyB z é ùéùéù ê úêúêú ê úêúêú =--==ê úêúêú ê úêúêú - ë ûëûëû
Use row operations to find the inverse of A, which is
1210 33939 181 1 33939 1177 33939 13210 1 1381. 39 13177 Aéù êú éù êú êú êú êú =ê---ú=--êú êú êú êú ëû êú êú ëû
Since –1 ,XAB = 132101 1 13819 39 131777 x y z éùéùéùêúêúêú êúêúêú =---êúêúêú êúêúêú ëûëûëû 391 1 782. 39 1173 éùéù êúêú ==êúêú êúêú êúêú ëûëû
The solution is (1, 2, 3).
63. 0.010.05200 , 0.040.03300 AD éùéù ==êúêú êúêú ëûëû 1 () XIAD=-
() 1
1.2660.12660.6751 0.12661.01270.40084 001.6667
Since –1 (), XIAD =1.2660.12660.6751500725.7 0.12661.01270.40084200305.9. 001.6667100166.7 X é ùéùéù ê úêúêú ê ==úêúêú ê úêúêú ê úêúêú ë ûëûëû 65. 27() 22() 3325() xyz1 xyz2 xyz3 ++= --= -+=-
(a) To solve the system by the echelon method, begin by eliminating x in equations (2) and (3).
122 133 27() 2RRR5312() 3RRR926() xyz1 yz4 yz5 ++= -+--=-+--=-
100.010.05 010.040.03
0.990.05 0.040.97 IA éùéù -=-êúêú êúêú ëûëû éùêú = êúëû
Use row operations to find the inverse of ,IAwhich is 1 1.01220.0522 (). 0.04171.0331 IAéù êú -= êú ëû
Since 1 (), XIAD=- the production matrix is
Eliminate y in equation (5). 233 27() 5312() 9R5RR2222() xyz1 yz4 z6 ++= --=-+=-
Make each leading coefficient equal 1. 1 22 5 1 33 22 27() 312 RR() 55 RR1() xyz1 yz7 z8 ++= -+= =-
1.01220.0522200 0.04171.0331300
218.1
318.3 X éùéù êúêú = êúêú ëûëû éù êú = êú ëû
64. () 1 0.20.10.3500 0.100.2,200 000.4100 AD XIADéùéù êúêú ==êúêú êúêú êúêú ëûëû =-
0.80.10.3 0.110.2 000.6 IA éù êú -=--êú êú êú ëû
Substitute 1 - for z in equation (7) to get 3. y = Substitute 1 - for z and 3 for y in equation (1) to get 2. x = The solution is ( ) 2,3,1. -
(b) The same system is to be solved using the Gauss-Jordan method. Write the augmented matrix and use row operations. 1217 2112 3325
ê ú ê ú ê ú ê ú ë û 122 133 1217 2RRR05312 3RRR09126 é ù ê ú ê ú -+--ê ú ê ú -+--ë û
2R5RR50111 05312
9R5RR002222 R22RR11000220
3R22RR01100330
0.230.320.05 0.320.050.14. 0.140.410.23
The corresponding system is 2 3 1. x y z = = =-
The solution is () 2,3,1 -
(c) The system can be written as a matrix equation AXB = by writing 1217 2112. 3325 x y z éùéùéù êúêúêú êúêúêú --= êúêúêú
(d) The inverse of the coefficient matrix A can be found by using row operations.
66. Use a table to organize the information.
2RRR053210 3RRR091301
2R5RR501120 053210
9R5RR0022395 R22RR1100025355
3R22RR0110035515
122 133 211 233 311 322 121100 |211010 332001 121100
Let x = the number of standard paper clips (in thousands), and y = the number of extra large paper clips (in thousands).
The given information leads to the system 11 43 11 23 4 6. xy xy += +=
Solve this system by any method to get 8, x = 6. y = The manufacturer can make 8 thousand (8000) standard and 6 thousand (6000) extra large paper clips.
67. Let x1 = the number of blankets, x2 = the number of rugs, and x3 = the number of skirts.
The given information leads to the system 123 123 123 243012306() 45359() 15189201.() xxx1 xxx2 xxx3 ++= ++= ++=
Simplify equations (1) and (3). 1 11123 6 123 1 33123 3 RR45251() 45359() RR56367() xxx4 xxx2 xxx5 ++= ++= ++=
Solve this system by the Gauss-Jordan method. Write the augmented matrix and use row operations. 122 313 45251 45359 56367 45251 1RRR0018 4R5RR01213 éù êú êú êú êú ëû éù êú êú -+ êú -+--êú ëû
Interchange the second and third rows. 211 45251 01213 0018 5RRR 4012116 01213 0018 éù êú êú êú êú ëû éù -+ êú êú êú êú ëû 311 322 1 11 4 12RRR40020 2RRR0103 0018 RR 1005 0103 0018 éù -+ êú êú + êú êú ëû éù êú êú êú êú ëû
The solution of the system is 5, 3, xy== 8. z = So, 5 blankets, 3 rugs, and 8 skirts can be made.
68. Let x = Tulsa's number of gallons, y = New Orleans’ number of gallons, and z = Ardmore’s number of gallons.
The system that may be written is 0.50.40.3219,000 0.20.40.4192,000 0.30.20.3144,000. xyzChicago xyzDallas xyzAtlanta ++= ++= ++=
The augmented matrix is 0.50.40.3219,000 0.20.40.4192,000. 0.30.20.3144,000 éù êú êú êú êú ëû
0.50.40.3219,000
2R(5)RR01.21.4522,000
3R(5)RR00.20.663,000
2RRR0.501.5345,000 01.21.4522,000
R6RR005900,000
0.3RRR0.50075, 14R50RR
122 133 311 233 311 322
+ -+ 11 1 22 60 1 33 5 000 060013,500,000 005900,000 2RR 100150,000 RR010225,000 001180,000 RR
Thus, 150,000 gal were produced at Tulsa, 225,000 gal at New Orleans, and 180,000 gal at Ardmore.
69. (a) Let x = the number of football jerseys y = the number of basketball jerseys z = the number of baseball jerseys
The system to be solved is Cutting1.50.5380 Sewing1.20.60.9330 xyz xyz ++= ++=
(b) Adding 4 times the first equation to 5 times the second yields 0.5130 so 1300.5. yz yz −−= =−
Substitute this value for y in the original cutting equation and solve for x in terms of z
1.50.5(1300.5)380 2100.5 xzz xz +−+= =−
The general solution is (2100.5,1300.5,). zzz
(c) For all quantities to be nonnegative we need 1300.50or0260 zz −≥≤≤
So the possible values for z are 0, 1, 2, … , 260.
(d) The greatest number of baseball jerseys is 260, which yields the solution (2100.5(260),1300.5(260),260) or(80,0,260)
The production will be 80 football jerseys, no basketball jerseys, and 260 baseball jerseys.
70. (a) Let x = the number of reserved seats
y = the number of box seats
z = the number of infield seats
The system to be solved is 8101247,000 5000 2 xyz yz xy ++= ++= =
Substitute 2y for x in the first two equations.
261247,000 5000 yz yz += +=
Add 12 times the second equation to the first equation.
1013,000 1300 2(1300)2600 5000130026001100 y y x z −=− = == =−−=
The stadium has 2600 reserved seats, 1300 box seats, and 1100 infield seats.
(b) The system to be solved is 8101222,264 3608 3 xyz xyz xz ++= ++= =
Substitute 3z for x in the first two equations. 103622,264 43608 yz yz += +=
Add 9 times the second equation to the first equation.
22,264(9)(3608)10,208 y =−=−
Since the solution yields a negative number of box seats, the outcome is impossible.
71. (a)
NEMWSW 3156180102Jan 3867209134Feb 3361237109Mar
NEMWSW 3051256120Jan 245726195Feb 305722892Mar
(b) 30512561203156180102 2457261953867209134 3057228923361237109 157618 14105239 34917
(c) The general trend is down, with only the South showing strongly improved sales averaged over the three months.
72. (a) High3170 Medium2360 Coated1800 é ù ê ú ê ú ê ú ê ú ë û (b) x y z é ù
(c) 10583170 12042360 01051800 x y z é ùéùéù ê úêúêú ê úêúêú = ê úêúêú ê úêúêú ë ûëûëû (d) 1 10583170 12042360 01051800 x y z - é ùéùéù ê úêúêú ê úêúêú = ê úêúêú ê úêúêú ë ûëûëû
0.1540.2120.07693170150 0.2310.1920.21542360110 0.4620.3850.2311800140 é ùéùéùê úêúêú ê =-=úêúêú ê úêúêú ê úêúêú ë ûëûëû
73. (a) The input-output matrix is 1 2 2 3 0 0 A éù êú êú = êú êú ëû
(b) 1 2 2 3 1 400 , 800 1 IAD é ùêú éù êú êú -== êú êú ëûêú ë û
Use row operations to find the inverse of ,IA - which is
33 24 1 3 2 (). 1 IAéù êú êú -= êú êú ëû
Since –1 (), XIAD =33 24 3 2 4001200 8001600 1 X éù êú éùéù êú ==êúêú êú êúêú ëûëû êú ëû
The production required is 1200 units of cheese and 1600 units of goats.
74. (a) Use a graphing calculator or a computer to find –1 (). IA - The solution, which may vary slightly, is
1.300.0450.5670.0120.0680.020
0.2041.0300.1830.0040.0220.006
0.1550.0381.1200.0200.1140.034
0.0180.0210.0281.0800.0160.033
0.5370.5250.4830.2791.7400.419
0.5730.3460.4970.5360.0871.940
Values have been rounded.
The value in row 2, column 1 of this matrix, 0.204, indicates that every $1 of increased demand for livestock will result in an increase of production demand of $0.204 in crops.
(b) Use a graphing calculator or computer to find –1 (). IAD - The solution, which may vary slightly, is 3855 1476 2726 . 1338 8439 10,256 éù êú êú êú êú êú êú êú êú êú êú êú ëû
Values have been rounded. In millions of dollars, produce $3855 in livestock, $1476 in crops, $2726 in food products, $1338 in mining and manufacturing, $8439 in households, and $10,256 in other business sectors.
75. The given information can be written as the following 43 ´ matrix. 888 1059 7107 897 éù êú êú êú êú êú êú êú ëû
76. (a) The X-ray passes through cells B and C, so the attenuation value for beam 3 is bc +
(b)
Beam1:0.8
Beam2:0.55
Beam3:0.65 ab ac bc += += += 1100.8 1010.55 0110.65 a b c é ùéùéù ê úêúêú ê úêúêú = ê úêúêú ê úêúêú ë ûëûëû 1 111 222 111 222 111 222 1100.8 1010.55 0110.65 0.8 0.55 0.65 a b céùéùéù êúêúêú êúêúêú = êúêúêú êúêúêú ëûëûëû éùêú éù êú êú êú êú =-êú êú êú êú êú ëû êúê ú ë û 0.35 0.45 0.2 é ù ê ú ê ú = ê ú ê ú ë û
The solution is (0.35,0.45,0.2), so A is tumorous, B is bone, and C is healthy.
(c) For patient X, 111 222 111 222 111 222 0.540.21 0.400.33. 0.520.19 a b c é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êúê ú ë û
A and C are healthy; B is tumorous. For patient Y, 111 222 111 222 111 222 0.650.35 0.800.3. 0.750.45 a b c é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êúê ú ë û
A and B are tumorous; C is bone. For patient Z, 111 222 111 222 111 222 0.510.28 0.490.23. 0.440.21 a b c é ùêú é ùéùéù êú ê úêúêú êú ê úêúêú =-= êú ê úêúêú êú ê úêúêú êú ë ûëûëû êúê ú ë û
A could be healthy or tumorous; B and C are healthy.
77. (a) 0.60() 0.75() 0.65() 0.70() ab1 cd2 ac3 bd4 += += += +=
The augmented matrix of the system is 133 11000.60 00110.75 10100.65 01010.70 11000.60 00110.75
1RRR 01100.05 01010.70
Interchange rows 2 and 4. 211 233 11000.60 01010.70 01100.05 00110.75
1RRR10010.10 01010.70 RRR00110.75 00110.75
Since R3 and R4 are identical, there will be infinitely many solutions. We do not have enough information to determine the values of a,b,c, and d
(b) i. If 0.33, d = the system of equations in part (a) becomes
Equation (2) gives 0.32, c = and equation (4) gives 0.27. b = Substituting 0.32 c = into equation (3) gives 0.33. a = Therefore, 0.33, 0.27, 0.32, and abc === 0.43. d =
Thus, A and C are tumorous, B could be healthy or tumorous, and D is bone.
(c) The original system now has two additional equations.
0.60() 0.75() 0.65() 0.70() 0.85() 0.50() ab1 cd2 ac3 bd4 bc5 ad6 += += += += += +=
The augmented matrix of this system is 11000.60 00110.75 10100.65 01010.70 01100.85 10010.50
0.60()
0.330.75() 0.65()
0.330.70.() ab1 c2 ac3 b4 += += += +=
Equation (2) gives 0.42, c = and equation (4) gives 0.37. b = Substituting 0.42 c = into equation (3) gives 0.23. a = Therefore, 0.23, 0.37, 0.42, and 0.33. abcd ==== Thus, A is healthy, B and D are tumorous, and C is bone.
ii. If 0.43, d = the system of equations in part (a) becomes 0.60()
Using the Gauss-Jordan method we obtain 10000.20 01000.40 00100.45 . 00010.30 00000 00000
Therefore, 0.20, 0.40, 0.45, abc === and 0.30. d = Thus, A is healthy, B and C are bone, and D is tumorous.
(d) As we saw in part (c), the six equations reduced to four independent equations. We need only four beams, correctly chosen, to obtain a solution. The four beams must pass through all four cells and must lead to independent equations. One such choice would be beams 1, 2, 3, and 6. Another choice would be beams 1, 2, 4, and 5.
0.430.70.() ab1 c2 ac3 b4 += += += +=
0.430.75() 0.65()
78. The matrix representing the rates per 1000 athleteexposures for specific injuries that caused a player wearing either shield to miss one or more events is
Since an equal number of players wear each type of shield and the total number of athlete-exposures for the league in a season is 8000, each type of shield is worn by 4000 players. Since the rates are given per 1000 athletic-exposures, the matrix representing the number of 1000 athlete-exposures for each type of shield is 4 4 éù êú êú ëû
The product of these matrices is 20 12 3 55 éù êú êú êú êú êú êú êú ëû
Values have been rounded.
There would be about 20 head and face injuries, 12 concussions, 3 neck injuries, and 55 other injuries.
79. 12 1 2 3 ()100() 2 0() WW1 WW2 += -=
Equation (2) gives 12.WW = Substitute W1 for W2 in equation (1). 11 1 1 1 3 ()100 2 3 (2)100 2 3100 1001003 58 3 3 WW W W W += = = ==»
Therefore, 12 58 lb. WW=»
80. 12 12 12 150() 22 32 0() 22 WW1 WW2 += -=
Adding equations (1) and (2) gives 1 13 150. 22 W æö ÷ ç ÷ ç += ÷ ç ÷ ÷ ç èø
Multiply by 2. 1 1 (13)300 300 110 13 W W += =» +
From equation (2), 12 21 32 22 3 2 WW WW = =
Substitute 300 13 + from above for W1 2 33003003 134 213(13)2 W =⋅=» ++
Therefore, 12 110 lb and 134 lb. WW»»
81. 2 Catbyc =++
Use the values for C from the table.
(a) Letting 0 t = in 1960, we get three equations for the coefficients a, b, and c 317 40020339 250050390 c abc abc = ++= ++=
Using the value for c in the first equation in the other equations we have the system 4002022 25005073 ab ab += +=
Adding 5 times the first equation to twice the second equation gives 2 300014611036. 36 So0.012 3000 22(400)(0.012) 0.86 20
Thus,0.0120.86317 a a b Ctt =−= == == =++
(b) The concentration will have doubled when 2 2 634. 0.0120.86317634 0.0120.863170 C Solve tt tt = ++= +−=
The roots are given by the quadratic formula:
2 2
−+−− = =
0.86(0.86)(4)(0.012)(317) 130.602 (2)(0.012)
0.86(0.86)(4)(0.012)(317) 202.269 (2)(0.012)
The positive root represents about 131 years after 1960, or 2091.
éùéùéù êúêúêú += êúêúêú ëûëûëû
éùéùéù êúêúêú += êúêúêú ëûëûëû
éùéù êúêú = êúêú + ëûëû
82. (a) 101 122 01 22 1 22 xy x xy x xy
Since corresponding elements must be equal, 1 x = and 22.xy+= Substituting 1 x = in the second equation gives 1 2 . y = Note that 1 x = and 1 2 y = are the values that balance the equation.
éùéùéùéù êúêúêúêú êúêúêúêú ++= êúêúêúêú êúêúêúêú ëûëûëûëû
éùéùéùéù
êúêúêúêú êúêúêúêú ++= êúêúêúêú êúêúêúêú ëûëûëûëû
éùéù + êúêú êúêú = êúêú êúêú + ëûëû
(b) 222 COHCOHO xyz++= 1010 0202 2011 00 0202 201 0 22 21 xyz xz y xz xz y xz
Since corresponding elements must be equal, 0, 22, and21.xzyxz +==+= Solving 22 gives1.yy== Solving the system 0 21 xz xz ì += ï ï í ï += ï î gives 1 and –1. xz== Thus, the values that balance the equation are 1, x = 1,and–1.yz==
83. Letthe number of boys andthe number of girls. x y = =
0.20.3500()
0.60.91500() xy1 xy2 += +=
The augmented matrix is
0.20.3500 . 0.60.91500
11.52500
5RR 0.60.91500
11
éù êú êú ëû éù êú êú ëû éù êú êú -+ ëû
11.52500
0.6RRR 000
122
Thus, 1.52500 25001.5. xy xy += =-
There are y girls and 25001.5 y - boys, where y is any even integer between 0 and 1666 since 0 y ³ and 25001.50 1.52500 1666.6. y y y -³ -³£
84. Letthe number of singles, the number of doubles, the number of triples, and the number of home runs hit by Ichiro Suzuki. x y z w = = = =
If the number of singles he hit was 11 more than four times the total of doubles and home runs, then 4()11, or 44 xywxyw =++-- 11. =
If the number of doubles he hit was 1 more than twice this total of triples and home runs, then 2()1, or 221.yzwzyw =++-+-=
If the total of singles and home runs he hit was 15 more than five times the total of doubles and triples, then 5()15, or xwyz+=++ 5515.xyzw--+=
Write the augmented matrix and use row operations to solve the system.
1111225 140411 01221 155115
133 144 1112250 01221 RRR 0515214 RRR 0660210 éù êú êú êú êú -+ êú êú -+ êú ëû 33 1111225 01221 RR0515214 0660210 éù êú êú êú êú - êú êú êú ëû
211 233 244 RRR 1033224 01221
5RRR 001115209 6RRR 001812204 éù -+ êú êú êú êú -+ êú êú + êú ëû 34 1033224 01221 001812204 RR 001115209 éù êú êú êú êú êú êú « êú ëû 1234 33 1833 1033224 01221 RR001 001115209 éù êú êú êú êú êú - êú êú ëû 311 271 33 322 234 33 23253 344 33 1001190 3RRR 010 2RRR 001 11RRR 000 éù -+ êú êúêú + êú êú êú êú êú -+ êú ëû 271 33 234 33 3 44 23 1001190 010 001 RR 000111 éù êú êúêú êú êú êú êú êú ëû 411 2 422 3 2 433 3 1RRR1000179 RRR010031 RRR00104 000111 éù -+ êú êú êú + êú êú -+ êú êú êú êú ëû
Ichiro Suzuki hit 179 singles, 31 doubles, 4 triples, and 11 home runs.
85. Letthe weight of a single chocolate wafer and the weight of a single layer of vanilla creme. x y = =
A serving of regular Oreo cookies is three cookies so that 3(2)34. xy+=
A serving of Double Stuf is two cookies so that 2(22)29. xy+=
Write the equations in proper form, obtain the augmented matrix, and use row operations to solve. 122 211 149 11 1212 119 22 66 6334 4429 6334
2R3RR 0619
1R2RR12049 0619 RR10 RR01 éù êú êú ëû éù êú êú -+ ëû éù -+ êú êú ëû éù êú êú êú êú ëû
The solution is () 4919 126,, or about (4.08,3.17).
A chocolate wafer weighs 4.08 g and a single layer of vanilla creme weighs 3.17g.
Extended Application: Contagion
1. 110111 10010 000010 00110 000000 11000 010100 100010 120211 010100 110121 PQ éù êú êú éù êú êú êú êú = êú êú êú êú êú ëû êú êú ëû éù êú êú = êú êú ë û
2. In the product PQ, 23 0, a = so there were no contacts.
3. In the product PQ, column 3 has all zeros. The third person had no contacts with the first group.
4. In the product PQ, the entries in columns 2 and 4 both have a sum of 4 while the entries in column 5 have a sum of 3. In Q, columns 2, 4, and 5 have a sum of 2, 2, and 3, respectively. Therefore, the second, fourth, and fifth persons in the third group each had a total of 6 first- and second-order contacts.