第11章連立方程式と連立不等式Systems of Equations and Inequalities

Answer Key解答

Solution 1解答 1

Not a solution.

解ではない。

Solution 2解答 2

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

この組の解は順序対 $(−5, 3)$ である。

A graph displays two intersecting lines. The blue line represents the equation y = (4/5)x + 7, and the orange line represents y = (2/5)x + 5. They intersect at the point (-5, 3).

Solution 3解答 3

$(−2, −5)$

Solution 4解答 4

$(−6, −2)$

Solution 5解答 5

$(10, −4)$

Solution 6解答 6

No solution. It is an inconsistent system.

解は無い。解を持たない組である。

Solution 7解答 7

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

この組は従属なので、 $(x, 2 x + 5)$ の形の解が無限にある。

Solution 8解答 8

700 children, 950 adults

子ども700人、大人950人

Solution 1解答 1

$(1, −1, 1)$

Solution 2解答 2

No solution.

解は無い。

Solution 3解答 3

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

$(x, 4 x −11, −5 x + 18)$ の形の解が無限にある

Solution 1解答 1

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

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

Solution 2解答 2

$(−1, 3)$

Solution 3解答 3

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

Solution 4解答 4

A graph illustrates a blue line and an orange parabola intersecting at (-1, 0) and (2, 3). The region bounded by these two curves between the intersection points is shaded light blue.

Solution 1解答 1

$\frac{3}{x −3} - \frac{2}{x −2}$

Solution 2解答 2

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

Solution 3解答 3

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

Solution 4解答 4

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

Solution 1解答 1

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

Solution 2解答 2

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

Solution 1解答 1

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

Solution 2解答 2

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

Solution 3解答 3

$(2, 1)$

Solution 4解答 4

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

Solution 5解答 5

$(1, 1, 1)$

Solution 6解答 6

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

7%で150,000ドル、8%で750,000ドル、10%で600,000ドル

Solution 1解答 1

\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}

Solution 2解答 2

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

Solution 3解答 3

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

Solution 4解答 4

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

Solution 1解答 1

$(3, - 7)$

Solution 2解答 2

$- 10$

Solution 3解答 3

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

11.1 Section Exercises11.1 節末問題

Solution 1解答 1

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

ありえない。解は零個か一個か無限個のどれかである。グラフを調べよ。

Solution 3解答 3

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

これは現実的な損益分岐点が無いということである。この会社は1台作った時点ですでに利益を出している。

Solution 5解答 5

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

代入( $x$ か $y$ を取り出す)、グラフ、または足し算で解ける。

Solution 7解答 7

Yes

そうである

Solution 9解答 9

Yes

そうである

Solution 11解答 11

$(−1, 2)$

Solution 13解答 13

$(−3, 1)$

Solution 15解答 15

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

Solution 17解答 17

No solutions exist.

解は無い。

Solution 19解答 19

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

Solution 21解答 21

$(6, −6)$

Solution 23解答 23

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

Solution 25解答 25

No solutions exist.

解は無い。

Solution 27解答 27

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

Solution 29解答 29

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

Solution 31解答 31

$(−4, 4)$

Solution 33解答 33

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

Solution 35解答 35

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

Solution 37解答 37

$(x, 2 (7 x −6))$

Solution 39解答 39

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

Solution 41解答 41

Consistent with one solution

解を持ち、解は一つ

Solution 43解答 43

Consistent with one solution

解を持ち、解は一つ

Solution 45解答 45

Dependent with infinitely many solutions

従属で解が無限にある

Solution 47解答 47

$(−3.08, 4.91)$

Solution 49解答 49

$(−1.52, 2.29)$

Solution 51解答 51

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

Solution 53解答 53

$(\frac{−1}{A - B}, \frac{A}{A - B})$

Solution 55解答 55

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

Solution 57解答 57

They never turn a profit.

けっして利益を出さない。

Solution 59解答 59

$(1, 250, 100, 000)$

Solution 61解答 61

The numbers are 7.5 and 20.5.

その数は7.5と20.5である。

Solution 63解答 63

24,000

Solution 65解答 65

790 second-year students, 805 first-year students

2年生790人、1年生805人

Solution 67解答 67

56 men, 74 women

男性56人、女性74人

Solution 69解答 69

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

10%の溶液10ガロン、60%の溶液15ガロン

Solution 71解答 71

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

スワン・ピーク: 750,000ドル、リバーサイド: 350,000ドル

Solution 73解答 73

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

一つ目の口座に12,500ドル、二つ目の口座に10,500ドル。

Solution 75解答 75

High-tops: 45, Low-tops: 15

ハイトップ: 45足、ロートップ: 15足

Solution 77解答 77

Infinitely many solutions. We need more information.

解は無限にある。もっと情報が要る。

11.2 Section Exercises11.2 節末問題

Solution 1解答 1

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

ありえない。解は一個か零個か無限個しかありえない。

Solution 3解答 3

Not necessarily. There could be zero, one, or infinitely many solutions. For example, $(0, 0, 0)$ is not a solution to the system below, but that does not mean that it has no solution.

そうとはかぎらない。解は零個か一個か無限個でありうる。たとえば $(0, 0, 0)$ は下の組の解ではないが、だからといってこの組に解が無いということにはならない。

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

Solution 5解答 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.

どの連立方程式も、グラフ、代入、足し算で解ける。だが三式の組はグラフで解くのがたいへん込み入るので、ふつうは他の方法のほうが望ましい。

Solution 7解答 7

そうでない

Solution 9解答 9

Yes

そうである

Solution 11解答 11

$(−1, 4, 2)$

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

$(4, −6, 1)$

Solution 19解答 19

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

Solution 21解答 21

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

Solution 23解答 23

No solutions exist

解は無い

Solution 25解答 25

$(0, 0, 0)$

Solution 27解答 27

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

Solution 29解答 29

$(7, 20, 16)$

Solution 31解答 31

$(−6, 2, 1)$

Solution 33解答 33

$(5, 12, 15)$

Solution 35解答 35

$(−5, −5, −5)$

Solution 37解答 37

$(10, 10, 10)$

Solution 39解答 39

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

Solution 41解答 41

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

Solution 43解答 43

$(2, 0, 0)$

Solution 45解答 45

$(1, 1, 1)$

Solution 47解答 47

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

Solution 49解答 49

$(6, −1, 0)$

Solution 51解答 51

24, 36, 48

Solution 53解答 53

70 grandparents, 140 parents, 190 children

祖父母70人、親140人、子190人

Solution 55解答 55

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

あなたの分は19.95ドル、シャニの分は40ドル、もう一人の同居人の分は22.05ドルだった。

Solution 57解答 57

There are infinitely many solutions; we need more information

解は無限にある。もっと情報が要る

Solution 59解答 59

500 students, 225 children, and 450 adults

学生500人、子ども225人、大人450人

Solution 61解答 61

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

BMWは49,636ドル、ジープは42,636ドル、トヨタは47,727ドルだった。

Solution 63解答 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.

利率3%の口座に400,000ドル、利率4%の口座に500,000ドル、利率2%の口座に100,000ドル。

Solution 65解答 65

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

世界の石油のうち、合衆国は26.3%、日本は7.1%、中国は6.4%を消費した。

Solution 67解答 67

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

サウジアラビアから16.8%、カナダから15.1%、メキシコから15.0%を輸入した

Solution 69解答 69

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

絶滅危惧種のうち鳥類は19.3%、魚類は18.6%、哺乳類は17.1%だった

11.3 Section Exercises11.3 節末問題

Solution 1解答 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.

一次でない組は、重なり合って二か所で交わる二つの円を表しうるので、解は二つになる。一次でない組は、放物線と円を表すこともでき、放物線の頂点が円に接し、枝も円と交わるなら、解は三つになる。

Solution 3解答 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.

ない。実行可能領域が必ずあるわけではない。二本の平行な直線で囲まれた組を考えよ。一方の不等式は上の直線より上の領域を表し、もう一方は下の直線より下の領域を表す。この場合、平面のどの点も両方の領域には入らない。ゆえに実行可能領域は無い。

Solution 5解答 5

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

各解のあいだの数を選び、 $C (x)$ と $R (x)$ に入れる。 $C (x) < R (x)$ なら利益が出る。

Solution 7解答 7

$(0, −3), (3, 0)$

Solution 9解答 9

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

Solution 11解答 11

$(−3, 0), (3, 0)$

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

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

Solution 19解答 19

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

Solution 21解答 21

$(5, 0)$

Solution 23解答 23

$(0, 0)$

Solution 25解答 25

$(3, 0)$

Solution 27解答 27

No Solutions Exist

解は無い

Solution 29解答 29

No Solutions Exist

解は無い

Solution 31解答 31

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

Solution 33解答 33

$(2, 0)$

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

A graph shows a downward-opening parabola centered on the y-axis, with its vertex at (0, 9). The parabola intersects the x-axis at x = -3 and x = 3. The region under the parabola is shaded in light blue, extending downwards to y = -10. The x-axis is labeled from -5 to 5, and the y-axis is labeled from -10 to 10.

Solution 41解答 41

A coordinate plane is displayed with an x-axis ranging from -5 to 2 and a y-axis ranging from -6 to 2. A light blue shaded region is enclosed by a dashed line. The shaded region is defined by two labeled points: an upper point at $(\sqrt{2}-1, 2(\sqrt{2}-1))$ and a lower point at $(-1-\sqrt{2}, -2(1+\sqrt{2}))$. The region extends from the lower left to the upper right, resembling a curved, elongated shape.

Solution 43解答 43

A graph on a Cartesian coordinate system shows two lens-shaped shaded regions, symmetric about the y-axis. The left region is between x=-2 and x=-5, and the right region is between x=2 and x=5. Four points are labeled.

Solution 45解答 45

A graph of a coordinate plane showing an x-axis and a y-axis. The region between two branches of a hyperbola is shaded light blue. The hyperbola opens upwards and downwards, with the shaded region resembling an hourglass shape. The vertices of the hyperbola are marked by the points (sqrt(19/10), sqrt(47/10)), (-sqrt(19/10), sqrt(47/10)), (sqrt(19/10), -sqrt(47/10)), and (-sqrt(19/10), -sqrt(47/10)). The boundaries of the shaded region are indicated by dashed lines, suggesting that the boundary itself is not included in the region.

Solution 47解答 47

A graph displays two curves and a shaded region on an x-y coordinate plane. An orange curve rises through (0,1). A blue curve falls through (1,0). A rectangle is shaded in the fourth quadrant, from x=0 to x=3 and y=0 to y=-3.

Solution 49解答 49

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

Solution 51解答 51

No Solution Exists

解は無い

Solution 53解答 53

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

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

Solution 55解答 55

12, 288

Solution 57解答 57

2–20 computers

計算機2〜20台

11.4 Section Exercises11.4 節末問題

Solution 1解答 1

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

できない。多項式の商は、分母が因数分解できるときにかぎって分解できる。たとえば $\frac{1}{x^{2} + 1}$ は、分母が因数分解できないので分解できない。

Solution 3解答 3

Graph both sides and ensure they are equal.

両辺のグラフを描き、等しいことを確かめる。

Solution 5解答 5

If we choose $x = −1,$ then the B-term disappears, letting us immediately know that $A = 3.$ We could alternatively plug in $x = - \frac{5}{3}$ , giving us a B-value of $−2.$

$x = −1$ を選べば B の項が消え、 $A = 3.$ がすぐに分かる。あるいは $x = - \frac{5}{3}$ を入れてもよく、そのとき B の値は $−2.$ になる。

Solution 7解答 7

$\frac{8}{x + 3} - \frac{5}{x −8}$

Solution 9解答 9

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

Solution 11解答 11

$\frac{3}{5 x −2} + \frac{4}{4 x −1}$

Solution 13解答 13

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

Solution 15解答 15

$\frac{3}{x + 2} + \frac{3}{x −2}$

Solution 17解答 17

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

Solution 19解答 19

$\frac{8}{x −3} - \frac{5}{x −2}$

Solution 21解答 21

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

Solution 23解答 23

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

Solution 25解答 25

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

Solution 27解答 27

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

Solution 29解答 29

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

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

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

Solution 41解答 41

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

Solution 43解答 43

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

Solution 45解答 45

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

Solution 47解答 47

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

Solution 49解答 49

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

Solution 51解答 51

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

Solution 53解答 53

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

Solution 55解答 55

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

Solution 57解答 57

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

Solution 59解答 59

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

11.5 Section Exercises11.5 節末問題

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

できない。型が同じでなければならない。例としては型の違う二つの行列が挙げられる。次の二つの行列は足せない。一つ目が $2 \times 2$ 行列、二つ目が $2 \times 3$ 行列だからである。 $[\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}] + [\begin{matrix} 6 & 5 & 4 \\ 3 & 2 & 1 \end{matrix}]$ に和は無い。

Solution 3解答 3

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

ありうる。 $A$ の型が $m \times n$ 、 $B$ の型が $n \times m$ なら、どちらの積も定められる。

Solution 5解答 5

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

そうとはかぎらない。 $A B$ を求めるには、 $A$ の1行目に $B$ の1列目を掛けて $A B$ の最初の成分を得る。 $B A$ を求めるには、 $B$ の1行目に $A$ の1列目を掛けて $B A$ の最初の成分を得る。よって、これらが等しくなければ、行列の掛け算は交換できない。

Solution 7解答 7

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

Solution 9解答 9

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

Solution 11解答 11

Undidentified; dimensions do not match

定められない。型が合わない

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

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

Solution 19解答 19

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

Solution 21解答 21

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

Solution 23解答 23

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

Solution 25解答 25

Undefined; dimensions do not match.

定められない。型が合わない。

Solution 27解答 27

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

Solution 29解答 29

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

Solution 31解答 31

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

Solution 33解答 33

Undefined; inner dimensions do not match.

定められない。内側の型が合わない。

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

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

Solution 41解答 41

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

Solution 43解答 43

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

Solution 45解答 45

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

Solution 47解答 47

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

Solution 49解答 49

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

Solution 51解答 51

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

Solution 53解答 53

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

Solution 55解答 55

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

Solution 57解答 57

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

Solution 59解答 59

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

式中の英語：odd：奇数

11.6 Section Exercises11.6 節末問題

Solution 1解答 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.

書ける。各行について、変数の係数を対応する行に横に書き、縦線を置く。そして定数を縦線の右に置く。

Solution 3解答 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 $R_{2} = R_{2} −9 R_{1} .$ (2) $R_{2} = R_{1} −9 R_{2} .$ Then divide row 1 by 9.

一つだけではない。行列に行の演算を使う正しいしかたは数多くある。ありうる二つのしかたは次である。(1) 1行目と2行目を入れ替える。すると $R_{2} = R_{2} −9 R_{1}$ 。(2) $R_{2} = R_{1} −9 R_{2}$ 。そして1行目を9で割る。

Solution 5解答 5

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

ありえない。一つの行の成分がすべて0である行列は、解が零個か無限個のどちらかになる。

Solution 7解答 7

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

Solution 9解答 9

$[\begin{array}{r} 1 & 5 & 8 \\ 12 & 3 & 0 \\ 3 & 4 & 9 \end{array} | \begin{array}{r} 19 \\ 4 \\ −7 \end{array}]$

Solution 11解答 11

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

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

No solutions

解は無い

Solution 19解答 19

$(−1, −2)$

Solution 21解答 21

$(6, 7)$

Solution 23解答 23

$(3, 2)$

Solution 25解答 25

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

Solution 27解答 27

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

Solution 29解答 29

$(3, 4)$

Solution 31解答 31

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

Solution 33解答 33

$(31, −42, 87)$

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

$(x, y, \frac{1}{2} (1 −2 x −3 y))$

Solution 41解答 41

$(x, - \frac{x}{2}, −1)$

Solution 43解答 43

$(125, −25, 0)$

Solution 45解答 45

$(8, 1, −2)$

Solution 47解答 47

$(1, 2, 3)$

Solution 49解答 49

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

Solution 51解答 51

No solutions exist.

解は無い。

Solution 53解答 53

860 red velvet, 1,340 chocolate

レッドベルベット860個、チョコレート1,340個

Solution 55解答 55

4% for account 1, 6% for account 2

口座1は4%、口座2は6%

Solution 57解答 57

$126

Solution 59解答 59

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

バナナは3%、パンプキンは7%、ロッキーロードは2%

Solution 61解答 61

100 almonds, 200 cashews, 600 pistachios

アーモンド100粒、カシューナッツ200粒、ピスタチオ600粒

11.7 Section Exercises11.7 節末問題

Solution 1解答 1

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

$A^{−1}$ が $A$ の逆なら、 $A A^{−1} = I$ すなわち単位行列である。 $A$ も $A^{−1}, A^{−1} A = I$ の逆だからである。 $2 \times 2$ 行列でこれを証明して確かめることもできる。

Solution 3解答 3

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

できない。 $a d$ と $b c$ がどちらも0なので $a d - b c = 0$ となり、公式で0で割ることになるからである。

Solution 5解答 5

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

ありうる。行列 $[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ を考えよ。逆は次の計算で求まる。 $A^{−1} = \frac{1}{0 (0) −1 (1)} [\begin{matrix} 0 & −1 \\ −1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$

Solution 7解答 7

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

Solution 9解答 9

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

Solution 11解答 11

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

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

There is no inverse

逆は無い

Solution 19解答 19

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

Solution 21解答 21

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

Solution 23解答 23

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

Solution 25解答 25

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

Solution 27解答 27

$(−5, 6)$

Solution 29解答 29

$(2, 0)$

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

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

Solution 37解答 37

$(5, 0, −1)$

Solution 39解答 39

$\frac{1}{34} (−35, −97, −154)$

Solution 41解答 41

$\frac{1}{690} (65, −1136, −229)$

Solution 43解答 43

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

Solution 45解答 45

$(\frac{10}{123}, −1, \frac{2}{5})$

Solution 47解答 47

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

Solution 49解答 49

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

Solution 51解答 51

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

Solution 53解答 53

Infinite solutions.

解は無限にある。

Solution 55解答 55

50% oranges, 25% bananas, 20% apples

みかん50%、バナナ25%、りんご20%

Solution 57解答 57

10 straw hats, 50 beanies, 40 cowboy hats

麦わら帽子10個、ニット帽50個、カウボーイ帽40個

Solution 59解答 59

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

ミカが6本、ジョーが3本、アルバートが3本食べた。

Solution 61解答 61

124 oranges, 10 lemons, 8 pomegranates

みかん124個、レモン10個、ざくろ8個

11.8 Section Exercises11.8 節末問題

Solution 1解答 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.

行列式は行列の成分の積の和なので、たとえ結果が0になるとしても、その積はつねに計算できる。

Solution 3解答 3

The inverse does not exist.

逆は存在しない。

Solution 5解答 5

$- 2$

Solution 7解答 7

$7$

Solution 9解答 9

$- 4$

Solution 11解答 11

$0$

Solution 13解答 13

$- 7, 990.7$

Solution 15解答 15

$3$

Solution 17解答 17

$- 1$

Solution 19解答 19

$224$

Solution 21解答 21

$15$

Solution 23解答 23

$- 17.03$

Solution 25解答 25

$(1, 1)$

Solution 27解答 27

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

Solution 29解答 29

$(2, 5)$

Solution 31解答 31

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

Solution 33解答 33

$(15, 12)$

Solution 35解答 35

$(1, 3, 2)$

Solution 37解答 37

$(- 1, 0, 3)$

Solution 39解答 39

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

Solution 41解答 41

$(2, 1, 4)$

Solution 43解答 43

Infinite solutions

解は無限にある

Solution 45解答 45

$24$

Solution 47解答 47

$1$

Solution 49解答 49

Yes; 18, 38

ある。18、38

Solution 51解答 51

Yes; 33, 36, 37

ある。33、36、37

Solution 53解答 53

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

一つ目の口座に7,000ドル、二つ目の口座に3,000ドル。

Solution 55解答 55

120 children, 1,080 adult

子ども120枚、大人1,080枚

Solution 57解答 57

4 gal yellow, 6 gal blue

黄色4ガロン、青6ガロン

Solution 59解答 59

13 green tomatoes, 17 red tomatoes

緑のトマト13個、赤のトマト17個

Solution 61解答 61

Strawberries 18%, oranges 9%, kiwi 10%

いちご18%、みかん9%、キウイ10%

Solution 63解答 63

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

映画1が100枚、映画2が230枚、映画3が312枚

Solution 65解答 65

300 almonds, 400 cranberries, 300 cashews

アーモンド300粒、クランベリー400粒、カシューナッツ300粒

Review Exercises復習問題

Solution 1解答 1

そうでない

Solution 3解答 3

$(- 2, 3)$

Solution 5解答 5

$(4, - 1)$

Solution 7解答 7

No solutions exist.

解は無い。

Solution 9解答 9

$(300, 60, 000)$

Solution 11解答 11

Infinite solutions

解は無限にある

Solution 13解答 13

No solutions exist.

解は無い。

Solution 15解答 15

$(- 1, - 2, 3)$

Solution 17解答 17

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

Solution 19解答 19

11, 17, 33

Solution 21解答 21

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

Solution 23解答 23

No solution

解は無い

Solution 25解答 25

No solution

解は無い

Solution 27解答 27

A coordinate plane with x and y axes ranging from -5 to 5. A light blue shaded ellipse is shown, centered at the origin (0,0). The ellipse has a dashed dark blue boundary. It extends horizontally from x = -4 to x = 4, and vertically from y = -2 to y = 2.

Solution 29解答 29

A two-dimensional graph displays an x-axis and a y-axis. A light blue shaded region is enclosed by two dashed lines. One boundary is a straight line segment connecting the points (1, 3) and (4, 0). The other boundary is a continuous curve that starts at (0, 3), passes through approximately (-0.5, 0) and (0, -1), and ends at (6, -2). The region extends across all four quadrants.

Solution 31解答 31

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

Solution 33解答 33

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

Solution 35解答 35

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

Solution 37解答 37

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

Solution 39解答 39

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

Solution 41解答 41

undefined; dimensions do not match

定められない。型が合わない

Solution 43解答 43

undefined; inner dimensions do not match

定められない。内側の型が合わない

Solution 45解答 45

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

Solution 47解答 47

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

Solution 49解答 49

undefined; inner dimensions do not match

定められない。内側の型が合わない

Solution 51解答 51

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

$\begin{array}{l} x - 3 z = 7 \\ y + 2 z = - 5 \end{array}$ で解は無限にある

Solution 53解答 53

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

Solution 55解答 55

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

Solution 57解答 57

No solutions exist.

解は無い。

Solution 59解答 59

No solutions exist.

解は無い。

Solution 61解答 61

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

Solution 63解答 63

No inverse exists.

逆は存在しない。

Solution 65解答 65

$(- 20, 40)$

Solution 67解答 67

$(- 1, 0.2, 0.3)$

Solution 69解答 69

17% oranges, 34% bananas, 39% apples

みかん17%、バナナ34%、りんご39%

Solution 71解答 71

Solution 73解答 73

Solution 75解答 75

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

Solution 77解答 77

(x, 5x + 3)

Solution 79解答 79

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

Practice Test実力テスト

Solution 1解答 1

Yes

そうである

Solution 3解答 3

No solutions exist.

解は無い。

Solution 5解答 5

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

Solution 7解答 7

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

Solution 9解答 9

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

Solution 11解答 11

This graph displays a coordinate plane with x and y axes ranging from -4 to 4. A light blue shaded region represents the solution set. This region is defined as being outside a dashed circle centered at the origin (0,0) with a radius of 2 units. The circle passes through points (2,0), (0,2), (-2,0), and (0,-2). The shaded region is also above a dashed parabola that opens upwards, with its vertex at the point (0,1). The parabola appears to pass through approximately (-1, 2) and (1, 2).

Solution 13解答 13

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

Solution 15解答 15

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

Solution 17解答 17

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

Solution 19解答 19

$- \frac{1}{8}$

Solution 21解答 21

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

Solution 23解答 23

No solutions exist.

解は無い。

Solution 25解答 25

$(100, 90)$

Solution 27解答 27

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

Solution 29解答 29

32 or more cell phones per day

一日に携帯電話32台以上