The document describes the process of analyzing a truss structure using the method of joints. It provides two examples of solving for the forces in each member of a truss given applied loads. In both examples, the document first calculates the support reactions, then analyzes the force in each member by examining the equilibrium of forces at each joint. It is able to determine the force magnitude and whether each member is in tension or compression.
The document describes the process of analyzing a truss structure using the method of joints. It provides two examples of solving for the forces in each member of a truss given applied loads. In both examples, the document first calculates the support reactions, then analyzes the force in each member by examining the equilibrium of forces at each joint. It is able to determine the force magnitude and whether each member is in tension or compression.
This document contrasts vector mechanics and variational formulations for structural analysis. It uses the example of a simply supported beam under a uniformly distributed load to illustrate the approaches.
[1] Using vector mechanics, the beam is modeled as discrete elements and equilibrium equations are written for each element and solved. [2] Alternatively, the variational approach introduces a potential energy function for the system and finds the shape that minimizes this function. [3] For the beam example, an approximate solution is first found using a simple shape function, then a more accurate solution is obtained using a shape function with more degrees of freedom that matches the exact solution.
- The document contains code and explanations for solving optimization problems using dynamic programming, including calculating minimum costs using a 2D array to store results.
- It describes applying dynamic programming to problems involving finding minimum costs for tasks that can be split into subtasks, with the overall cost determined by combining subtask costs.
- The code provided shows initializing a 2D array and using nested for loops to iterate through values, calculate minimum costs based on previous results, and store them in the 2D array to build up an optimal solution.
This document contains 20 multi-part engineering problems involving the calculation of shear and moment diagrams for beams and shafts. The problems include beams under various loading conditions such as point loads, distributed loads, overhanging sections, and compound sections. Shear and moment diagrams are drawn and the shear and moment values are calculated as functions of position along the members.
This document discusses various factors used to compare alternatives that provide the same service over an extended period of time when interest is involved. It defines single payment present worth (P/F) and future worth (F/P) factors, uniform series present worth (P/A) and capital recovery (A/P) factors, sinking fund (A/F) and uniform series future worth (F/A) factors, and how to find factor values using interpolation when exact interest rates or time periods are not listed in tables.
The document discusses independent and dependent events. Independent events are those where the occurrence of one event does not affect the other event, such as flipping a coin or rolling a die. Dependent events are those where one event affects the other, like drawing cards from a deck without replacement. The probability of dependent events decreases as the sample space decreases with each subsequent event, while independent events have an unchanged sample space and probability.
The document discusses benefit-cost analysis for public sector projects. It covers calculating benefit-cost ratios for single and multiple alternatives, using cost-effectiveness analysis for service sector projects, and some key differences between public and private projects like size, life, funding sources, and selection criteria. Ethical considerations are also prevalent in public policy and planning where engineers are involved.
Bisection Method is a Derivative Based Method for Optimization.
It is one of the classical optimization techniques.
Numerical on Bisection method is discussed in this Presentation
This document contains 8 assignment sheets related to mechanical engineering concepts including:
1. Free body diagrams and reactions at supports
2. Internal reaction diagrams for beams
3. Axially loaded bars including stresses and deflections
4. Bending of bars including stresses, deflections, and internal reaction diagrams
5. Torsion of bars including shear stresses and angles of twist
6. Thin walled pressure containers including stress components and allowable pressures
7. Stress transformation including Mohr's circle and principal stresses
8. A problem involving stresses in a thin walled steel pressure container
The assignments cover a range of load cases and ask students to calculate stresses, deflections, reactions and other mechanical properties.
This document contrasts vector mechanics and variational formulations for structural analysis. It uses the example of a simply supported beam under a uniformly distributed load to illustrate the approaches.
[1] Using vector mechanics, the beam is modeled as discrete elements and equilibrium equations are written for each element and solved. [2] Alternatively, the variational approach introduces a potential energy function for the system and finds the shape that minimizes this function. [3] For the beam example, an approximate solution is first found using a simple shape function, then a more accurate solution is obtained using a shape function with more degrees of freedom that matches the exact solution.
- The document contains code and explanations for solving optimization problems using dynamic programming, including calculating minimum costs using a 2D array to store results.
- It describes applying dynamic programming to problems involving finding minimum costs for tasks that can be split into subtasks, with the overall cost determined by combining subtask costs.
- The code provided shows initializing a 2D array and using nested for loops to iterate through values, calculate minimum costs based on previous results, and store them in the 2D array to build up an optimal solution.
This document contains 20 multi-part engineering problems involving the calculation of shear and moment diagrams for beams and shafts. The problems include beams under various loading conditions such as point loads, distributed loads, overhanging sections, and compound sections. Shear and moment diagrams are drawn and the shear and moment values are calculated as functions of position along the members.
This document discusses various factors used to compare alternatives that provide the same service over an extended period of time when interest is involved. It defines single payment present worth (P/F) and future worth (F/P) factors, uniform series present worth (P/A) and capital recovery (A/P) factors, sinking fund (A/F) and uniform series future worth (F/A) factors, and how to find factor values using interpolation when exact interest rates or time periods are not listed in tables.
The document discusses independent and dependent events. Independent events are those where the occurrence of one event does not affect the other event, such as flipping a coin or rolling a die. Dependent events are those where one event affects the other, like drawing cards from a deck without replacement. The probability of dependent events decreases as the sample space decreases with each subsequent event, while independent events have an unchanged sample space and probability.
The document discusses benefit-cost analysis for public sector projects. It covers calculating benefit-cost ratios for single and multiple alternatives, using cost-effectiveness analysis for service sector projects, and some key differences between public and private projects like size, life, funding sources, and selection criteria. Ethical considerations are also prevalent in public policy and planning where engineers are involved.
Bisection Method is a Derivative Based Method for Optimization.
It is one of the classical optimization techniques.
Numerical on Bisection method is discussed in this Presentation
This document contains 8 assignment sheets related to mechanical engineering concepts including:
1. Free body diagrams and reactions at supports
2. Internal reaction diagrams for beams
3. Axially loaded bars including stresses and deflections
4. Bending of bars including stresses, deflections, and internal reaction diagrams
5. Torsion of bars including shear stresses and angles of twist
6. Thin walled pressure containers including stress components and allowable pressures
7. Stress transformation including Mohr's circle and principal stresses
8. A problem involving stresses in a thin walled steel pressure container
The assignments cover a range of load cases and ask students to calculate stresses, deflections, reactions and other mechanical properties.
4. 第6 章 聯立方程式 123
6
Chapter 6 聯立方程式
趨勢分析
主題簡介 二階與三階行列式、聯立方程式的解法、克拉瑪公式。
最常考題型 行列式的運算與化簡、行列式方程式的解。
次重要題型 聯立方程式的解,克拉瑪公式的應用。
綜合分析 本單元的命題以「三階行列式」的運算、化簡與方程式為主。
行列式
重點整理 行列式的意義
1. 二階行列式:
形如a b
c d
的式子稱為「二階行列式」,且規定a b
ad bc
c d
= − 。
2. 三階行列式:
形如
a b c
d e f
g h i
的式子稱為「三階行列式」,且規定
a b c
d e f
g h i
= aei + dhc + gfb − ceg − bdi − ahf。
6-1
QRcode 影音解題
(蘋果系列行動裝置無法觀看)
5. 124 第 6 章 聯立方程式
試求下列的值:
(1)
1 2
4 5
(2)
1 0 1
1 2 3
2 1 3
−
。
(1) 所求 1 5 4 2 3= × − × = −
(2) 所求 ( )1 2 3 1 1 1 2 3 0= − × × + × × + × ×
( )1 2 2 0 1 3 1 1 3− × × − × × − − × ×
6= −
試求下列的值:
(1)
3 4
1 2
(2)
1 2 1
2 4 1
0 5 2
−
−
−
。
(1) 所求 3 2 1 4 2= × − × =
(2) 所求 ( ) ( ) ( )1 4 2 2 5 1 0 1 2= × × − + − × × − + × ×
( ) ( ) ( )1 4 0 2 2 2 1 5 1− − × × − × − × − − × ×
11= −
設
1 3 5
1 3 49
5 1
x
x
= 的解為α 與β ,則α β+ =?
原式⇒ 2
5 6 21 49x x− + =
⇒ 2
5 6 28 0x x− − =
兩根和: 6 6
5 5
α β −
+ = − =
設
1 4 6
1 4 64
6 1
x
x
= 的解為α 與β ,則αβ =?
原式⇒ 2
6 8 61 64x x− + =
⇒ 2
6 8 3 0x x− − =
兩根積: 3 1
6 2
αβ −
= = −
重點整理 三階行列式的降階
1. 三階行列式可對某一行(列)降成二階行列式展開。
a b c
e f d f d e b c a c a b b c a c a b
d e f a b c d e f g h i
h i g i g h h i g i g h e f d f d e
g h i
= − + = − + − = − +
e f b c b c d f a c a c d e a b a b
a d g b e h c f i
h i h i e f g i g i d f g h g h d e
= − + = − + − = − +
其中係數的正負由
+ − +
− + −
+ − +
決定,a 的二階行列式取法為 e
a b c
d f
g h i
。
2. 若某行(列)有2 個0,則對其降階可以簡化計算。如:
5 7 9
4 6
0 4 6 5
3 8
0 3 8
= ,
5 7 9
4 6
4 0 6 7
3 8
3 0 8
= − 。
2
1
(依第一列展開) (依第二列展開)
(依第一行展開) (依第二行展開)
行列式的定義
☆
行列式方程式
☆☆
(依第三列展開)
(依第三行展開)
7. 126 第 6 章 聯立方程式
(7) 對某一行(列),可拆成兩個行列式的和:1 2 1 0 2 1 2 0 2
3 4 2 1 4 2 4 1 4
+
= = +
+
。
2. 三階行列式的性質:
與二階行列式的性質相同。
若
1
1 3
1
a d
b e
c f
= ,試求下列的值:
(1)
5 2
5 2
5 2
a d
b e
c f
−
−
−
(2)
2 8 2
3 12 3
5 20 5
a d
b e
c f− − −
。
(1) 所求 ( )
1
5 2 1
1
a d
b e
c f
= × − ×
( )5 2 3 30= × − × = −
(2) 所求 ( )
1
2 3 5 4 1
1
a d
b e
c f
= × × − × ×
1 1
120 1 120 1
1 1
a d a d
b e b e
c f c f
= − =
120 3 360= × =
若 5
1 1 1
a b c
d e f = ,試求下列的值:
(1)
4 4 4
3 3 3
a b c
d e f (2)
6 12 3
12 24 6
2 4
a b c
d e f
− − −
。
(1) 所求 4 3
1 1 1
a b c
d e f= × ×
4 3 5 60= × × =
(2) 所求 ( )
1 1 1
3 6 2 4 a b c
d e f
= − × × × ×
1 1 1
144 a b c
d e f
= − 144
1 1 1
a b c
d e f= −
144 5 720= − × = −
試求下列的值:
(1)
11 33 43
13 39 49
15 45 57
(2)
11 9 7
22 19 15
55 47 35
。
(1) 前兩行成比例,所求 0=
(2) 所求=
1 9 7
11 2 19 15
5 47 35
×
1 0 0
11 2 1 1
5 2 0
= ×
1 1
11 22
2 0
= × = −
試求下列的值:
(1)
19 23 28
24 25 26
48 50 52
(2)
70 33 70
40 20 41
80 40 81
−
−
−
。
(1) 後兩列成比例,所求 0=
(2) 所求
7 33 70
10 4 20 41
8 40 81
= − ×
7 2 0
10 4 0 1
8 0 1
−
= − ×
4 1
10 2 80
8 1
= − × × =
5
6
行列式的性質
☆☆
行列式的化簡
☆☆☆
( )7× −
( )9× −
( )10× −
( )5× −
8. 第6 章 聯立方程式 127
6
若
1 1 1
1 1 1
1 1 1
x
x
x
+
+
+
0= ,試求x 的值。
把後兩行加到第一行
3 1 1
3 1 1 0
3 1 1
x
x x
x x
+
+ + =
+ +
( )
1 1 1
3 1 1 1 0
1 1 1
x x
x
⇒ + + =
+
( )
1 0 0
3 1 0 0
1 0
x x
x
⇒ + =
( )
2
3 0x x⇒ + = 3x⇒ =− 或0
若
1 2 3
1 2 3 0
1 2 3
x
x
x
+
+ =
+
,試求x 的值。
把後兩行加到第一行
6 2 3
6 2 3 0
6 2 3
x
x x
x x
+
+ + =
+ +
( )
1 2 3
6 1 2 3 0
1 2 3
x x
x
⇒ + + =
+
( )
1 0 0
6 1 0 0
1 0
x x
x
⇒ + =
( )
2
6 0x x⇒ + =
6x⇒ =− 或0
若
2
2
2
1
1 16
1
a a
b b
c c
= 且
3
3
3
1
1 240
1
a a
b b
c c
= ,則
( )
( )
( )
2
2
2
1 1 1
1 1 1
1 1 1
a a a
b b b
c c c
+ +
+ + =
+ +
?
所求第一行乘( )1− 加到第二行
( )
( )
( )
2 3 2
2 3 2
2 3 2
1 1 1
1 1 1
1 1 1
a a a a a a
b b b b b b
c c c c c c
+ +
+ = +
+ +
2 3
2 3
2 3
1 1
1 1
1 1
a a a a
b b b b
c c c c
= +
16 240 256= + =
若
1 0
9 3 5
4 2
x
y
z
= ,則
1 0
9 3
4 2 6
x
y
z
=
+
?
所求
1 0 1 0 0
9 3 9 3 0
4 2 4 2 6
x
y
z
= +
3 0
5
2 6
= +
5 18= +
23=
7
8
行列式的化簡
☆☆☆
行列式的分解
☆☆
( )1× −
( )1× −
( )3× −
( )2× −
9. 128 第 6 章 聯立方程式
1. 試求下列的值:1 3
5 7
= 8− , sin10 cos10
cos10 sin10
° − °
=
° °
1 。
2. 已知 1
1
4 3
x
= − ,則 1 1
3 5
x x+ −
= 10 。
3. 試求下列的值:
1 2 1
2 3 1
1 1 3
= 3− ,
3 2 1
2 1 0
1 3 2
−
− =
−
7 。
4. 設
1 1 2
1 1 3
2 1
x
x
−
− = 的解為α 與β ,則α β+ = 1− ,αβ = 2− 。
5. 若
2 3 7
5 6 4 6 4 5
4 5 6 2 7
1 8 9 8 9 1
9 1 8
a= × + × + ×
3 7 2 7 2 3
4 5
1 8 9 8 9 1
b= − × + × + × ,則
a = 3− ,b = 6− 。
6. 若
3 0 0
2 6
4 2 6
5 7
1 5 7
a= × 與
3 2 6
2 6
4 0 0
5 7
1 5 7
b= × ,則a = 3 ,b = 4− 。
7. 若
1
1 2
1
a d
b e
c f
= − ,則
3 2 1
3 2 1
3 2 1
a d
b e
c f
= 12− ,
2 8 2
4
3 12 3
a d
b e
c f
= 48 。
8. 試求下列的值:
21 22 23
24 25 26
27 28 29
= 0 ,
15 5 9
30 11 21
45 17 31
= 30− 。
*9. 若
1 1 1
1 1 1 0
1 1 1
x
x
x
−
− =
−
,則x = 1 2− 或 。
10. 若
1
1 2
1
a p
b q
c r
= 且
1
1 3
1
a x
b y
c z
= ,則
1 2
1 2
1 2
a p x
b q y
c r z
−
− =
−
4− 。
實力測驗1
10. 第6 章 聯立方程式 129
6
聯立方程式與克拉瑪公式
重點整理 二元一次方程組的克拉瑪公式
在 1 1 1
2 2 2
a x b y c
a x b y c
+ =⎧
⎨
+ =⎩
之中,令 1 1
2 2
a b
a b
Δ = , 1 1
2 2
x
c b
c b
Δ = , 1 1
2 2
y
a c
a c
Δ = ,
(1) 當 0Δ ≠ 時,方程組恰有一組解 x
x
Δ
=
Δ
, y
y
Δ
=
Δ
。
(2) 當 x yΔ = Δ = Δ 0= 時,方程組有無限多組解。
(3) 當 0Δ = ,而 0xΔ ≠ 或 0yΔ ≠ 時,方程組無解。
試解方程組 2 2
3 9
x y
x y
+ =⎧
⎨
+ = −⎩
。
2 2
3 9
x y
x y
+ =⎧
⎨
+ = −⎩
1
2
3× − 5 15 3y y⇒ = ⇒ =
3y = 代入 得 4x = −
故 4x = − , 3y =
試解方程組 2 3 1
3 2 4
x y
x y
+ =⎧
⎨
+ =⎩
。
2 3 1
3 2 4
x y
x y
+ =⎧
⎨
+ =⎩
1
2
3× , 2×
6 9 3
6 4 8
x y
x y
+ =⎧
⇒ ⎨
+ =⎩
3
4
− 得5 5 1y y=− ⇒ =−
1y = − 代入 得2 4 2x x= ⇒ =
故 2x = , 1y = −
試解方程組
5 3
21
5 2
11
x y
x y
⎧
+ =⎪
⎪
⎨
⎪ − =
⎪⎩
。
令 1
X
x
= , 1
Y
y
=
原式 5 3 21
5 2 11
X Y
X Y
+ =⎧
⇒ ⎨
− =⎩
1
2
− 5 10 2Y Y⇒ = ⇒ =
2Y = 代入 得5 15X = 3X⇒ =
故 1
3
x = , 1
2
y =
試解方程組
1 1
2
3 5
2
x y
x y
⎧
+ =⎪
⎪
⎨
⎪ − = −
⎪⎩
。
令 1
X
x
= , 1
Y
y
=
原式 2
3 5 2
X Y
X Y
+ =⎧
⇒ ⎨
− = −⎩
1
2
3× − 8 8 1Y Y⇒ = ⇒ =
1Y = 代入 得 1X =
故 1x = , 1y =
1
聯立方程式的解
☆
6-2
2
聯立方程式的解
☆☆
11. 130 第 6 章 聯立方程式
若 2 7
3
x y
ax by
− =⎧
⎨
+ = −⎩
與 2 0
3 8
ax by
x y
+ =⎧
⎨
+ =⎩
有共同解,試
求a 、b 的值。
2 7
3 8
x y
x y
− =⎧
⎨
+ =⎩
的解為共同解
3x⇒ = , 1y = −
則 3 3
6 0
a b
a b
− = −⎧
⎨
− =⎩
1a⇒ = , 6b =
若 2 3
13
x y
ax by
− =⎧
⎨
+ =⎩
與 7
3 16
ax by
x y
− =⎧
⎨
+ =⎩
有相同的解,試
求a 、b 的值。
2 3
3 16
x y
x y
− =⎧
⎨
+ =⎩
的解為相同的解
5x⇒ = , 1y =
則 5 13
5 7
a b
a b
+ =⎧
⎨
− =⎩
2a⇒ = , 3b =
在 3
4
ax by
cx dy
+ =⎧
⎨
+ =⎩
之中,若 2
a b
c d
= ,3
8
4
b
d
= − ,
3
6
4
a
c
= ,試求x 、y 的值。
由克拉瑪公式
3
4 8
4
2
b
d
x
a b
c d
−
= = = − ,
3
4 6
3
2
a
c
y
a b
c d
= = =
在 1
2
ax by
cx dy
+ =⎧
⎨
+ =⎩
之中,若 3
a b
c d
= ,1
6
2
b
d
= ,
1
12
2
a
c
= − ,試求x 、y 的值。
由克拉瑪公式
1
2 6
2
3
b
d
x
a b
c d
= = = ,
1
2 12
4
3
a
c
y
a b
c d
−
= = = −
重點整理 三元一次聯立方程式的克拉瑪公式
在
1 1 1 1
2 2 2 2
3 3 3 3
a x b y c z d
a x b y c z d
a x b y c z d
+ + =⎧
⎪
+ + =⎨
⎪ + + =⎩
之中,令
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
Δ = ,
1 1 1
2 2 2
3 3 3
x
d b c
d b c
d b c
Δ = ,
1 1 1
2 2 2
3 3 3
y
a d c
a d c
a d c
Δ = ,
1 1 1
2 2 2
3 3 3
z
a b d
a b d
a b d
Δ = ,
(1) 當 0Δ ≠ 時,方程組恰有一組解 x
x
Δ
=
Δ
, y
y
Δ
=
Δ
, z
z
Δ
=
Δ
。
(2) 當 0Δ = 時,方程組有無限多組解或無解。
當三元一次齊次方程組有異於( )0, 0, 0 的解,則有無限多組解。
3
4
聯立方程式的解
☆☆
克拉瑪公式
☆
12. 第6 章 聯立方程式 131
6
利用克拉瑪公式解
1
2 3 4
4 9 16
x y z
x y z
x y z
+ + =⎧
⎪
+ + =⎨
⎪ + + =⎩
。
1 1 1
2 3 1 3 18 4 12 9 2 2
4 9 1
Δ = = + + − − − =
1 1 1
4 3 1 3 36 16 48 9 4 6
16 9 1
xΔ = = + + − − − = −
1 1 1
2 4 1 4 32 4 16 16 2 6
4 16 1
yΔ = = + + − − − =
1 1 1
2 3 4 48 18 16 12 36 32 2
4 9 16
z
Δ = = + + − − − =
6
3
2
x
x
Δ −
= = = −
Δ
, 6
3
2
y
y
Δ
= = =
Δ
,
2
1
2
z
z
Δ
= = =
Δ
利用克拉瑪公式解
2 3
2 6
3 5 10 11
x y z
x y z
x y z
+ + =⎧
⎪
− + =⎨
⎪ + + =⎩
。
1 1 2
2 1 1 10 20 3 6 5 20 6
3 5 10
Δ = − = − + + + − − = −
3 1 2
6 1 1 30 60 11 22 15 60 12
11 5 10
xΔ = − = − + + + − − = −
1 3 2
2 6 1 60 44 9 36 11 60 6
3 11 10
yΔ = = + + − − − =
1 1 3
2 1 6 11 30 18 9 30 22 6
3 5 11
z
Δ = − = − + + + − − = −
12
2
6
x
x
Δ −
= = =
Δ −
, 6
1
6
y
y
Δ
= = = −
Δ −
,
6
1
6
z
z
Δ −
= = =
Δ −
重點整理 齊次方程組
若 1 1 1
2 2 2
0
0
a x b y c z
a x b y c z
+ + =⎧
⎨
+ + =⎩
,其中x 、y 、z 不全為0,則
1 1 1 1 1 1
2 2 2 2 2 2
: : : :
b c c a a b
x y z
b c c a a b
= 。
若 2 4 5 0
2 0
x y z
x y z
+ − =⎧
⎨
− + =⎩
,其中 0xyz ≠ ,試求
: :x y z 。
所求 4 5 5 2 2 4
: :
2 1 1 1 1 2
− −
=
− −
( ) ( ) ( )6 : 7 : 8= − − −
6:7 :8=
若 2 2 0
5 3 2 0
x y z
x y z
− + =⎧
⎨
− − =⎩
,其中 0xyz ≠ ,試求
: :x y z 。
所求 2 1 1 2 2 2
: :
3 2 2 5 5 3
− −
=
− − − −
7 :9: 4=
5
6
克拉瑪公式
☆☆
齊次方程組
☆☆
13. 132 第 6 章 聯立方程式
1. 若2 6
1
x y
+ = 且4 3
3
x y
− = − ,則x = 2− ,y = 3 。
2. 已知 2 7 2 3 7 0x y x y+ − + − + = ,則x = 1 ,y = 3 。
3. 若方程組
2 5 0
2 0
3 0
x y
x y
x y k
+ − =⎧
⎪
+ − =⎨
⎪ − + =⎩
只有一組解,其中k 為實數,則k = 6− 。
4. 若 2 3
2 9
x y
ax by
− = −⎧
⎨
+ =⎩
與 3
6
ax by
x y
+ =⎧
⎨
+ =⎩
有共同解,則a = 2 ,b = 1− 。
5. 在 5
6
ax by
cx dy
+ =⎧
⎨
+ =⎩
之 中 , 若 2
a b
c d
= , 5
6
6
b
d
= , 5
4
6
a
c
= , 則 x = 3 ,
y = 2 。
6. 若方程組 2
5
ax y c
x by
− =⎧
⎨
+ =⎩
有無限多組解,則 4
1
b
a
= 6 。
*7. 若 1 1 1
2 2 2
a x b y c
a x b y c
+ =⎧
⎨
+ =⎩
的解為 4x = 、 1y = ,則 1 1 1
2 2 2
2 3
2 3
a x b y c
a x b y c
+ =⎧
⎨
+ =⎩
的解為 x = 6 ,
y = 3 。
8. 在
1 1 1 1
2 2 2 2
3 3 3 3
a x b y c z d
a x b y c z d
a x b y c z d
+ + =⎧
⎪
+ + =⎨
⎪ + + =⎩
之中,
1 1 1
2 2 2
3 3 3
3
a b c
a b c
a b c
= 、
1 1 1
2 2 2
3 3 3
1
d b c
d b c
d b c
= 、
1 1 1
2 2 2
3 3 3
4
a d c
a d c
a d c
= 、
1 1 1
2 2 2
3 3 3
7
a b d
a b d
a b d
= ,
則x y z+ + = 4 。
9. 在
4
2 3 3
2 6
x y z
x y z
x y z
− + =⎧
⎪
+ + =⎨
⎪ − + =⎩
之中, 5Δ = − ,則 xΔ = 10− ,x = 2 。
10. 若 2 0
2 0
x y z
x y z
− + =⎧
⎨
+ − =⎩
,其中 0xyz ≠ ,則 : :x y z = 1:3:5 。
11. 甜在心水果店推出3種綜合水果禮盒:第一種每盒有3顆蘋果與3顆水梨,售價210 元;第二
種每盒有4 顆蘋果與4 顆石榴,售價260 元;第三種每盒有5顆水梨與5顆石榴,售價375元。
則蘋果、水梨、石榴每顆的售價分別為 30 40 35、 、 元。
實力測驗2
14. 第6 章 聯立方程式 133
6
一、基礎題
( B )1. 已知a 、b 為整數且3
4
5
a
b
= ,試求a b+ 的值? (A)11 (B)12 (C)13 (D)14。
( C )2. 試求行列式cos15 sin15
sin15 cos15
° °
=
° °
(A)0 (B)1
2
(C) 3
2
(D)1。
( D )3. 設 6
a b
c d
= ,則下列何者正確? (A) 6
a c
b d
= − (B) 6
a c
b d
−
= −
−
(C)
5 5
30
5 5
a c
b d
=
(D)
5
30
5
a a b
c c d
+
=
+
。
( A )4. 設 5
a b
c d
= ,則4 24
6
a b
c d
= (A)120 (B)100 (C)90 (D)80。
( B )5. 設 3
a b
c d
= ,則3 2 4
3 2 4
a b a
c d c
−
=
−
(A)12 (B)24 (C) 12− (D) 24− 。
( C )6. 試求行列式996 997
998 999
= (A)2 (B)0 (C) 2− (D) 4− 。
( A )7. 試求行列式
1 2 1
5 5 1
4 2 4
=
− −
(A)0 (B)10 (C)20 (D)40。
( D )8. 若
2 0 1
1 2 1
3 1 3
x− = ,則x = (A)5 (B)4 (C)3 (D)2。
( C )9. 設 3
a b c
x y z
p q r
= ,則
2 2 2
2 2 2
2 2 2
a b c
x y z
p q r
= (A)6 (B)12 (C)24 (D)48。
( A )10. 設 5
a b c
b c a
c a b
= ,則
a b c b c
a b c c a
a b c a b
+ +
+ + =
+ +
(A)5 (B)10 (C)15 (D)20。
二、進階題
( B )11. 設α 、β 為 2
2 1 0x x+ − = 的兩根,試求 1 1
1 1
α β
β α
+ +
=
− − +
(A)2 (B)4 (C)6 (D)8。
綜合實力評量
綜合實力評量
15. 134 第 6 章 聯立方程式
( D )12. 若
1 2
1 2
1 2
x
x
x
展開後為多項式 ( )f x ,則下列何者錯誤? (A) ( )deg 3f x = (B) ( )1 0f =
(C) ( )2 0f = (D) ( )5 0f = 。
( A )13. 若
1 2
2 1 0
1 2
x x x
x x x
x x x
+ +
+ + =
+ +
,則x = (A) 1− (B)0 (C)1 (D)2。
( A )14. 若
2 1
3 0 2 1
1 1
a
b
−
= ,則
2 1 1
3 0 4
1 1
a
b
− +
= (A) 2− (B) 1− (C)0 (D)1。
( D )15. 設 0xyz ≠ ,則
2
2
2
1
1
1
x x
y y
z z
= (A)3xyz (B)( )( )( )x y y z z x+ + +
(C)( )( )( )x y y z x z− − − (D)( )( )( )x y y z z x− − − 。
( D )16. 已知 2 3
2
x y
x y
α
β
= +⎧
⎨
= +⎩
,若令 x a b
y c d
α β
α β
= +⎧
⎨
= +⎩
,則b c+ = (A) 7− (B) 6− (C) 5− (D) 4− 。
( C )17. 若方程組
5 10
1
2 3
1
x y x y
x y x y
⎧
+ = −⎪ + −⎪
⎨
⎪ − =
⎪ + −⎩
的解為x α= ,y β= ,試求2α β+ = (A)1 (B)3 (C)5
(D)7。
( B )18. 甲、乙兩人同解方程組 4 5
2 5
ax y
x by
+ =⎧
⎨
+ =⎩
,甲看錯a ,解得 2x = , 1y = ;乙看錯b ,解得
1x = − , 2y = 。試問方程組的正確解為何? (A) 3x = , 1y = (B) 3x = , 1y = −
(C) 3x = − , 1y = (D) 3x = − , 1y = − 。
( B )19. 若
1
2
3
kx y z
x ky z
x y kz
+ + =⎧
⎪
+ + =⎨
⎪ + + =⎩
恰有一組解答,則 (A) 1k = 或 2k = − (B) 1k ≠ 且 2k ≠ −
(C) 2 1k− ≤ ≤ (D)k 為任何實數。
( C )20. 設 0xyz ≠ ,若 ( )
2
4 3 3 2 0x y z x y z− − + − − = ,則 2 2 2
xy yz zx
x y z
+ +
=
+ +
(A)1
7
(B)1
6
(C)1
5
(D)1
4
。
16. 第6 章 聯立方程式 135
6
( C )1.
1 3 5
3 5 1 0
5 1 3
x x x
x x x
x x x
+ + +
+ + + =
+ + +
,則 (A) 1x = (B) 1x = − (C) 3x = − (D) 5x = − 。
( A )2. 解聯立方程式
3 6
2 3 22
13 22
x y z
x y z
x y z
− + = −⎧
⎪
− − = −⎨
⎪ + − = −⎩
,可得y = (A)80 (B)104 (C)210 (D)240。
( D )3. 滿足
1 1 1
1 1 1 0
1 1 1
x
x
x
+
+ =
+
之所有x 解的和為 (A)0 (B) 1− (C) 2− (D) 3− 。
( D )4. 解方程組
1 1 2
3
1 1 5
6
1 1 5
6
x y y z
y z z x
z x x y
⎧
+ =⎪ + +⎪
⎪
+ =⎨
+ +⎪
⎪
+ =⎪
+ +⎩
,則x y z+ + = (A)1 (B)2 (C)3 (D)4。
( A )5. 設聯立方程組 2 3 1
4
x y
ax by
+ = −⎧
⎨
+ =⎩
與 3 2 5
2 1
x y
ax by
− =⎧
⎨
+ = −⎩
有共同解,求a b+ 之值為 (A) 14−
(B)14 (C)4 (D) 4− 。
( C )6. 若 1 1 1
2a b c
a b c
+ = + = + = ,則行列式
2
2
2
1
1
1
a b c
a b c
a b c
+
+
+
的值為 (A)0 (B) abc
(C)4abc (D)8abc 。
( A )7. 若
1 2
1 2 4 0
2 4 7
x
x
x
− =
−
,則x = (A) 1− (B)0 (C)1 (D)2。 【92 統測】
( A )8. 設
1 2 3
1 2 36
3 1
x
x
= 的解為a 與b ,則a b+ = (A) 4
3
(B)4 (C) 20
3
(D) 28
3
。【93 統測】
( D )9. 設k 為自然數,若行列式
1 2 3
1 2 3 0
1 2 3
k
k
k
−
− =
−
,則k = (A)3 (B)4 (C)5 (D)6。
【94 統測】
( B )10. 試求
1 0
2 1
2 3 4
1
1 1
x
x
x
x
− =
−
−
之解為何? (A) 2
7
(B) 2
7
− (C)7
2
(D) 7
2
− 。【94 統測補】
精選考題觀摩
17. 136 第 6 章 聯立方程式
( C )11. 設a ,b ,c 為實數,若
2
2
2
1
1 12
1
a a
b b
c c
= 且
3
3
3
1
1 156
1
a a
b b
c c
= ,則
( )
( )
( )
2
2
2
1 1 1
1 1 1
1 1 1
a a a
b b b
c c c
+ +
+ + =
+ +
(A)13
(B)144 (C)168 (D)1872。 【95 統測】
( C )12. 若
1
1 2
1
a d
b e
c f
= ,則
2 3 4
2 3 4
10 15 20
a d
b e
c f
−
− =
− −
(A)120 (B) 120− (C)240 (D) 240− 。
【96 統測】
( A )13. 行列式
1 10 20
5 50 1
10 1 5
= (A) 2
99− (B) 2
100− (C) 2
99 (D) 2
100 。 【97 統測】
( C )14. 若a,b 為方程式
2
9 5
1 2 7 2 0
3 1
x
x
x
+ = 的二根,則 2 2
a b+ = (A)9 (B)11 (C)13 (D)15。
【98 統測 B】
( A )15. 設α 、β 為
2
2 4 6
1 2 4 0
2 5 7
x
x
+ =
+
的兩個根,則α β+ = (A) 1
2
− (B)1
2
(C) 3
2
(D)5
2
。
【99 統測 B】
( D )16. 設二元一次方程組
3 7 11
3 7 11
x y
y x
− =⎧
⎨
− =⎩
,則其解為何? (A)無解 (B)無限多組解
(C) 6x = , 1y = (D) 11
4
x = − ,
11
4
y = − 。 【100 統測 B】
( B )17. 某餐廳有 A、 B 及C 三種套餐,今志志訂 2 個 A套餐,2 個 B 套餐,總共 2000 元;
敏敏訂 3 個 A套餐,1 個 B 套餐,總共 2400 元;耀耀訂 1 個 A套餐,1 個 B 套餐,2
個C 套餐,總共 3200 元。若訂 6 個 A套餐,4 個 B 套餐及 2 個C 套餐,則總共為多
少元? (A)7400 (B)7600 (C)7800 (D)8000。 【100 統測 B】
( D )18. 已知方程組
1 2 1 1
4 5 2
x y x y+ + − +
= = 的解為 ( ),a b ,求 a b− 之值為 (A) 2− (B) 1−
(C)0 (D)1。 【102 統測 B】
( C )19. 求 二 次 方 程 式
1 2 3
1 6 0
1 4
x
x
− − = 的 解 集 合 為 (A) { }1, 2 (B) { }1, 2− (C) { }1, 2−
(D){ }1, 2− − 。 【102 統測 B】
( B )20. 若三階行列式
13 16
11 14 17
12 15 18
x
之值為 3 ,則三階行列式
2 13 16
11 14 17
12 15 18
x +
之值為何?
(A) 9− (B) 3− (C)3 (D)9。 【102 統測 C】