Contents

Preface
1Prerequisites
2Equations and Inequalities
3Functions
4Linear Functions
5Polynomial and Rational Functions
6Exponential and Logarithmic Functions
7The Unit Circle: Sine and Cosine Functions
8Periodic Functions
9Trigonometric Identities and Equations
10Further Applications of Trigonometry
11Systems of Equations and Inequalities
12Analytic Geometry
13Sequences, Probability, and Counting Theory
AProofs, Identities, and Toolkit Functions
Answer Key
Index

Chapter 11

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Not a solution.

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

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$(−2, −5) (−2, −5)$

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$(−6, −2) (−6, −2)$

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$(10, −4) (10, −4)$

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No solution. It is an inconsistent system.

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

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700 children, 950 adults

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$(1, −1, 1) (1, −1, 1)$

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No solution.

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

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$(- \frac{1}{2}, \frac{1}{2}) (- \frac{1}{2}, \frac{1}{2})$ and $(2, 8) (2, 8)$

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$(−1, 3) (−1, 3)$

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

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$\frac{3}{x −3} - \frac{2}{x −2} \frac{3}{x −3} - \frac{2}{x −2}$

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$\frac{6}{x −1} - \frac{5}{{(x −1)}^{2}} \frac{6}{x −1} - \frac{5}{{(x −1)}^{2}}$

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$\frac{3}{x −1} + \frac{2 x −4}{x^{2} + 1} \frac{3}{x −1} + \frac{2 x −4}{x^{2} + 1}$

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$\frac{x −2}{x^{2} −2 x + 3} + \frac{2 x + 1}{{(x^{2} −2 x + 3)}^{2}} \frac{x −2}{x^{2} −2 x + 3} + \frac{2 x + 1}{{(x^{2} −2 x + 3)}^{2}}$

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$A + B = [\begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \begin{matrix} 6 \\ 0 \\ −3 \end{matrix}] + [\begin{matrix} 3 \\ 1 \\ −4 \end{matrix} \begin{matrix} −2 \\ 5 \\ 3 \end{matrix}] A + B = [\begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \begin{matrix} \end{matrix}] + []$
$= [\begin{matrix} 2 + 3 \\ 1 + 1 \\ 1 + (−4) \end{matrix} \begin{matrix} 6 + (−2) \\ 0 + 5 \\ −3 + 3 \end{matrix}] = [\begin{matrix} 5 \\ 2 \\ −3 \end{matrix} \begin{matrix} 4 \\ 5 \\ 0 \end{matrix}] = [\begin{matrix} 2 \\ 1 \\ 1 + (−4) \end{matrix}] = [\begin{matrix} −3 \end{matrix}]$

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$−2 B = [\begin{matrix} −8 & −2 \\ −6 & −4 \end{matrix}] −2 B = [\begin{matrix} −8 & −2 \\ −6 & −4 \end{matrix}]$

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$[\begin{matrix} 4 & −3 \\ 3 & 2 \end{matrix} | \begin{matrix} 11 \\ 4 \end{matrix}] [\begin{matrix} 4 & −3 \\ 3 & 2 \end{matrix} | \begin{matrix} 11 \\ 4 \end{matrix}]$

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$\begin{matrix} x - y + z = 5 \\ 2 x - y + 3 z = 1 \\ y + z = −9 \end{matrix} \begin{matrix} x - y + z = 5 \\ 2 x - y + 3 z = 1 \\ y + z = −9 \end{matrix}$

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$(2, 1) (2,)$

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$[\begin{matrix} 1 & - \frac{5}{2} & \frac{5}{2} \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} \frac{17}{2} \\ 9 \\ 2 \end{matrix}] [\begin{matrix} 1 & - \frac{5}{2} & \frac{5}{2} \\ 0 & 1 & 5 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} \frac{17}{2} \\ 9 \\ 2 \end{matrix}]$

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

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$150,000 at 7%, $750,000 at 8%, $600,000 at 10%

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\begin{array}{l} A B = [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] = [\begin{array}{r} 1 (−3) + 4 (1) & 1 (−4) + 4 (1) \\ −1 (−3) + −3 (1) & −1 (−4) + −3 (1) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \\ B A = [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] = [\begin{array}{r} −3 (1) + −4 (−1) & −3 (4) + −4 (−3) \\ 1 (1) + 1 (−1) & 1 (4) + 1 (−3) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \end{array} \begin{array}{l} A B = [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] = [\begin{array}{r} 1 (−3) + 4 (1) & 1 (−4) + 4 (1) \\ −1 (−3) + −3 (1) & −1 (−4) + −3 (1) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \\ B A = [\begin{array}{r} −3 & −4 \\ 1 & 1 \end{array}] \begin{array}{r}  \end{array} [\begin{array}{r} 1 & 4 \\ −1 & −3 \end{array}] = [\begin{array}{r} −3 (1) + −4 (−1) & −3 (4) + −4 (−3) \\ 1 (1) + 1 (−1) & 1 (4) + 1 (−3) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] \end{array}

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$A^{−1} = [\begin{matrix} \frac{3}{5} & \frac{1}{5} \\ - \frac{2}{5} & \frac{1}{5} \end{matrix}] A^{−1} = [\begin{matrix} \frac{3}{5} & \frac{1}{5} \\ - \frac{2}{5} & \frac{1}{5} \end{matrix}]$

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$A^{−1} = [\begin{matrix} 1 & 1 & 2 \\ 2 & 4 & −3 \\ 3 & 6 & −5 \end{matrix}] A^{−1} = [\begin{matrix} 1 & 1 \\ 2 & 4 & −3 \\ 3 & 6 & −5 \end{matrix}]$

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$X = [\begin{matrix} 4 \\ 38 \\ 58 \end{matrix}] X = [\begin{matrix} 4 \\ 38 \\ 58 \end{matrix}]$

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$(3, - 7) (3, - 7)$

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$- 10 - 10$

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$(- 2, \frac{3}{5}, \frac{12}{5}) (- 2, \frac{3}{5}, \frac{12}{5})$

11.1 Section Exercises

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No, you can either have zero, one, or infinitely many. Examine graphs.

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This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.

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You can solve by substitution (isolating $x x$ or $y y$ ), graphically, or by addition.

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Yes

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Yes

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11.

$(−1, 2) (−1, 2)$

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13.

$(−3, 1) (−3, 1)$

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15.

$(- \frac{3}{5}, 0) (- \frac{3}{5}, 0)$

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17.

No solutions exist.

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19.

$(\frac{72}{5}, \frac{132}{5}) (\frac{72}{5}, \frac{132}{5})$

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21.

$(6, −6) (6, −6)$

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23.

$(- \frac{1}{2}, \frac{1}{10}) (- \frac{1}{2}, \frac{1}{10})$

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25.

No solutions exist.

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27.

$(- \frac{1}{5}, \frac{2}{3}) (- \frac{1}{5}, \frac{2}{3})$

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29.

$(x, \frac{x + 3}{2}) (x, \frac{x + 3}{2})$

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31.

$(−4, 4) (−4, 4)$

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33.

$(\frac{1}{2}, \frac{1}{8}) (\frac{1}{2}, \frac{1}{8})$

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35.

$(\frac{1}{6}, 0) (\frac{1}{6}, 0)$

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37.

$(x, 2 (7 x −6)) (x, 2 (7 x −6))$

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39.

$(- \frac{5}{6}, \frac{4}{3}) (- \frac{5}{6}, \frac{4}{3})$

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41.

Consistent with one solution

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43.

Consistent with one solution

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45.

Dependent with infinitely many solutions

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47.

$(−3.08, 4.91) (−3.08, 4.91)$

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49.

$(−1.52, 2.29) (−1.52, 2.29)$

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51.

$(\frac{A + B}{2}, \frac{A - B}{2}) (\frac{A + B}{2}, \frac{A - B}{2})$

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53.

$(\frac{−1}{A - B}, \frac{A}{A - B}) (\frac{−1}{A - B}, \frac{A}{A - B})$

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55.

$(\frac{C E - B F}{B D - A E}, \frac{A F - C D}{B D - A E}) (\frac{C E - B F}{B D - A E}, \frac{A F - C D}{B D - A E})$

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57.

They never turn a profit.

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59.

$(1, 250, 100, 000) (1, 250, 100, 000)$

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61.

The numbers are 7.5 and 20.5.

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63.

24,000

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65.

790 sophomores, 805 freshman

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67.

56 men, 74 women

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69.

10 gallons of 10% solution, 15 gallons of 60% solution

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71.

Swan Peak: $750,000, Riverside: $350,000

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73.

$12,500 in the first account, $10,500 in the second account.

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75.

High-tops: 45, Low-tops: 15

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77.

Infinitely many solutions. We need more information.

11.2 Section Exercises

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No, there can be only one, zero, or infinitely many solutions.

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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.

$\begin{array}{l} 2x+3y−6z=1 \\ −4 x −6 y + 12 z = −2 \\ x+2y+5z=10 \end{array} \begin{array}{l} −4 x −6 y + 12 z = −2 \end{array}$

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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.

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Yes

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11.

$(−1, 4, 2) (−1, 4, 2)$

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13.

$(- \frac{85}{107}, \frac{312}{107}, \frac{191}{107}) (- \frac{85}{107}, \frac{312}{107}, \frac{191}{107})$

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15.

$(1, \frac{1}{2}, 0) (1, \frac{1}{2}, 0)$

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17.

$(4, −6, 1) (4, −6, 1)$

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19.

$(x, \frac{1}{27} (65 −16 x), \frac{x + 28}{27}) (x, \frac{1}{27} (65 −16 x), \frac{x + 28}{27})$

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21.

$(- \frac{45}{13}, \frac{17}{13}, −2) (- \frac{45}{13}, \frac{17}{13}, −2)$

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23.

No solutions exist

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25.

$(0, 0, 0) (0, 0, 0)$

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27.

$(\frac{4}{7}, - \frac{1}{7}, - \frac{3}{7}) (\frac{4}{7}, - \frac{1}{7}, - \frac{3}{7})$

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29.

$(7, 20, 16) (7, 20, 16)$

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31.

$(−6, 2, 1) (−6, 2, 1)$

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33.

$(5, 12, 15) (5, 12, 15)$

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35.

$(−5, −5, −5) (−5, −5, −5)$

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37.

$(10, 10, 10) (10, 10, 10)$

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39.

$(\frac{1}{2}, \frac{1}{5}, \frac{4}{5}) (\frac{1}{2}, \frac{1}{5}, \frac{4}{5})$

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41.

$(\frac{1}{2}, \frac{2}{5}, \frac{4}{5}) (\frac{1}{2}, \frac{2}{5}, \frac{4}{5})$

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43.

$(2, 0, 0) (2, 0, 0)$

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45.

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

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47.

$(\frac{128}{557}, \frac{23}{557}, \frac{28}{557}) (\frac{128}{557}, \frac{23}{557}, \frac{28}{557})$

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49.

$(6, −1, 0) (6, −1, 0)$

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51.

24, 36, 48

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53.

70 grandparents, 140 parents, 190 children

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55.

Your share was $19.95, Sarah’s share was $40, and your other roommate’s share was $22.05.

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57.

There are infinitely many solutions; we need more information

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59.

500 students, 225 children, and 450 adults

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61.

The BMW was $49,636, the Jeep was $42,636, and the Toyota was $47,727.

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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.

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65.

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

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67.

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

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69.

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

11.3 Section Exercises

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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.

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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.

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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.

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$(0, −3), (3, 0) (0, −3), (3, 0)$

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$(- \frac{3 \sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}), (\frac{3 \sqrt{2}}{2}, - \frac{3 \sqrt{2}}{2}) (- \frac{3 \sqrt{2}}{2}, \frac{3 \sqrt{2}}{2}), (\frac{3 \sqrt{2}}{2}, - \frac{3 \sqrt{2}}{2})$

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11.

$(−3, 0), (3, 0) (−3, 0), (3, 0)$

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13.

$(\frac{1}{4}, - \frac{\sqrt{62}}{8}), (\frac{1}{4}, \frac{\sqrt{62}}{8}) (\frac{1}{4}, - \frac{\sqrt{62}}{8}), (\frac{1}{4}, \frac{\sqrt{62}}{8})$

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15.

$(- \frac{\sqrt{398}}{4}, \frac{199}{4}), (\frac{\sqrt{398}}{4}, \frac{199}{4}) (- \frac{\sqrt{398}}{4}, \frac{199}{4}), (\frac{\sqrt{398}}{4}, \frac{199}{4})$

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17.

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

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19.

$(- \sqrt{\frac{1}{2} (\sqrt{5} −1)}, \frac{1}{2} (1 - \sqrt{5})), (\sqrt{\frac{1}{2} (\sqrt{5} −1)}, \frac{1}{2} (1 - \sqrt{5})) (- \sqrt{\frac{1}{2} (\sqrt{5} −1)}, \frac{1}{2} (1 - \sqrt{5})), (\sqrt{\frac{1}{2} (\sqrt{5} −1)}, \frac{1}{2} (1 - \sqrt{5}))$

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21.

$(5, 0) (5, 0)$

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23.

$(0, 0) (0, 0)$

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25.

$(3, 0) (3, 0)$

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27.

No Solutions Exist

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29.

No Solutions Exist

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31.

$(- \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}), (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}), (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) (- \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}), (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}), (\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}), (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$

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33.

$(2, 0) (2, 0)$

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35.

$(- \sqrt{7}, −3), (- \sqrt{7}, 3), (\sqrt{7}, −3), (\sqrt{7}, 3) (- \sqrt{7}, −3), (- \sqrt{7}, 3), (\sqrt{7}, −3), (\sqrt{7}, 3)$

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37.

$(- \sqrt{\frac{1}{2} (\sqrt{73} −5)}, \frac{1}{2} (7 - \sqrt{73})), (\sqrt{\frac{1}{2} (\sqrt{73} −5)}, \frac{1}{2} (7 - \sqrt{73})) (- \sqrt{\frac{1}{2} (\sqrt{73} −5)}, \frac{1}{2} (7 - \sqrt{73})), (\sqrt{\frac{1}{2} (\sqrt{73} −5)}, \frac{1}{2} (7 - \sqrt{73}))$

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39.

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41.

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43.

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45.

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47.

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49.

$(−2 \sqrt{\frac{70}{383}}, −2 \sqrt{\frac{35}{29}}), (−2 \sqrt{\frac{70}{383}}, 2 \sqrt{\frac{35}{29}}), (2 \sqrt{\frac{70}{383}}, −2 \sqrt{\frac{35}{29}}), (2 \sqrt{\frac{70}{383}}, 2 \sqrt{\frac{35}{29}}) (−2 \sqrt{\frac{70}{383}}, −2 \sqrt{\frac{35}{29}}), (−2 \sqrt{\frac{70}{383}}, 2 \sqrt{\frac{35}{29}}), (2 \sqrt{\frac{70}{383}}, −2 \sqrt{\frac{35}{29}}), (2 \sqrt{\frac{70}{383}}, 2 \sqrt{\frac{35}{29}})$

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51.

No Solution Exists

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53.

$x = 0, y > 0 x = 0, y > 0$ and $0 < x < 1, \sqrt{x} < y < \frac{1}{x} 0 < x < 1, \sqrt{x} < y < \frac{1}{x}$

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55.

12, 288

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57.

2–20 computers

11.4 Section Exercises

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No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, $\frac{1}{x^{2} + 1} \frac{1}{x^{2} + 1}$ cannot be decomposed because the denominator cannot be factored.

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Graph both sides and ensure they are equal.

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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 $x = - \frac{5}{3} x = - \frac{5}{3}$ , giving us a B-value of $−2. −2.$

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$\frac{8}{x + 3} - \frac{5}{x −8} \frac{8}{x + 3} - \frac{5}{x −8}$

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$\frac{1}{x + 5} + \frac{9}{x + 2} \frac{1}{x + 5} + \frac{9}{x + 2}$

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11.

$\frac{3}{5 x −2} + \frac{4}{4 x −1} \frac{3}{5 x −2} + \frac{4}{4 x −1}$

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13.

$\frac{5}{2 (x + 3)} + \frac{5}{2 (x −3)} \frac{5}{2 (x + 3)} + \frac{5}{2 (x −3)}$

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15.

$\frac{3}{x + 2} + \frac{3}{x −2} \frac{3}{x + 2} + \frac{3}{x −2}$

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17.

$\frac{9}{5 (x + 2)} + \frac{11}{5 (x −3)} \frac{9}{5 (x + 2)} + \frac{11}{5 (x −3)}$

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19.

$\frac{8}{x −3} - \frac{5}{x −2} \frac{8}{x −3} - \frac{5}{x −2}$

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21.

$\frac{1}{x −2} + \frac{2}{{(x −2)}^{2}} \frac{1}{x −2} + \frac{2}{{(x −2)}^{2}}$

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23.

$- \frac{6}{4 x + 5} + \frac{3}{{(4 x + 5)}^{2}} - \frac{6}{4 x + 5} + \frac{3}{{(4 x + 5)}^{2}}$

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25.

$- \frac{1}{x −7} - \frac{2}{{(x −7)}^{2}} - \frac{1}{x −7} - \frac{2}{{(x −7)}^{2}}$

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27.

$\frac{4}{x} - \frac{3}{2 (x + 1)} + \frac{7}{2 {(x + 1)}^{2}} \frac{4}{x} - \frac{3}{2 (x + 1)} + \frac{7}{2 {(x + 1)}^{2}}$

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29.

$\frac{4}{x} + \frac{2}{x^{2}} - \frac{3}{3 x + 2} + \frac{7}{2 {(3 x + 2)}^{2}} \frac{4}{x} + \frac{2}{x^{2}} - \frac{3}{3 x + 2} + \frac{7}{2 {(3 x + 2)}^{2}}$

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31.

$\frac{x + 1}{x^{2} + x + 3} + \frac{3}{x + 2} \frac{x + 1}{x^{2} + x + 3} + \frac{3}{x + 2}$

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33.

$\frac{4 −3 x}{x^{2} + 3 x + 8} + \frac{1}{x −1} \frac{4 −3 x}{x^{2} + 3 x + 8} + \frac{1}{x −1}$

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35.

$\frac{2 x −1}{x^{2} + 6 x + 1} + \frac{2}{x + 3} \frac{2 x −1}{x^{2} + 6 x + 1} + \frac{2}{x + 3}$

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37.

$\frac{1}{x^{2} + x + 1} + \frac{4}{x −1} \frac{1}{x^{2} + x + 1} + \frac{4}{x −1}$

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39.

$\frac{2}{x^{2} −3 x + 9} + \frac{3}{x + 3} \frac{2}{x^{2} −3 x + 9} + \frac{3}{x + 3}$

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41.

$- \frac{1}{4 x^{2} + 6 x + 9} + \frac{1}{2 x −3} - \frac{1}{4 x^{2} + 6 x + 9} + \frac{1}{2 x −3}$

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43.

$\frac{1}{x} + \frac{1}{x + 6} - \frac{4 x}{x^{2} −6 x + 36} \frac{1}{x} + \frac{1}{x + 6} - \frac{4 x}{x^{2} −6 x + 36}$

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45.

$\frac{x + 6}{x^{2} + 1} + \frac{4 x + 3}{{(x^{2} + 1)}^{2}} \frac{x + 6}{x^{2} + 1} + \frac{4 x + 3}{{(x^{2} + 1)}^{2}}$

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47.

$\frac{x + 1}{x + 2} + \frac{2 x + 3}{{(x + 2)}^{2}} \frac{x + 1}{x + 2} + \frac{2 x + 3}{{(x + 2)}^{2}}$

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49.

$\frac{1}{x^{2} + 3 x + 25} - \frac{3 x}{{(x^{2} + 3 x + 25)}^{2}} \frac{1}{x^{2} + 3 x + 25} - \frac{3 x}{{(x^{2} + 3 x + 25)}^{2}}$

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51.

$\frac{1}{8 x} - \frac{x}{8 (x^{2} + 4)} + \frac{10 - x}{2 {(x^{2} + 4)}^{2}} \frac{1}{8 x} - \frac{x}{8 (x^{2} + 4)} + \frac{10 - x}{2 {(x^{2} + 4)}^{2}}$

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53.

$- \frac{16}{x} - \frac{9}{x^{2}} + \frac{16}{x −1} - \frac{7}{{(x −1)}^{2}} - \frac{16}{x} - \frac{9}{x^{2}} + \frac{16}{x −1} - \frac{7}{{(x −1)}^{2}}$

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55.

$\frac{1}{x + 1} - \frac{2}{{(x + 1)}^{2}} + \frac{5}{{(x + 1)}^{3}} \frac{1}{x + 1} - \frac{2}{{(x + 1)}^{2}} + \frac{5}{{(x + 1)}^{3}}$

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57.

$\frac{5}{x −2} - \frac{3}{10 (x + 2)} + \frac{7}{x + 8} - \frac{7}{10 (x −8)} \frac{5}{x −2} - \frac{3}{10 (x + 2)} + \frac{7}{x + 8} - \frac{7}{10 (x −8)}$

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59.

$- \frac{5}{4 x} - \frac{5}{2 (x + 2)} + \frac{11}{2 (x + 4)} + \frac{5}{4 (x + 4)} - \frac{5}{4 x} - \frac{5}{2 (x + 2)} + \frac{11}{2 (x + 4)} + \frac{5}{4 (x + 4)}$

11.5 Section Exercises

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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 $2 \times 2 2 \times 2$ matrix and the second is a $2 \times 3 2 \times 3$ matrix. $[\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] + [\begin{matrix} 6 & 5 & 4 \\ 3 & 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] + [\begin{matrix} 6 & 5 & 4 \\ 3 & 2 & 1 \end{matrix}]$ has no sum.

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Yes, if the dimensions of $A A$ are $m \times n m \times n$ and the dimensions of $B B$ are $n \times m, n \times m,$ both products will be defined.

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

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$[\begin{matrix} 11 & 19 \\ 15 & 94 \\ 17 & 67 \end{matrix}] [\begin{matrix} 11 & 19 \\ 15 & 94 \\ 17 & 67 \end{matrix}]$

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$[\begin{matrix} −4 & 2 \\ 8 & 1 \end{matrix}] [\begin{matrix} −4 & 2 \\ 8 & 1 \end{matrix}]$

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11.

Undidentified; dimensions do not match

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13.

$[\begin{matrix} 9 & 27 \\ 63 & 36 \\ 0 & 192 \end{matrix}] [\begin{matrix} 9 & 27 \\ 63 & 36 \\ 0 & 192 \end{matrix}]$

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15.

$[\begin{matrix} −64 & −12 & −28 & −72 \\ −360 & −20 & −12 & −116 \end{matrix}] [\begin{matrix} −64 & −12 & −28 & −72 \\ −360 & −20 & −12 & −116 \end{matrix}]$

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17.

$[\begin{matrix} 1, 800 & 1, 200 & 1, 300 \\ 800 & 1, 400 & 600 \\ 700 & 400 & 2, 100 \end{matrix}] [\begin{matrix} 1, 800 & 1, 200 & 1, 300 \\ 800 & 1, 400 & 600 \\ 700 & 400 & 2, 100 \end{matrix}]$

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19.

$[\begin{matrix} 20 & 102 \\ 28 & 28 \end{matrix}] [\begin{matrix} 20 & 102 \\ 28 & 28 \end{matrix}]$

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21.

$[\begin{matrix} 60 & 41 & 2 \\ −16 & 120 & −216 \end{matrix}] [\begin{matrix} 60 & 41 & 2 \\ −16 & 120 & −216 \end{matrix}]$

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23.

$[\begin{matrix} −68 & 24 & 136 \\ −54 & −12 & 64 \\ −57 & 30 & 128 \end{matrix}] [\begin{matrix} −68 & 24 & 136 \\ −54 & −12 & 64 \\ −57 & 30 & 128 \end{matrix}]$

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25.

Undefined; dimensions do not match.

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27.

$[\begin{matrix} −8 & 41 & −3 \\ 40 & −15 & −14 \\ 4 & 27 & 42 \end{matrix}] [\begin{matrix} −8 & 41 & −3 \\ 40 & −15 & −14 \\ 4 & 27 & 42 \end{matrix}]$

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29.

$[\begin{matrix} −840 & 650 & −530 \\ 330 & 360 & 250 \\ −10 & 900 & 110 \end{matrix}] [\begin{matrix} −840 & 650 & −530 \\ 330 & 360 & 250 \\ −10 & 900 & 110 \end{matrix}]$

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31.

$[\begin{matrix} −350 & 1, 050 \\ 350 & 350 \end{matrix}] [\begin{matrix} −350 & 1, 050 \\ 350 & 350 \end{matrix}]$

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33.

Undefined; inner dimensions do not match.

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35.

$[\begin{matrix} 1, 400 & 700 \\ −1, 400 & 700 \end{matrix}] [\begin{matrix} 1, 400 & 700 \\ −1, 400 & 700 \end{matrix}]$

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37.

$[\begin{matrix} 332, 500 & 927, 500 \\ −227, 500 & 87, 500 \end{matrix}] [\begin{matrix} 332, 500 & 927, 500 \\ −227, 500 & 87, 500 \end{matrix}]$

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39.

$[\begin{matrix} 490, 000 & 0 \\ 0 & 490, 000 \end{matrix}] [\begin{matrix} 490, 000 & 0 \\ 0 & 490, 000 \end{matrix}]$

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41.

$[\begin{matrix} −2 & 3 & 4 \\ −7 & 9 & −7 \end{matrix}] [\begin{matrix} −2 & 3 & 4 \\ −7 & 9 & −7 \end{matrix}]$

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43.

$[\begin{matrix} −4 & 29 & 21 \\ −27 & −3 & 1 \end{matrix}] [\begin{matrix} −4 & 29 & 21 \\ −27 & −3 & 1 \end{matrix}]$

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45.

$[\begin{matrix} −3 & −2 & −2 \\ −28 & 59 & 46 \\ −4 & 16 & 7 \end{matrix}] [\begin{matrix} −3 & −2 & −2 \\ −28 & 59 & 46 \\ −4 & 16 & 7 \end{matrix}]$

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47.

$[\begin{matrix} 1 & −18 & −9 \\ −198 & 505 & 369 \\ −72 & 126 & 91 \end{matrix}] [\begin{matrix} 1 & −18 & −9 \\ −198 & 505 & 369 \\ −72 & 126 & 91 \end{matrix}]$

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49.

$[\begin{matrix} 0 & 1.6 \\ 9 & −1 \end{matrix}] [\begin{matrix} 0 & 1.6 \\ 9 & −1 \end{matrix}]$

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51.

$[\begin{matrix} 2 & 24 & −4.5 \\ 12 & 32 & −9 \\ −8 & 64 & 61 \end{matrix}] [\begin{matrix} 2 & 24 & −4.5 \\ 12 & 32 & −9 \\ −8 & 64 & 61 \end{matrix}]$

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53.

$[\begin{matrix} 0.5 & 3 & 0.5 \\ 2 & 1 & 2 \\ 10 & 7 & 10 \end{matrix}] [\begin{matrix} 0.5 & 3 & 0.5 \\ 2 & 1 & 2 \\ 10 & 7 & 10 \end{matrix}]$

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55.

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

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57.

$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

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59.

$B^{n} = {\begin{cases} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], n even, \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}], n odd . \end{cases} B^{n} = {\begin{cases} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}], n \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}], n \end{cases}$

11.6 Section Exercises

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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.

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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 $R_{2} = R_{2} −9 R_{1} . R_{2} = R_{2} −9 R_{1} .$ (2) $R_{2} = R_{1} −9 R_{2} . R_{2} = R_{1} −9 R_{2} .$ Then divide row 1 by 9.

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

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$[\begin{array}{r} 0 & 16 \\ 9 & −1 \end{array} | \begin{array}{r} 4 \\ 2 \end{array}] [\begin{array}{r} 0 & 16 \\ 9 & −1 \end{array} | \begin{array}{r} 4 \\ 2 \end{array}]$

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$[\begin{array}{r} 1 & 5 & 8 \\ 12 & 3 & 0 \\ 3 & 4 & 9 \end{array} | \begin{array}{r} 16 \\ 4 \\ −7 \end{array}] [\begin{array}{r} 1 & 5 & 8 \\ 12 & 3 & 0 \\ 3 & 4 & 9 \end{array} | \begin{array}{r} 16 \\ 4 \\ −7 \end{array}]$

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11.

$\begin{array}{l} −2 x + 5 y = 5 \\ 6 x −18 y = 26 \end{array} \begin{array}{l} −2 x + 5 y = 5 \\ 6 x −18 y = 26 \end{array}$

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13.

$\begin{array}{l} 3 x + 2 y = 3 \\ - x −9 y + 4 z = −1 \\ 8 x + 5 y + 7 z = 8 \end{array} \begin{array}{l} 3 x + 2 y = 3 \\ - x −9 y + 4 z = −1 \\ 8 x + 5 y + 7 z = 8 \end{array}$

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15.

$\begin{array}{l} 4 x + 5 y −2 z = 12 \\ y + 58 z = 2 \\ 8 x + 7 y −3 z = −5 \end{array} \begin{array}{l} 4 x + 5 y −2 z = 12 \\ y + 58 z = 2 \\ 8 x + 7 y −3 z = −5 \end{array}$

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17.

No solutions

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19.

$(−1, −2) (−1, −2)$

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21.

$(6, 7) (6, 7)$

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23.

$(3, 2) (3, 2)$

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25.

$(\frac{1}{5}, \frac{1}{2}) (\frac{1}{5}, \frac{1}{2})$

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27.

$(x, \frac{4}{15} (5 x + 1)) (x, \frac{4}{15} (5 x + 1))$

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29.

$(3, 4) (3, 4)$

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31.

$(\frac{196}{39}, - \frac{5}{13}) (\frac{196}{39}, - \frac{5}{13})$

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33.

$(31, −42, 87) (31, −42, 87)$

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35.

$(\frac{21}{40}, \frac{1}{20}, \frac{9}{8}) (\frac{21}{40}, \frac{1}{20}, \frac{9}{8})$

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37.

$(\frac{18}{13}, \frac{15}{13}, - \frac{15}{13}) (\frac{18}{13}, \frac{15}{13}, - \frac{15}{13})$

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39.

$(x, y, \frac{1}{2} (1 −2 x −3 y)) (x, y, \frac{1}{2} (1 −2 x −3 y))$

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41.

$(x, - \frac{x}{2}, −1) (x, - \frac{x}{2}, −1)$

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43.

$(125, −25, 0) (125, −25, 0)$

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45.

$(8, 1, −2) (8, 1, −2)$

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47.

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

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49.

$(x, \frac{31}{28} - \frac{3 x}{4}, \frac{1}{28} (−7 x −3)) (x, \frac{31}{28} - \frac{3 x}{4}, \frac{1}{28} (−7 x −3))$

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51.

No solutions exist.

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53.

860 red velvet, 1,340 chocolate

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55.

4% for account 1, 6% for account 2

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57.

$126

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59.

Banana was 3%, pumpkin was 7%, and rocky road was 2%

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61.

100 almonds, 200 cashews, 600 pistachios

11.7 Section Exercises

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If $A^{−1} A^{−1}$ is the inverse of $A, A,$ then $A A^{−1} = I, A A^{−1} = I,$ the identity matrix. Since $A A$ is also the inverse of $A^{−1}, A^{−1} A = I . A^{−1}, A^{−1} A = I .$ You can also check by proving this for a $2 \times 2 2 \times 2$ matrix.

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No, because $a d a d$ and $b c b c$ are both 0, so $a d - b c = 0, a d - b c = 0,$ which requires us to divide by 0 in the formula.

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Yes. Consider the matrix $[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] . [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] .$ The inverse is found with the following calculation: $A^{−1} = \frac{1}{0 (0) −1 (1)} [\begin{matrix} 0 & −1 \\ −1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] . A^{−1} = \frac{1}{0 (0) −1 (1)} [\begin{matrix} 0 & −1 \\ −1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] .$

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$A B = B A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I A B = B A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

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$A B = B A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I A B = B A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I$

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11.

$A B = B A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I A B = B A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I$

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13.

$\frac{1}{29} [\begin{matrix} 9 & 2 \\ −1 & 3 \end{matrix}] \frac{1}{29} [\begin{matrix} 9 & 2 \\ −1 & 3 \end{matrix}]$

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15.

$\frac{1}{69} [\begin{matrix} −2 & 7 \\ 9 & 3 \end{matrix}] \frac{1}{69} [\begin{matrix} −2 & 7 \\ 9 & 3 \end{matrix}]$

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17.

There is no inverse

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19.

$\frac{4}{7} [\begin{matrix} 0.5 & 1.5 \\ 1 & −0.5 \end{matrix}] \frac{4}{7} [\begin{matrix} 0.5 & 1.5 \\ 1 & −0.5 \end{matrix}]$

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21.

$\frac{1}{17} [\begin{matrix} −5 & 5 & −3 \\ 20 & −3 & 12 \\ 1 & −1 & 4 \end{matrix}] \frac{1}{17} [\begin{matrix} −5 & 5 & −3 \\ 20 & −3 & 12 \\ 1 & −1 & 4 \end{matrix}]$

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23.

$\frac{1}{209} [\begin{matrix} 47 & −57 & 69 \\ 10 & 19 & −12 \\ −24 & 38 & −13 \end{matrix}] \frac{1}{209} [\begin{matrix} 47 & −57 & 69 \\ 10 & 19 & −12 \\ −24 & 38 & −13 \end{matrix}]$

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25.

$[\begin{matrix} 18 & 60 & −168 \\ −56 & −140 & 448 \\ 40 & 80 & −280 \end{matrix}] [\begin{matrix} 18 & 60 & −168 \\ −56 & −140 & 448 \\ 40 & 80 & −280 \end{matrix}]$

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27.

$(−5, 6) (−5, 6)$

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29.

$(2, 0) (2, 0)$

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31.

$(\frac{1}{3}, - \frac{5}{2}) (\frac{1}{3}, - \frac{5}{2})$

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33.

$(- \frac{2}{3}, - \frac{11}{6}) (- \frac{2}{3}, - \frac{11}{6})$

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35.

$(7, \frac{1}{2}, \frac{1}{5}) (7, \frac{1}{2}, \frac{1}{5})$

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37.

$(5, 0, −1) (5, 0, −1)$

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39.

$\frac{1}{34} (−35, −97, −154) \frac{1}{34} (−35, −97, −154)$

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41.

$\frac{1}{690} (65, −1136, −229) \frac{1}{690} (65, −1136, −229)$

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43.

$(- \frac{37}{30}, \frac{8}{15}) (- \frac{37}{30}, \frac{8}{15})$

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45.

$(\frac{10}{123}, −1, \frac{2}{5}) (\frac{10}{123}, −1, \frac{2}{5})$

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47.

$\frac{1}{2} [\begin{array}{r} 2 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & - 1 \\ 0 & - 1 & 1 & 1 \\ 0 & 1 & - 1 & 1 \end{array}] \frac{1}{2} [\begin{array}{r} 2 & 1 & - 1 & - 1 \\ 0 & 1 & 1 & - 1 \\ 0 & - 1 & 1 & 1 \\ 0 & 1 & - 1 & 1 \end{array}]$

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49.

$\frac{1}{39} [\begin{array}{r} 3 & 2 & 1 & - 7 \\ 18 & - 53 & 32 & 10 \\ 24 & - 36 & 21 & 9 \\ - 9 & 46 & - 16 & - 5 \end{array}] \frac{1}{39} [\begin{array}{r} 3 & 2 & 1 & - 7 \\ 18 & - 53 & 32 & 10 \\ 24 & - 36 & 21 & 9 \\ - 9 & 46 & - 16 & - 5 \end{array}]$

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51.

$[\begin{array}{r} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ - 1 & - 1 & - 1 & - 1 & - 1 & 1 \end{array}] [\begin{array}{r} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ - 1 & - 1 & - 1 & - 1 & - 1 & 1 \end{array}]$

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53.

Infinite solutions.

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55.

50% oranges, 25% bananas, 20% apples

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57.

10 straw hats, 50 beanies, 40 cowboy hats

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59.

Tom ate 6, Joe ate 3, and Albert ate 3.

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61.

124 oranges, 10 lemons, 8 pomegranates

11.8 Section Exercises

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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.

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The inverse does not exist.

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$- 2 - 2$

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$77$

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$- 4 - 4$

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11.

$00$

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13.

$- 7, 990.7 - 7, 990.7$

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15.

$33$

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17.

$- 1 - 1$

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19.

$224224$

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21.

$1515$

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23.

$- 17.03 - 17.03$

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25.

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

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27.

$(\frac{1}{2}, \frac{1}{3}) (\frac{1}{2}, \frac{1}{3})$

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29.

$(2, 5) (2, 5)$

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31.

$(- 1, - \frac{1}{3}) (- 1, - \frac{1}{3})$

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33.

$(15, 12) (15, 12)$

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35.

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

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37.

$(- 1, 0, 3) (- 1, 0, 3)$

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39.

$(\frac{1}{2}, 1, 2) (\frac{1}{2}, 1, 2)$

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41.

$(2, 1, 4) (2, 1, 4)$

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43.

Infinite solutions

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45.

$2424$

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47.

$11$

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49.

Yes; 18, 38

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51.

Yes; 33, 36, 37

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53.

$7,000 in first account, $3,000 in second account.

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55.

120 children, 1,080 adult

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57.

4 gal yellow, 6 gal blue

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59.

13 green tomatoes, 17 red tomatoes

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61.

Strawberries 18%, oranges 9%, kiwi 10%

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63.

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

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65.

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

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67.

300 almonds, 400 cranberries, 300 cashews

Review Exercises

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$(- 2, 3) (- 2, 3)$

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$(4, - 1) (4, - 1)$

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No solutions exist.

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

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11.

Infinite solutions

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13.

No solutions exist.

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15.

$(- 1, - 2, 3) (- 1, - 2, 3)$

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17.

$(x, \frac{8 x}{5}, \frac{14 x}{5}) (x, \frac{8 x}{5}, \frac{14 x}{5})$

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19.

11, 17, 33

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21.

$(2, - 3), (3, 2) (2, - 3), (3, 2)$

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23.

No solution

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25.

No solution

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27.

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29.

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31.

$\frac{2}{x + 2}, \frac{- 4}{x + 1} \frac{2}{x + 2}, \frac{- 4}{x + 1}$

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33.

$\frac{7}{x + 5}, \frac{- 15}{{(x + 5)}^{2}} \frac{7}{x + 5}, \frac{- 15}{{(x + 5)}^{2}}$

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35.

$\frac{3}{x - 5}, \frac{- 4 x + 1}{x^{2} + 5 x + 25} \frac{3}{x - 5}, \frac{- 4 x + 1}{x^{2} + 5 x + 25}$

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37.

$\frac{x - 4}{(x^{2} - 2)}, \frac{5 x + 3}{{(x^{2} - 2)}^{2}} \frac{x - 4}{(x^{2} - 2)}, \frac{5 x + 3}{{(x^{2} - 2)}^{2}}$

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39.

$[\begin{matrix} - 16 & 8 \\ - 4 & - 12 \end{matrix}] [\begin{matrix} - 16 & 8 \\ - 4 & - 12 \end{matrix}]$

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41.

undefined; dimensions do not match

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43.

undefined; inner dimensions do not match

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45.

$[\begin{matrix} 113 & 28 & 10 \\ 44 & 81 & - 41 \\ 84 & 98 & - 42 \end{matrix}] [\begin{matrix} 113 & 28 & 10 \\ 44 & 81 & - 41 \\ 84 & 98 & - 42 \end{matrix}]$

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47.

$[\begin{matrix} - 127 & - 74 & 176 \\ - 2 & 11 & 40 \\ 28 & 77 & 38 \end{matrix}] [\begin{matrix} - 127 & - 74 & 176 \\ - 2 & 11 & 40 \\ 28 & 77 & 38 \end{matrix}]$

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49.

undefined; inner dimensions do not match

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51.

$\begin{array}{l} x - 3 z = 7 \\ y + 2 z = - 5 \end{array} \begin{array}{l} x - 3 z = 7 \\ y + 2 z = - 5 \end{array}$ with infinite solutions

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53.

$[\begin{array}{r} - 2 & 2 & 1 \\ 2 & - 8 & 5 \\ 19 & - 10 & 22 \end{array} | \begin{array}{r} 7 \\ 0 \\ 3 \end{array}] [\begin{array}{r} - 2 & 2 & 1 \\ 2 & - 8 & 5 \\ 19 & - 10 & 22 \end{array} | \begin{array}{r} 7 \\ 0 \\ 3 \end{array}]$

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55.

$[\begin{array}{r} 1 & 0 & 3 \\ −1 & 4 & 0 \\ 0 & 1 & 2 \end{array} | \begin{array}{r} 12 \\ 0 \\ −7 \end{array}] [\begin{array}{r} 1 & 0 & 3 \\ −1 & 4 & 0 \\ 0 & 1 & 2 \end{array} | \begin{array}{r} 12 \\ 0 \\ −7 \end{array}]$

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57.

No solutions exist.

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59.

No solutions exist.

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61.

$\frac{1}{8} [\begin{matrix} 2 & 7 \\ 6 & 1 \end{matrix}] \frac{1}{8} [\begin{matrix} 2 & 7 \\ 6 & 1 \end{matrix}]$

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63.

No inverse exists.

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65.

$(- 20, 40) (- 20, 40)$

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67.

$(- 1, 0.2, 0.3) (- 1, 0.2, 0.3)$

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69.

17% oranges, 34% bananas, 39% apples

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71.

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73.

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75.

$(6, \frac{1}{2}) (6, \frac{1}{2})$

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77.

(x, 5x + 3)

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79.

$(0, 0, - \frac{1}{2}) (0, 0, - \frac{1}{2})$

Practice Test

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Yes

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No solutions exist.

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$\frac{1}{20} (10, 5, 4) \frac{1}{20} (10, 5, 4)$

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$(x, \frac{16 x}{5} - \frac{13 x}{5}) (x, \frac{16 x}{5} - \frac{13 x}{5})$

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$(- 2 \sqrt{2}, - \sqrt{17}), (- 2 \sqrt{2}, \sqrt{17}), (2 \sqrt{2}, - \sqrt{17}), (2 \sqrt{2}, \sqrt{17}) (- 2 \sqrt{2}, - \sqrt{17}), (- 2 \sqrt{2}, \sqrt{17}), (2 \sqrt{2}, - \sqrt{17}), (2 \sqrt{2}, \sqrt{17})$

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11.

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13.

$\frac{5}{3 x + 1} - \frac{2 x + 3}{{(3 x + 1)}^{2}} \frac{5}{3 x + 1} - \frac{2 x + 3}{{(3 x + 1)}^{2}}$

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15.

$[\begin{matrix} 17 & 51 \\ - 8 & 11 \end{matrix}] [\begin{matrix} 17 & 51 \\ - 8 & 11 \end{matrix}]$

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17.

$[\begin{matrix} 12 & - 20 \\ - 15 & 30 \end{matrix}] [\begin{matrix} 12 & - 20 \\ - 15 & 30 \end{matrix}]$

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19.

$- \frac{1}{8} - \frac{1}{8}$

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21.

$[\begin{array}{r} 14 & - 2 & 13 \\ - 2 & 3 & - 6 \\ 1 & - 5 & 12 \end{array} | \begin{array}{r} 140 \\ - 1 \\ 11 \end{array}] [\begin{array}{r} 14 & - 2 & 13 \\ - 2 & 3 & - 6 \\ 1 & - 5 & 12 \end{array}]$

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23.

No solutions exist.

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25.

$(100, 90) (100, 90)$

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27.

$(\frac{1}{100}, 0) (\frac{1}{100}, 0)$

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29.

32 or more cell phones per day