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Chapter 1 Systems of Linear Equations 感感 感感感感 感感感感感 感感感感感 感感 Linear Algebra

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Page 1: Linear algebra

Chapter 1Systems of Linear Equations

感謝 大葉大學 資訊工程系黃鈴玲老師 提供

Linear Algebra

Page 2: Linear algebra

Ch1_2

Definition• An equation such as x+3y=9 is called a linear equation.

( 線性方程式 )

• The graph of this equation is a straight line in the x-y plane.

• A pair of values of x and y that satisfy the equation is called a solution.

1.1 Matrices and Systems of Linear Equations

system of linear equations ( 線性聯立方程式 )

Page 3: Linear algebra

Ch1_3

Definition A linear equation in n variables ( 變數 ) x1, x2, x3, …, xn has

the form

a1 x1 + a2 x2 + a3 x3 + … + an xn = b

where the coefficients ( 係數 ) a1, a2, a3, …, an and b are real numbers ( 實數 ).

常見數系的英文名稱: natural number ( 自然數 ), integer ( 整數 ), rational number ( 有理數 ), real number ( 實數 ), complex number ( 複數 ) positive ( 正 ), negative ( 負 )

Page 4: Linear algebra

Ch1_4

Figure 1.2No solution ( 無解 )–2x + y = 3–4x + 2y = 2 Lines are parallel.No point of intersection. No solutions.

Solutions for system of linear equations

Figure 1.1Unique solution ( 唯一解 ) x + 3y = 9–2x + y = –4 Lines intersect at (3, 2)Unique solution: x = 3, y = 2.

Figure 1.3Many solution ( 無限多解 )4x – 2y = 66x – 3y = 9 Both equations have the same graph. Any point on the graph is a solution. Many solutions.

Page 5: Linear algebra

Ch1_5

A linear equation in three variables corresponds to a plane in three-dimensional ( 三維 ) space.

Unique solution

※ Systems of three linear equations in three variables:

Page 6: Linear algebra

Ch1_6

No solutions

Many solutions

Page 7: Linear algebra

Ch1_7

How to solve a system of linear equations?

Gauss-Jordan elimination. ( 高斯 - 喬登消去法 )

1.2 節會介紹

Page 8: Linear algebra

Ch1_8

Definition• A matrix ( 矩陣 ) is a rectangular array of numbers. • The numbers in the array are called the elements ( 元素 ) of

the matrix.

Matrices

1298

520

653

C

38

50

17

B 157

432A

注意矩陣左右兩邊是中括號不是直線,直線表示的是行列式。

Page 9: Linear algebra

Ch1_9

Submatrix (子矩陣 )

A matrix

215

032

471

A

Row (列 ) and Column (行 )

3column 2column 1column 2 row 1 row

1

4

5

3

7

2 157 432

157

432

A

A of ssubmatrice

25

41

1

3

7

15

32

71

RQP

Page 10: Linear algebra

Ch1_10

Identity Matrices (單位矩陣 ) diagonal ( 對角線 ) 上都是 1 ,其餘都是 0 , I 的下標表示 size

100010001

1001

32 II

Location

7 ,4 157

4322113

aaAaij 表示在 row i, column j 的元素值也寫成 location (1,3) = 4

Size and Type

matrixcolumn a matrix row a matrix square a

matrix 13 matrix 41 matrix 33 32 :Size

2

3

8

5834

853

109

752

542

301

Page 11: Linear algebra

Ch1_11

matrix of coefficient and augmented matrix

62

3 32

2

321

321

321

xxx

xxx

xxx

Relations between system of linear equations and matrices

係數矩陣 擴大矩陣

隨堂作業: 5(f)

tcoefficien ofmatrix

211

132

111

matrix augmented

6211

3132

2111

Page 12: Linear algebra

Ch1_12

給定聯立方程式後,不會改變解的一些轉換

Elementary Transformation

1. Interchange two equations.

2. Multiply both sides of an equation by a nonzero constant.

3. Add a multiple of one equation to another equation.

將左邊的轉換對應到矩陣上Elementary Row Operation ( 基本列運算 )

1. Interchange two rows of a matrix. ( 兩列交換 )

2. Multiply the elements of a row by a nonzero constant. ( 某列的元素同乘一非零常數 )

3. Add a multiple of the elements of one row to the corresponding elements of another row. ( 將一個列的倍數加進另一列裡 )

Elementary Row Operations of Matrices

Page 13: Linear algebra

Ch1_13

Example 1Solving the following system of linear equation.

623322

321

321

321

xxxxxxxxx

621131322111

623322

321

321

321

xxxxxxxxx

SolutionEquation MethodInitial system:

Analogous Matrix MethodAugmented matrix:

1

2

32

321

xx

xxx

Eq2+(–2)Eq1

Eq3+(–1)Eq1

832011102111

R2+(–2)R1R3+(–1)R1

符號表示 row equivalent

832 32 xx

Page 14: Linear algebra

Ch1_14

1051

32

3

32

31

xxxxx

0150011103201

21

32

3

32

31

xxxxx

210011103201

21

1

3

2

1

xxx

2100

1010

1001

Eq1+(–1)Eq2

Eq3+(2)Eq2

(–1/5)Eq3

Eq1+(–2)Eq3

Eq2+Eq3

The solution is .2 ,1 ,1 321 xxx The solution is

.2 ,1 ,1 321 xxx

8321

2

32

32

321

xxxxxxx

832011102111

R1+(–1)R2 R3+(2)R2

(–1/5)R3

R1+(–2)R3 R2+R3

隨堂作業: 7(d)

Page 15: Linear algebra

Ch1_15

Example 2Solving the following system of linear equation.

83318521242

321

321

321

xxxxxxxxx

Solution (請先自行練習 )

83311851212421

41106330

12421

R23

1

41102110

12421

620021108201

3100

2110

8201

R1R3

2)R1(R2

R2)1(R3

(2)R2R1

R32

1

3100

1010

2001

R3R2

2)R3(R1

.

3

1

2

solution

3

2

1

x

x

x

Page 16: Linear algebra

Ch1_16

Example 3Solve the system

7 2

328 63

441284

21

321

321

xx

xxx

xxx

7012

32863

441284

7012

32863

11321

15630

1100

11321

1100

5210

11321

1100

5210

1101

.

1100

3010

2001

.1 ,3 ,2 issolution The 321 xxx

R23

1

R12R3

3)R1(R2

R3R2

1100

15630

11321

R14

1

2)R2(R1

2R3R2

1)R3(R1

Solution (請先自行練習 )

隨堂作業: 10(d)(f)

Page 17: Linear algebra

Ch1_17

Summary

7012

32863

441284

]:[ BA

A BUse row operations to [A: B] :

.

1100

3010

2001

7012

32863

441284]:[]:[ XIBA ni.e.,

Def. [In : X] is called the reduced echelon form ( 簡化梯式 ) of [A : B].Note. 1. If A is the matrix of coefficients of a system of n equations

in n variables that has a unique solution, then A is row equivalent to In (A In).

2. If A In, then the system has unique solution.

7 2

328 63

441284

21

321

321

xx

xxx

xxx

Page 18: Linear algebra

Ch1_18

Example 4 Many SystemsSolving the following three systems of linear equation, all of which have the same matrix of coefficients.

3321

2321

1321

42

for 42

3

bxxx

bxxx

bxxx

in turn 433

,210

,11118

3

2

1

bbb

Solution

42114213111412308311

.

2

1

2

,

1

3

0

,

2

1

1

3

2

1

3

2

1

3

2

1

x

x

x

x

x

x

x

x

x

123110315210308311

212100

315210

013101

212100

131010

201001

R2+(–2)R1 R3+R1

1)R2(R3

R2R1

R32R2

R3)1(R1

隨堂作業: 13(b)The solutions to the three systems are

Page 19: Linear algebra

Ch1_19

Homework

Exercise 1.1 :1, 2, 4, 5, 6, 7, 10, 13

Page 20: Linear algebra

Ch1_20

1-2 Gauss-Jordan EliminationDefinitionA matrix is in reduced echelon form ( 簡化梯式 ) if

1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix.

2. The first nonzero element of each other row is 1. This element is called a leading 1.

3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row.

4. All other elements in a column that contains a leading 1 are zero.

Page 21: Linear algebra

Ch1_21

Examples for reduced echelon form

10000

04300

03021

9100

3010

7001

3100

0000

4021

000

210

801

() ()() ()

利用一連串的 elementary row operations ,可讓任何矩陣變成 reduced echelon form

The reduced echelon form of a matrix is unique.

隨堂作業: 2(b)(d)(h)

Page 22: Linear algebra

Ch1_22

Gauss-Jordan Elimination

System of linear equations augmented matrix reduced echelon from solution

Page 23: Linear algebra

Ch1_23

41200

22200

43111

4)R1(R3

610001110052011

R2)2(R3R2R1

Example 1Use the method of Gauss-Jordan elimination to find reduced echelon form of the following matrix.

1211244129333

22200

Solution

1211244

22200

129333

R2R1

pivot ( 樞軸,未來的 leading 1)

12112442220043111

R13

1

pivot

6100050100

170011

R3R22)R3(R1

The matrix is the reduced echelon form of the given matrix.

41200

11100

43111

R221

Page 24: Linear algebra

Ch1_24

Example 2Solve, if possible, the system of equations

7537429333

321

321

321

xxxxxxxxx

Solution (請先自行練習 )

715374123111

715374129333

R131

242012103111

3)R1(R3

2)R1(R2

000012104301

R2R3

R2R11243

1243

32

31

32

31

xxxx

xxxx

The general solution to the system is

.parameter) a (callednumber real is where,

12

43

3

2

1

rrx

rx

rx

隨堂作業: 5(c)

Page 25: Linear algebra

Ch1_25

Example 3Solve the system of equations

44694214232

58101242

4321

4321

4321

xxxxxxxxxxxx

Solution(自行練習 )

446942142321295621

44694214232158101242

R121

144300153300295621

2)R1(R3

R1R2

14430051100

295621

R231

110005110011021

3R2R3

6)R2(R1

110006010020021

R3R2

1)R3(R1

. somefor ,

1

6

22

1

6

22

4

3

2

1

4

3

21

r

x

x

rx

rx

x

x

xx

變數個數 >方程式個數 many sol.

Page 26: Linear algebra

Ch1_26

Example 4Solve the system of equations

432636242232

54321

54321

54321

xxxxxxxxxxxxxxx

Solution(自行練習 )

642000210000213121

431121636242213121

R1R3

2)R1(R2

210000

642000213121

R3R2

210000321000213121

2R21

210000

321000750121

3)R2(R1

210000101000300121

2)R3(R2

5R3R1

. and somefor ,2

,1 , ,32

21

32

5

432

1

5

4

321

srx

xsxrxsrx

xx

xxx

Page 27: Linear algebra

Ch1_27

Example 5This example illustrates a system that has no solution. Let us try to solve the system

12232

452232

32

321

321

321

xxxxxxxxxxx

Solution(自行練習 )

1220210063303211

R3R2

1220210021103211

R231

5400210021101101

2)R2(R4

R2R1

13000210040103001

4)R3R(R4

R3R2

1)R3(R1

1000210040103001

R4131

The system has no solution.

1220633021003211

1220312145223211

1)R1(R3

2)R1(R2

0x1+0x2+0x3=1隨堂作業: 5(d)

Page 28: Linear algebra

Ch1_28

Homogeneous System of linear Equations

Definition A system of linear equations is said to be homogeneous ( 齊

次 ) if all the constant terms ( 等號右邊的常數項 ) are zeros.

Example:

0632

052

321

321

xxx

xxx

Observe that is a solution. 0 ,0 ,0 321 xxx

Theorem 1.1

A system of homogeneous linear equations in n variables always has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called the trivial solution.

Page 29: Linear algebra

Ch1_29

Homogeneous System of linear Equations

Theorem 1.2

A system of homogeneous linear equations that has more variables than equations has many solutions.

Note. 除 trivial solution 外,可能還有其他解。

隨堂作業: 8(e)

0410

0301

0632

0521

0632

052

321

321

xxx

xxx

The system has other nontrivial solutions.

rxrxrx 321 ,4 ,3

Example:

Page 30: Linear algebra

Ch1_30

Homework

Exercise 1.2:2, 5, 8, 14

(Section 1.3 跳過 )