Transcript Document

CHAPTER 1
SYSTEMS OF LINEAR
EQUATIONS
1.1 Introduction to Systems of Linear Equations
1.2 Gaussian Elimination and Gauss-Jordan Elimination
1.3 Applications of Systems of Linear Equations
Elementary Linear Algebra
R. Larson (7 Edition)
投影片設計製作者
淡江大學 電機系 翁慶昌 教授
CH 1 Linear Algebra Applied
Balancing Chemical Equations (p.4)
Global Positioning System (p.16)
Airspeed of a Plane (p.11)
Traffic Flow (p.28)
Electrical Network Analysis (p.30)
2/46
1.1 Introduction to Systems of Linear Equations

a linear equation in n variables:
a1,a2,a3,…,an, b: real number
a1: leading coefficient
x1: leading variable

Notes:
(1) Linear equations have no products or roots of variables and
no variables involved in trigonometric, exponential, or
logarithmic functions.
(2) Variables appear only to the first power.
Elementary Linear Algebra: Section 1.1, p.2
3/46

Ex 1: (Linear or Nonlinear)
1
x  y  z 
Linear ( a ) 3 x  2 y  7
(b)
Linear ( c ) x1  2 x 2  10 x 3  x 4  0
( d ) ( sin
2
Linear
2

2
) x1  4 x 2  e
2
Linear
Exponentia l
Nonlinear ( e ) xy  z  2
( f )e 2y  4
x
Nonlinear
not the first power
Nonlinear ( g ) sin x1  2 x 2  3 x 3  0
trigonomet ric functions
Elementary Linear Algebra: Section 1.1, p.2
(h )
1
x

1
y
4
Nonlinear
not the first power
4/46

a solution of a linear equation in n variables:
a 1 x1  a 2 x 2  a 3 x 3    a n x n  b
x 1  s1 , x 2  s 2 , x 3  s 3 ,  , x n  s n
such a 1 s1  a 2 s 2  a 3 s 3    a n s n  b
that

Solution set:
the set of all solutions of a linear equation
Elementary Linear Algebra: Section 1.1, p.3
5/46

Ex 2： (Parametric representation of a solution set)
x1  2 x 2  4
a solution: (2, 1), i.e. x1  2 , x 2  1
If you solve for x1 in terms of x2, you obtain
x1  4  2 x 2 ,
By letting x 2  t you can represent the solution set as
x1  4  2 t
And the solutions are ( 4  2 t , t ) | t  R  or ( s , 2  12 s ) | s  R 
Elementary Linear Algebra: Section 1.1, p.3
6/46

a system of m linear equations in n variables:
a 11 x1

a 12 x 2

a 13 x 3



a1 n x n

b1
a 21 x1

a 22 x 2

a 23 x 3



a2n xn

b2
a 31 x1

a 32 x 2

a 33 x 3



a3n xn

b3
a m 3 x3



a mn x n

bm

a m 1 x1


am 2 x2

Consistent:
A system of linear equations has at least one solution.

Inconsistent:
A system of linear equations has no solution.
Elementary Linear Algebra: Section 1.1, pp.4-5
7/46

Notes:
Every system of linear equations has either
(1) exactly one solution,
(2) infinitely many solutions, or
(3) no solution.
Elementary Linear Algebra: Section 1.1, p.5
8/46

Ex 4: (Solution of a system of linear equations)
(1)
x 
x 
y  3
y  1
exactly one solution
two intersecti ng lines
(2)
x  y  3
2x  2 y  6
two coincident
(3)
x 
x 
inifinite number
lines
y  3
y  1
no solution
two parallel lines
Elementary Linear Algebra: Section 1.1, p.5
9/46

Ex 5: (Using back substitution to solve a system in row echelon form)
x  2y 
5
y  2
(1)
(2)
Sol: By substituting y   2 into (1), you obtain
x  2(2)  5
x  1
The system has exactly one solution: x  1, y   2
Elementary Linear Algebra: Section 1.1, p.6
10/46

Ex 6: (Using back substitution to solve a system in row echelon form)
x  2 y  3z  9
y  3z  5
z  2
(1)
(2)
(3)
Sol: Substitute z  2 into (2)
y  3( 2 ) 
5
y  1
and substitute y   1 and z  2 into (1)
x  2 (  1)  3 ( 2 )  9
x  1
The system has exactly one solution:
x  1, y   1, z  2
Elementary Linear Algebra: Section 1.1, p.6
11/46

Equivalent:
Two systems of linear equations are called equivalent
if they have precisely the same solution set.

Notes:
Each of the following operations on a system of linear
equations produces an equivalent system.
(1) Interchange two equations.
(2) Multiply an equation by a nonzero constant.
(3) Add a multiple of an equation to another equation.
Elementary Linear Algebra: Section 1.1, pp.6-7
12/46

Ex 7: Solve a system of linear equations (consistent system)
x  2 y  3z 
9
 x  3y
 4
2 x  5 y  5 z  17
Sol:
(1)
(2)
(3)
(1)  (2)  (2)
x  2 y  3z  9
y  3z  5
2 x  5 y  5 z  17
(4)
(1)  (  2)  (3)  (3)
x 
2 y  3z 
9
y  3z 
5
 y 
z  1
Elementary Linear Algebra: Section 1.1, p.7
(5)
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(4)  (5)  (5)
x  2y 
y 
3z  9
3z  5
2z  4
(6)
(6)  12  ( 6 )
x  2 y  3z  9
y  3z  5
z  2
So the solution is x  1, y   1, z  2 (only one solution)
Elementary Linear Algebra: Section 1.1, p.7
14/46

Ex 8: Solve a system of linear equations (inconsistent system)
x1  3 x 2
2 x1 
x2
x1  2 x 2

x3 
1
 2 x3 
2
 3 x3   1
(1)
(2)
(3)
Sol: (1)  (  2)  (2)  (2)
(1)  (  1)  (3)  (3)
x1  3 x 2
5 x2
5 x2

x3 
1
 4 x3 
0
 4 x3   2
Elementary Linear Algebra: Section 1.1, p.8
(4)
(5)
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( 4 )  (  1)  ( 5 )  ( 5 )
x1  3 x 2
5 x2

x3 
1
 4 x3 
0
0  2
(a false statement)
So the system has no solution (an inconsistent system).
Elementary Linear Algebra: Section 1.1, p.8
16/46

Ex 9: Solve a system of linear equations (infinitely many solutions)
x2
x1
 x1  3 x 2

x3 
0
 3 x3   1

1
(1)
(2)
(3)
 3 x3   1

x3 
0

1
(1)
(2)
(3)
Sol: (1)  ( 2 )
x1
 x1
x2
 3x2
(1)  (3)  (3)
x1
x2
3x2
 3 x3   1

x3 
0
 3 x3 
0
Elementary Linear Algebra: Section 1.1, p.9
(4)
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x1
x2
 x2  x3 ,

3 x3

1

x3

0
x1   1  3 x 3
let x 3  t
then x 1  3 t  1,
x2  t,
t R
x3  t,
So this system has infinitely many solutions.
Elementary Linear Algebra: Section 1.1, p.9
18/46
Key Learning in Section 1.1

Recognize a linear equation in n variables.

Find a parametric representation of a solution set.


Determine whether a system of linear equations is consistent or
inconsistent.
Use back-substitution and Gaussian elimination to solve a
system of linear equations.
19/46
Keywords in Section 1.1

linear equation: 線性方程式

system of linear equations: 線性方程式系統

leading coefficient: 領先係數

leading variable: 領先變數

solution: 解

solution set: 解集合

parametric representation: 參數化表示

consistent: 一致性(有解)

inconsistent: 非一致性(無解、矛盾)

equivalent: 等價
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1.2 Gaussian Elimination and Gauss-Jordan Elimination

mn matrix:
 a 11 a 12 a 13
 a 21 a 22 a 23
a
a 32 a 33
31



a m1 a m 2 a m 3
a1n 
a2n 
a 3n 


 a mn 



m rows
n columns

Notes:
(1) Every entry aij in a matrix is a number.
(2) A matrix with m rows and n columns is said to be of size mn .
(3) If m  n , then the matrix is called square of order n.
(4) For a square matrix, the entries a11, a22, …, ann are called
the main diagonal entries.
Elementary Linear Algebra: Section 1.2, p.13
21/46

Ex 1:
Matrix
[2 ]
11
0 0
 0 0 
22
1

1 3 0

2 
1 4
 e
 2
  7

Size
 
2
4 
3 2
Note:
One very common use of matrices is to represent a system
of linear equations.
Elementary Linear Algebra: Section 1.2, p.13
22/46

a system of m equations in n variables:
a 11 x1

a 12 x 2

a 13 x 3



a1 n x n

b1
a 21 x1

a 22 x 2

a 23 x 3



a2n xn

b2
a 31 x1

a 32 x 2

a 33 x 3



a3n xn

b3



a mn x n

bm

a m 1 x1

am 2 x2

a m 3 x3
Ax  b
Matrix form:
 a 11 a 12 a 13
 a 21 a 22 a 23
A   a 31 a 32 a 33



a m1 a m 2 a m 3
a1n 
a2n 
a 3n 


 a mn 



Elementary Linear Algebra: Section 1.2, p.13
 x1 
 x2 
x
  
x 
 n
 b1 
 b2 
b
  
b 
 m
23/46

Augmented matrix:
 a 11 a 12 a 13
 a 21 a 22 a 23
a
a 32 a 33
31



a m1 a m 2 a m 3




a1n
a2n
a 3n
 a mn
b1 
b2 
b3   [ A b ]


bm 
Coefficient matrix:
 a 11 a 12 a 13
 a 21 a 22 a 23
a
a 32 a 33
31



a m1 a m 2 a m 3
a1n 
a2n 
a 3n   A


 a mn 



Elementary Linear Algebra: Section 1.2, p.13
24/46

Elementary row operation:
(1) Interchange two rows.
rij : R i  R j
(2) Multiply a row by a nonzero constant.
ri
(k )
: ( k ) Ri  Ri
(3) Add a multiple of a row to another row. rij( k ) : ( k ) R i  R j  R j

Row equivalent:
Two matrices are said to be row equivalent if one can be obtained
from the other by a finite sequence of elementary row operation.
Elementary Linear Algebra: Section 1.2, p.14
25/46

Ex 2: (Elementary row operation)
1 3 4
 0
2 0 3
 1
 2  3 4 1
6  2
2  4
3 3
0
1
1
2 
 5  2
3
1 2  4
 0 3  2  1
5  2 
 2 1
2 0 3
 1
1 3 4
 0
 2  3 4 1
r12
( 12 )
3  1
1  2
3 3
0
1
1
2 
 5  2
( 2 )
2 4
3
1
3  2  1
0
 0  3 13  8 
r1
r13
Elementary Linear Algebra: Section 1.2, p.14
26/46
26/9

Ex 3: Using elementary row operations to solve a system
Associated
Augemented Matrix
Linear System
x

2y

x

3y
2x

5y

x

2y


9

4
5z

17
3z

9
3z
y

3z

5
2x

5y

5z

17
x

2y

3z

9
y

3z

5
y

z

1

Elementary Linear Algebra: Section 1.2, p.15
 1

1

 2
2
3
3
0
5
5
1

0

 2
2
3
1

0

 0
1
3
5
5
2
3
1
3
1
1
Elementary
Row Operation
9

4

17 
9

5

17 
9

5

 1
r12 : (1) R1  R 2  R 2
(1 )
( 2 )
r13
: (  2 ) R1  R 3  R 3
27/46
Linear System
x
x


2y

3z

9
y

3z

5
2z

4
3z

9
2y
y


3z

5
z

2
x
y
Elementary
Row Operation
Associated
Augemented Matrix
1

0

 0
2
1

0

 0
3
1
3
0
2
2
3
1
3
0
1
9

5

4 
9

5

2 
r23 : (1) R 2  R 3  R 3
(1 )
1
( )
2
3
r
1
: ( ) R3  R3
2

1
 1
z 
2
Elementary Linear Algebra: Section 1.2, p.15
28/46

Row-echelon form: (1, 2, 3)

Reduced row-echelon form: (1, 2, 3, 4)
(1) All row consisting entirely of zeros occur at the bottom
of the matrix.
(2) For each row that does not consist entirely of zeros,
the first nonzero entry is 1 (called a leading 1).
(3) For two successive (nonzero) rows, the leading 1 in the higher
row is farther to the left than the leading 1 in the lower row.
(4) Every column that has a leading 1 has zeros in every position
above and below its leading 1.
Elementary Linear Algebra: Section 1.2, p.15
29/46

Ex 4: (Row-echelon form or reduced row-echelon form)
4
1 2  1
0
3
0 1
1  2 
 0 0
(row - echelon
form)
0 1 0 5
 0 0 1 3
 0 0 0 0 
0  1
0
2
1
3
0
0 
3
1  5 2  1
0 1
3  2  (row - echelon
0
0 0
1
4  form)
0
0 0
0
1
 0
1
0
0
 0
4
1 2  3
1  1
0 2
1  3 
 0 0
2
1 2  1
0
0
0 0
2  4 
 0 1
Elementary Linear Algebra: Section 1.2, p.16
0
1
0
0
(reduced row echelon form)
(reduced row echelon form)
30/46

Gaussian elimination:
The procedure for reducing a matrix to a row-echelon form.

Gauss-Jordan elimination:
The procedure for reducing a matrix to a reduced row-echelon
form.

Notes:
(1) Every matrix has an unique reduced row echelon form.
(2) A row-echelon form of a given matrix is not unique.
(Different sequences of row operations can produce
different row-echelon forms.)
Elementary Linear Algebra: Section 1.2, pp.17-19
31/46
Ex: (Procedure of Gaussian elimination and Gauss-Jordan elimination)

Produce leading 1
0

2

 2
0
2
0
8
8
6
4
12
4
5
6
5
12 

28

4 
r12
2

0

 2
8
6
4
12
0
2
0
8
4
5
6
5
28 

12

4 
The first nonzero column
( 12 )
r1
leading 1
1

0

 2
4
3
2
6
0
2
0
8
4
5
6
5
14 

12

4 
( 2 )
r13
Zeros elements below leading 1
Elementary Linear Algebra: Section 1.2, Addition
1

0

 0
4
3
0
0
Produce leading 1
2
6
2
0
8
5
0
 17
14 

12

 24 
The first nonzero Submatrix
column
32/46
leading 1
1

0

 0
(  12 )
r2
4
3
2
6
0
1
0
4
0
5
0
 17
14 

6

 24 
( 5 )
r23
1

0

 0
4
3
2
6
0
1
0
4
0
0
0
3
Submatrix
Zeros elements below leading 1
Produce leading 1
Zeros elsewhere
(
1
3
r3
)
1

0

 0
4
3
2
6
0
1
0
4
0
0
0
1
14 

6

2 
(row - echelon form)
(4)
r32
1

0

 0
4
3
2
0
0
1
0
0
0
0
0
1
( 6 )
r31
1

0

 0
leading 1
2

2

2 
(row - echelon form)
Elementary Linear Algebra: Section 1.2, Addition
14 

6

6 
4
3
2
0
0
1
0
4
0
0
0
1
2

6

2 
(row - echelon form)
(3)
r21
1

0

 0
4
0
2
0
0
1
0
0
0
0
0
1
8

2

2 
(reduced row - echelon form)
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Ex 7: Solve a system by Gauss-Jordan elimination method
(only one solution)

x  2 y  3z 
9
 x  3y
 4
2 x  5 y  5 z  17
Sol:
augmented
matrix
9  r (1 ) , r (  2 )
 1 2 3
12
13
3 0  4
 1
 2  5 5 17 
(
r3
1
2
)
(2)
( 3)
 1  2 3 9  r21 , r32
1 3 5
0
0 1 2 
 0
( 9 )
, r31
(row - echelon form)
Elementary Linear Algebra: Section 1.2, p.19
1

0

 0
2
3
1
3
1
1
9

5

 1
(1 )
r23
1
1 0 0
 0 1 0  1
2 
 0 0 1
1

0

 0
2
3
1
3
0
2
x
y
9

5

4 

1
 1
z 
2
(reduced row - echelon form)
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
Ex 8：Solve a system by Gauss-Jordan elimination method
(infinitely many solutions)
2 x1  4 x1  2 x 3  0
3 x1  5 x 2
 1
Sol: augmented
matrix
2 4  2 0
0 1
 3 5
( 12 )
( 3)
r1 , r12
( 1)
, r2
( 2 )
, r21
5
2
1 0
 0 1  3  1
the correspond ing system of equations
x1

5 x3 
2
x2
 3 x3   1
leading
(reduced row echelon form)
is
variable ：x1 , x 2
free variable ： x 3
Elementary Linear Algebra: Section 1.2, p.20
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x1 
2  5 x3
x2   1  3 x3
Let x 3  t
x1  2  5 t ,
x 2   1  3t ,
t R
x3  t,
So this system has infinitely many solutions.
Elementary Linear Algebra: Section 1.2, p.20
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
Homogeneous systems of linear equations:
A system of linear equations is said to be homogeneous
if all the constant terms are zero.
a 11 x 1 
a 21 x 1 
a 31 x 1 
a 12 x 2  a 13 x 3
a 22 x 2  a 23 x 3
a 32 x 2  a 33 x 3

a m 1 x1  a m 2 x 2  a m 3 x 3
Elementary Linear Algebra: Section 1.2, p.21
  
  
  
a1n x n  0
a2n xn  0
a 3n x n  0
   a mn x n  0
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
Trivial solution:
x1  x 2  x 3    x n  0


Nontrivial solution:
other solutions
Notes:
(1) Every homogeneous system of linear equations is consistent.
(2) If the homogenous system has fewer equations than variables,
then it must have an infinite number of solutions.
(3) For a homogeneous system, exactly one of the following is true.
(a) The system has only the trivial solution.
(b) The system has infinitely many nontrivial solutions in
addition to the trivial solution.
Elementary Linear Algebra: Section 1.2, p.21
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
Ex 9: Solve the following homogeneous system
Sol:
x1

x2

3 x3

0
2 x1

x2

3 x3

0
augmented
matrix
 1  1 3 0
1 3 0 
 2
leading
( 2 )
12
r
( 13 )
(1 )
, r2 , r21
2 0
1 0
 0 1  1 0 
(reduced row echelon form)
variable ：x1 , x 2
free variable ： x 3
Let x 3  t
x1   2 t , x 2  t , x 3  t , t  R
When t  0 , x1  x 2  x 3  0 (trivial solution)
Elementary Linear Algebra: Section 1.2, p.21
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Key Learning in Section 1.2





Determine the size of a matrix .
Write an augmented or coefficient matrix from a system of
linear equations.
Use matrices and Gaussian elimination with back-substitution to
solve a system of linear equations.
Use matrices and Gauss-Jordan elimination to solve a system of
linear equations.
Solve a homogeneous system of linear equations.
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Key Learning in Section 1.3


Set up and solve a system of equations to fit a polynomial
function to a set of data points.
Set up and solve a system of equations to represent a network.
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Keywords in Section 1.2

matrix: 矩陣

row: 列

column: 行

entry: 元素

size: 大小

square matrix: 方陣

order: 階

main diagonal: 主對角線

augmented matrix: 增廣矩陣

coefficient matrix: 係數矩陣
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Keywords in Section 1.2

elementary row operation: 基本列運算

row equivalent: 列等價

row-echelon form: 列梯形形式

reduced row-echelon form: 列簡梯形形式

leading 1: 領先1

Gaussian elimination: 高斯消去法

Gauss-Jordan elimination: 高斯-喬登消去法

free variable: 自由變數

leading variable: 領先變數

homogeneous system: 齊次系統

trivial solution: 顯然解

nontrivial solution: 非顯然解
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1.1 Linear Algebra Applied

Balancing Chemical Equations
In a chemical reaction, atoms reorganize in one or more
substances. For instance, when methane gas(CH4)
combines with oxygen(O2) and burns, carbon dioxide
(CO2) and water (H2O) form. Chemists represent this
process by a chemical equation of the form
(x1)CH4 + (x2)O2 → (x3)CO2 + (x4)H2O
Because a chemical reaction can neither create nor
destroy atoms, all of the atoms represented on the left
side of the arrow must be accounted for on the right side
of the arrow. This is called balancing the chemical
equation. In the given example, chemists can use a
system of linear equations to find values of x1, x2, x3,
and x4 that will balance the chemical equation.
Elementary Linear Algebra: Section 1.1, p.4
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1.2 Linear Algebra Applied

Global Positioning System
The Global Positioning System (GPS) is a network of 24
satellites originally developed by the U.S. military as a
navigational tool. Today, GPS receivers are used in a
wide variety of civilian applications, such as determining
directions, locating vessels lost at sea, and monitoring
earthquakes. A GPS receiver works by using satellite
readings to calculate its location. In three dimensions, the
receiver uses signals from at least four satellites to
“trilaterate” its position. In a simplified mathematical
model, a system of three linear equations in four
unknowns (three dimensions and time) is used to
determine the coordinates of the receiver as functions of
time.
Elementary Linear Algebra: Section 1.2, p.16
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1.3 Linear Algebra Applied

Traffic Flow
Researchers in Italy studying the acoustical noise levels
from vehicular traffic at a busy three-way intersection
on a college campus used a system of linear equations to
model the traffic flow at the intersection. To help
formulate the system of equations, “operators” stationed
themselves at various locations along the intersection
and counted the numbers of vehicles going by.
Elementary Linear Algebra: Section 1.3, p.28
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