# 13.2.9: Chapter 9

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### 9.1 Systems of Linear Equations: Two Variables

1.

Not a solution.

2.

The solution to the system is the ordered pair (−5,3).(−5,3).

3.

(−2,−5)(−2,−5)

4.

(−6,−2)(−6,−2)

5.

(10,−4)(10,−4)

6.

No solution. It is an inconsistent system.

7.

The system is dependent so there are infinite solutions of the form (x,2x+5).(x,2x+5).

8.

### 9.2 Systems of Linear Equations: Three Variables

1.

(1,−1,1)(1,−1,1)

2.

No solution.

3.

Infinite number of solutions of the form (x,4x−11,−5x+18).(x,4x−11,−5x+18).

### 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables

1.

(−12,12)(−12,12) and (2,8)(2,8)

2.

(−1,3)(−1,3)

3.

{(1,3),(1,−3),(−1,3),(−1,−3)}{ (1,3),(1,−3),(−1,3),(−1,−3) }

4.

### 9.4 Partial Fractions

1.

3x−3−2x−23x−3−2x−2

2.

6x−1−5(x−1)26x−1−5(x−1)2

3.

3x−1+2x−4x2+13x−1+2x−4x2+1

4.

x−2x2−2x+3+2x+1(x2−2x+3)2x−2x2−2x+3+2x+1(x2−2x+3)2

### 9.5 Matrices and Matrix Operations

1.

A+B=⎡⎣⎢2116​​​0−3⎤⎦⎥+⎡⎣⎢31−4−253⎤⎦⎥A+B=[ 2116​​​0−3 ]+[ 31−4−253 ]
=⎡⎣⎢2+31+11+(−4)6+(−2)0+5−3+3⎤⎦⎥=⎡⎣⎢52−3450⎤⎦⎥=[ 2+31+11+(−4)6+(−2)0+5−3+3 ]=[ 52−3450 ]

2.

−2B=[−8−6−2−4]−2B=[ −8−2−6−4 ]

### 9.6 Solving Systems with Gaussian Elimination

1.

[43−32∣∣∣114][ 4−332|114 ]

2.

x−y+z=52x−y+3z=1y+z=−9x−y+z=52x−y+3z=1y+z=−9

3.

(2,1)(2,1)

4.

⎡⎣⎢1​00−52105251∣∣∣∣∣17292⎤⎦⎥[ 1−5252​015001|17292 ]

5.

(1,1,1)(1,1,1)

6.

$150,000 at 7%,$750,000 at 8%, $600,000 at 10% ### 9.7 Solving Systems with Inverses 1. AB=[1−14−3][−31−41]=[1(−3)+4(1)−1(−3)+−3(1)1(−4)+4(1)−1(−4)+−3(1)]=[1001]BA=[−31−41][1−14−3]=[−3(1)+−4(−1)1(1)+1(−1)−3(4)+−4(−3)1(4)+1(−3)]=[1001]AB=[ 14−1−3 ][ −3−411 ]=[ 1(−3)+4(1)1(−4)+4(1)−1(−3)+−3(1)−1(−4)+−3(1) ]=[ 1001 ]BA=[ −3−411 ][ 14−1−3 ]=[ −3(1)+−4(−1)−3(4)+−4(−3)1(1)+1(−1)1(4)+1(−3) ]=[ 1001 ] 2. A−1=[35−251515]A−1=[ 3515−2515 ] 3. A−1=⎡⎣⎢1231462−3−5⎤⎦⎥A−1=[ 11224−336−5 ] 4. X=⎡⎣⎢43858⎤⎦⎥X=[ 43858 ] ### 9.8 Solving Systems with Cramer's Rule 1. (3,−7)(3,−7) 2. −10−10 3. (−2,35,125)(−2,35,125) ### 9.1 Section Exercises 1. No, you can either have zero, one, or infinitely many. Examine graphs. 3. This means there is no realistic break-even point. By the time the company produces one unit they are already making profit. 5. You can solve by substitution (isolating xx or yy ), graphically, or by addition. 7. Yes 9. Yes 11. (−1,2)(−1,2) 13. (−3,1)(−3,1) 15. (−35,0)(−35,0) 17. No solutions exist. 19. (725,1325)(725,1325) 21. (6,−6)(6,−6) 23. (−12,110)(−12,110) 25. No solutions exist. 27. (−15,23)(−15,23) 29. (x,x+32)(x,x+32) 31. (−4,4)(−4,4) 33. (12,18)(12,18) 35. (16,0)(16,0) 37. (x,2(7x−6))(x,2(7x−6)) 39. (−56,43)(−56,43) 41. Consistent with one solution 43. Consistent with one solution 45. Dependent with infinitely many solutions 47. (−3.08,4.91)(−3.08,4.91) 49. (−1.52,2.29)(−1.52,2.29) 51. (A+B2,A−B2)(A+B2,A−B2) 53. (−1A−B,AA−B)(−1A−B,AA−B) 55. (CE−BFBD−AE,AF−CDBD−AE)(CE−BFBD−AE,AF−CDBD−AE) 57. They never turn a profit. 59. (1,250,100,000)(1,250,100,000) 61. The numbers are 7.5 and 20.5. 63. 24,000 65. 790 sophomores, 805 freshman 67. 56 men, 74 women 69. 10 gallons of 10% solution, 15 gallons of 60% solution 71. Swan Peak:$750,000, Riverside: $350,000 73.$12,500 in the first account, $10,500 in the second account. 75. High-tops: 45, Low-tops: 15 77. Infinitely many solutions. We need more information. ### 9.2 Section Exercises 1. No, there can be only one, zero, or infinitely many solutions. 3. Not necessarily. There could be zero, one, or infinitely many solutions. For example, (0,0,0)(0,0,0) is not a solution to the system below, but that does not mean that it has no solution. 2x+3y−6z=1−4x−6y+12z=−2x+2y+5z=102x+3y−6z=1−4x−6y+12z=−2x+2y+5z=10 5. Every system of equations can be solved graphically, by substitution, and by addition. However, systems of three equations become very complex to solve graphically so other methods are usually preferable. 7. No 9. Yes 11. (−1,4,2)(−1,4,2) 13. (−85107,312107,191107)(−85107,312107,191107) 15. (1,12,0)(1,12,0) 17. (4,−6,1)(4,−6,1) 19. (x,127(65−16x),x+2827)(x,127(65−16x),x+2827) 21. (−4513,1713,−2)(−4513,1713,−2) 23. No solutions exist 25. (0,0,0)(0,0,0) 27. (47,−17,−37)(47,−17,−37) 29. (7,20,16)(7,20,16) 31. (−6,2,1)(−6,2,1) 33. (5,12,15)(5,12,15) 35. (−5,−5,−5)(−5,−5,−5) 37. (10,10,10)(10,10,10) 39. (12,15,45)(12,15,45) 41. (12,25,45)(12,25,45) 43. (2,0,0)(2,0,0) 45. (1,1,1)(1,1,1) 47. (128557,23557,28557)(128557,23557,28557) 49. (6,−1,0)(6,−1,0) 51. 24, 36, 48 53. 70 grandparents, 140 parents, 190 children 55. Your share was$19.95, Sarah’s share was $40, and your other roommate’s share was$22.05.

57.

59.

500 students, 225 children, and 450 adults

61.

The BMW was $49,636, the Jeep was$42,636, and the Toyota was $47,727. 63.$400,000 in the account that pays 3% interest, $500,000 in the account that pays 4% interest, and$100,000 in the account that pays 2% interest.

65.

The United States consumed 26.3%, Japan 7.1%, and China 6.4% of the world’s oil.

67.

Saudi Arabia imported 16.8%, Canada imported 15.1%, and Mexico 15.0%

69.

Birds were 19.3%, fish were 18.6%, and mammals were 17.1% of endangered species

### 9.3 Section Exercises

1.

A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches also intersect the circle, hence three solutions.

3.

No. There does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region.

5.

Choose any number between each solution and plug into C(x)C(x) and R(x).R(x). If C(x)<R(x),C(x)<R(x), then there is profit.

7.

(0,−3),(3,0)(0,−3),(3,0)

9.

(−32√2,32√2),(32√2,−32√2)(−322,322),(322,−322)

11.

(−3,0),(3,0)(−3,0),(3,0)

13.

(14,−62√8),(14,62√8)(14,−628),(14,628)

15.

(−398√4,1994),(398√4,1994)(−3984,1994),(3984,1994)

17.

(0,2),(1,3)(0,2),(1,3)

19.

(−12(5–√−1)−−−−−−−−√,12(1−5–√)),(12(5–√−1)−−−−−−−−√,12(1−5–√))(−12(5−1),12(1−5)),(12(5−1),12(1−5))

21.

(5,0)(5,0)

23.

(0,0)(0,0)

25.

(3,0)(3,0)

27.

No Solutions Exist

29.

No Solutions Exist

31.

(−2√2,−2√2),(−2√2,2√2),(2√2,−2√2),(2√2,2√2)(−22,−22),(−22,22),(22,−22),(22,22)

33.

(2,0)(2,0)

35.

(−7–√,−3),(−7–√,3),(7–√,−3),(7–√,3)(−7,−3),(−7,3),(7,−3),(7,3)

37.

(−12(73−−√−5)−−−−−−−−−√,12(7−73−−√)),(12(73−−√−5)−−−−−−−−−√,12(7−73−−√))(−12(73−5),12(7−73)),(12(73−5),12(7−73))

39.

41.

43.

45.

47.

49.

(−270383−−−√,−23529−−√),(−270383−−−√,23529−−√),(270383−−−√,−23529−−√),(270383−−−√,23529−−√)(−270383,−23529),(−270383,23529),(270383,−23529),(270383,23529)

51.

No Solution Exists

53.

x=0,y>0x=0,y>0 and 0<x<1,x−−√<y<1x0<x<1,x<y<1x

55.

12, 288

57.

2–20 computers

### 9.4 Section Exercises

1.

No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, 1x2+11x2+1 cannot be decomposed because the denominator cannot be factored.

3.

Graph both sides and ensure they are equal.

5.

If we choose x=−1,x=−1, then the B-term disappears, letting us immediately know that A=3.A=3. We could alternatively plug in x=−53x=−53, giving us a B-value of −2.−2.

7.

8x+3−5x−88x+3−5x−8

9.

1x+5+9x+21x+5+9x+2

11.

35x−2+44x−135x−2+44x−1

13.

52(x+3)+52(x−3)52(x+3)+52(x−3)

15.

3x+2+3x−23x+2+3x−2

17.

95(x+2)+115(x−3)95(x+2)+115(x−3)

19.

8x−3−5x−28x−3−5x−2

21.

1x−2+2(x−2)21x−2+2(x−2)2

23.

−64x+5+3(4x+5)2−64x+5+3(4x+5)2

25.

−1x−7−2(x−7)2−1x−7−2(x−7)2

27.

4x−32(x+1)+72(x+1)24x−32(x+1)+72(x+1)2

29.

4x+2x2−33x+2+72(3x+2)24x+2x2−33x+2+72(3x+2)2

31.

x+1x2+x+3+3x+2x+1x2+x+3+3x+2

33.

4−3xx2+3x+8+1x−14−3xx2+3x+8+1x−1

35.

2x−1x2+6x+1+2x+32x−1x2+6x+1+2x+3

37.

1x2+x+1+4x−11x2+x+1+4x−1

39.

2x2−3x+9+3x+32x2−3x+9+3x+3

41.

−14x2+6x+9+12x−3−14x2+6x+9+12x−3

43.

1x+1x+6−4xx2−6x+361x+1x+6−4xx2−6x+36

45.

x+6x2+1+4x+3(x2+1)2x+6x2+1+4x+3(x2+1)2

47.

x+1x+2+2x+3(x+2)2x+1x+2+2x+3(x+2)2

49.

1x2+3x+25−3x(x2+3x+25)21x2+3x+25−3x(x2+3x+25)2

51.

18x−x8(x2+4)+10−x2(x2+4)218x−x8(x2+4)+10−x2(x2+4)2

53.

−16x−9x2+16x−1−7(x−1)2−16x−9x2+16x−1−7(x−1)2

55.

1x+1−2(x+1)2+5(x+1)31x+1−2(x+1)2+5(x+1)3

57.

5x−2−310(x+2)+7x+8−710(x−8)5x−2−310(x+2)+7x+8−710(x−8)

59.

−54x−52(x+2)+112(x+4)+54(x+4)−54x−52(x+2)+112(x+4)+54(x+4)

### 9.5 Section Exercises

1.

No, they must have the same dimensions. An example would include two matrices of different dimensions. One cannot add the following two matrices because the first is a 2×22×2 matrix and the second is a 2×32×3 matrix. [1324]+[635241][ 1234 ]+[ 654321 ] has no sum.

3.

Yes, if the dimensions of AA are m×nm×n and the dimensions of BB are n×m,n×m, both products will be defined.

5.

Not necessarily. To find AB,AB, we multiply the first row of AA by the first column of BB to get the first entry of AB.AB. To find BA,BA, we multiply the first row of BB by the first column of AA to get the first entry of BA.BA. Thus, if those are unequal, then the matrix multiplication does not commute.

7.

⎡⎣⎢111517199467⎤⎦⎥[ 111915941767 ]

9.

[−4821][ −4281 ]

11.

Undidentified; dimensions do not match

13.

⎡⎣⎢96302736192⎤⎦⎥[ 92763360192 ]

15.

[−64−360−12−20−28−12−72−116][ −64−12−28−72−360−20−12−116 ]

17.

⎡⎣⎢1,8008007001,2001,4004001,3006002,100⎤⎦⎥[ 1,8001,2001,3008001,4006007004002,100 ]

19.

[202810228][ 201022828 ]

21.

[60−16411202−216][ 60412−16120−216 ]

23.

⎡⎣⎢−68−54−5724−123013664128⎤⎦⎥[ −6824136−54−1264−5730128 ]

25.

Undefined; dimensions do not match.

27.

⎡⎣⎢−840441−1527−3−1442⎤⎦⎥[ −841−340−15−1442742 ]

29.

⎡⎣⎢−840330−10650360900−530250110⎤⎦⎥[ −840650−530330360250−10900110 ]

31.

[−3503501,050350][ −3501,050350350 ]

33.

Undefined; inner dimensions do not match.

35.

[1,400−1,400700700][ 1,400700−1,400700 ]

37.

[332,500−227,500927,50087,500][ 332,500927,500−227,50087,500 ]

39.

[490,00000490,000][ 490,00000490,000 ]

41.

[−2−7394−7][ −234−79−7 ]

43.

[−4−2729−3211][ −42921−27−31 ]

45.

⎡⎣⎢−3−28−4−25916−2467⎤⎦⎥[ −3−2−2−285946−4167 ]

47.

⎡⎣⎢1−198−72−18505126−936991⎤⎦⎥[ 1−18−9−198505369−7212691 ]

49.

[091.6−1][ 01.69−1 ]

51.

⎡⎣⎢212−8243264−4.5−961⎤⎦⎥[ 224−4.51232−9−86461 ]

53.

⎡⎣⎢0.52103170.5210⎤⎦⎥[ 0.530.521210710 ]

55.

⎡⎣⎢100010001⎤⎦⎥[ 100010001 ]

57.

⎡⎣⎢100010001⎤⎦⎥[ 100010001 ]

59.

Bn=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎡⎣⎢100010001⎤⎦⎥, neven,⎡⎣⎢100001010⎤⎦⎥, nodd.Bn={ [ 100010001 ], neven,[ 100001010 ], nodd.

### 9.6 Section Exercises

1.

Yes. For each row, the coefficients of the variables are written across the corresponding row, and a vertical bar is placed; then the constants are placed to the right of the vertical bar.

3.

No, there are numerous correct methods of using row operations on a matrix. Two possible ways are the following: (1) Interchange rows 1 and 2. Then R2=R2−9R1.R2=R2−9R1. (2) R2=R1−9R2.R2=R1−9R2. Then divide row 1 by 9.

5.

No. A matrix with 0 entries for an entire row would have either zero or infinitely many solutions.

7.

[0916−1∣∣∣42][ 0169−1|42 ]

9.

⎡⎣⎢1123534809∣∣∣∣164−7⎤⎦⎥[ 1581230349|164−7 ]

11.

−2x+5y=56x−18y=26−2x+5y=56x−18y=26

13.

3x+2y=3−x−9y+4z=−18x+5y+7z=83x+2y=3−x−9y+4z=−18x+5y+7z=8

15.

4x+5y−2z=12 y+58z=28x+7y−3z=−54x+5y−2z=12 y+58z=28x+7y−3z=−5

17.

No solutions

19.

(−1,−2)(−1,−2)

21.

(6,7)(6,7)

23.

(3,2)(3,2)

25.

(15,12)(15,12)

27.

(x,415(5x+1))(x,415(5x+1))

29.

(3,4)(3,4)

31.

(19639,−513)(19639,−513)

33.

(31,−42,87)(31,−42,87)

35.

(2140,120,98)(2140,120,98)

37.

(1813,1513,−1513)(1813,1513,−1513)

39.

(x,y,12(1−2x−3y))(x,y,12(1−2x−3y))

41.

(x,−x2,−1)(x,−x2,−1)

43.

(125,−25,0)(125,−25,0)

45.

(8,1,−2)(8,1,−2)

47.

(1,2,3)(1,2,3)

49.

(x,3128−3x4,128(−7x−3))(x,3128−3x4,128(−7x−3))

51.

No solutions exist.

53.

860 red velvet, 1,340 chocolate

55.

4% for account 1, 6% for account 2

57.

$126 59. Banana was 3%, pumpkin was 7%, and rocky road was 2% 61. 100 almonds, 200 cashews, 600 pistachios ### 9.7 Section Exercises 1. If A−1A−1 is the inverse of A,A, then AA−1=I,AA−1=I, the identity matrix. Since AA is also the inverse of A−1,A−1A=I.A−1,A−1A=I. You can also check by proving this for a 2×22×2 matrix. 3. No, because adad and bcbc are both 0, so ad−bc=0,ad−bc=0, which requires us to divide by 0 in the formula. 5. Yes. Consider the matrix [0110].[ 0110 ]. The inverse is found with the following calculation: A−1=10(0)−1(1)[0−1−10]=[0110].A−1=10(0)−1(1)[ 0−1−10 ]=[ 0110 ]. 7. AB=BA=[1001]=IAB=BA=[ 1001 ]=I 9. AB=BA=[1001]=IAB=BA=[ 1001 ]=I 11. AB=BA=⎡⎣⎢100010001⎤⎦⎥=IAB=BA=[ 100010001 ]=I 13. 129[9−123]129[ 92−13 ] 15. 169[−2973]169[ −2793 ] 17. There is no inverse 19. 47[0.511.5−0.5]47[ 0.51.51−0.5 ] 21. 117⎡⎣⎢−52015−3−1−3124⎤⎦⎥117[ −55−320−3121−14 ] 23. 1209⎡⎣⎢4710−24−57193869−12−13⎤⎦⎥1209[ 47−57691019−12−2438−13 ] 25. ⎡⎣⎢18−564060−14080−168448−280⎤⎦⎥[ 1860−168−56−1404484080−280 ] 27. (−5,6)(−5,6) 29. (2,0)(2,0) 31. (13,−52)(13,−52) 33. (−23,−116)(−23,−116) 35. (7,12,15)(7,12,15) 37. (5,0,−1)(5,0,−1) 39. 134(−35,−97,−154)134(−35,−97,−154) 41. 1690(65,−1136,−229)1690(65,−1136,−229) 43. (−3730,815)(−3730,815) 45. (10123,−1,25)(10123,−1,25) 47. 12⎡⎣⎢⎢⎢⎢200011−11−111−1−1−111⎤⎦⎥⎥⎥⎥12[ 21−1−1011−10−11101−11 ] 49. 139⎡⎣⎢⎢⎢⎢31824−92−53−364613221−16−7109−5⎤⎦⎥⎥⎥⎥139[ 321−718−53321024−36219−946−16−5 ] 51. ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢10000−101000−100100−100010−100001−1000001⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥⎥[ 100000010000001000000100000010−1−1−1−1−11 ] 53. Infinite solutions. 55. 50% oranges, 25% bananas, 20% apples 57. 10 straw hats, 50 beanies, 40 cowboy hats 59. Tom ate 6, Joe ate 3, and Albert ate 3. 61. 124 oranges, 10 lemons, 8 pomegranates ### 9.8 Section Exercises 1. A determinant is the sum and products of the entries in the matrix, so you can always evaluate that product—even if it does end up being 0. 3. The inverse does not exist. 5. −2−2 7. 77 9. −4−4 11. 00 13. −7,990.7−7,990.7 15. 33 17. −1−1 19. 224224 21. 1515 23. −17.03−17.03 25. (1,1)(1,1) 27. (12,13)(12,13) 29. (2,5)(2,5) 31. (−1,−13)(−1,−13) 33. (15,12)(15,12) 35. (1,3,2)(1,3,2) 37. (−1,0,3)(−1,0,3) 39. (12,1,2)(12,1,2) 41. (2,1,4)(2,1,4) 43. Infinite solutions 45. 2424 47. 11 49. Yes; 18, 38 51. Yes; 33, 36, 37 53.$7,000 in first account, \$3,000 in second account.

55.

57.

4 gal yellow, 6 gal blue

59.

13 green tomatoes, 17 red tomatoes

61.

Strawberries 18%, oranges 9%, kiwi 10%

63.

100 for movie 1, 230 for movie 2, 312 for movie 3

65.

20–29: 2,100, 30–39: 2,600, 40–49: 825

67.

300 almonds, 400 cranberries, 300 cashews

### Review Exercises

1.

No

3.

(−2,3)(−2,3)

5.

(4,−1)(4,−1)

7.

No solutions exist.

9.

(300,60,000)(300,60,000)

11.

Infinite solutions

13.

No solutions exist.

15.

(−1,−2,3)(−1,−2,3)

17.

(x,8x5,14x5)(x,8x5,14x5)

19.

11, 17, 33

21.

(2,−3),(3,2)(2,−3),(3,2)

23.

No solution

25.

No solution

27.

29.

31.

2x+2,−4x+12x+2,−4x+1

33.

7x+5,−15(x+5)27x+5,−15(x+5)2

35.

3x−5,−4x+1x2+5x+253x−5,−4x+1x2+5x+25

37.

x−4(x2−2),5x+3(x2−2)2x−4(x2−2),5x+3(x2−2)2

39.

[−16−48−12][ −168−4−12 ]

41.

undefined; dimensions do not match

43.

undefined; inner dimensions do not match

45.

⎡⎣⎢113448428819810−41−42⎤⎦⎥[ 11328104481−418498−42 ]

47.

⎡⎣⎢−127−228−7411771764038⎤⎦⎥[ −127−74176−21140287738 ]

49.

undefined; inner dimensions do not match

51.

x−3z=7y+2z=−5x−3z=7y+2z=−5 with infinite solutions

53.

⎡⎣⎢−22192−8−101522∣∣∣∣703⎤⎦⎥[ −2212−8519−1022|703 ]

55.

⎡⎣⎢1−10041302∣∣∣∣120−7⎤⎦⎥[ 103−140012|120−7 ]

57.

No solutions exist.

59.

No solutions exist.

61.

18[2671]18[ 2761 ]

63.

No inverse exists.

65.

(−20,40)(−20,40)

67.

(−1,0.2,0.3)(−1,0.2,0.3)

69.

17% oranges, 34% bananas, 39% apples

71.

0

73.

6

75.

(6,12)(6,12)

77.

(x, 5x + 3)

79.

(0,0,−12)(0,0,−12)

### Practice Test

1.

Yes

3.

No solutions exist.

5.

120(10,5,4)120(10,5,4)

7.

(x,16x5−13x5)(x,16x5−13x5)

9.

(−22–√,−17−−√),(−22–√,17−−√),(22–√,−17−−√),(22–√,17−−√)(−22,−17),(−22,17),(22,−17),(22,17)

11.

13.

53x+1−2x+3(3x+1)253x+1−2x+3(3x+1)2

15.

[17−85111][ 1751−811 ]

17.

[12−15−2030][ 12−20−1530 ]

19.

−18−18

21.

⎡⎣⎢14−21−23−513−612∣∣∣∣140−111⎤⎦⎥[ 14−213−23−61−512|140−111 ]

23.

No solutions exist.

25.

(100,90)(100,90)

27.

(1100,0)(1100,0)

29.

32 or more cell phones per day

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