Lecture 2 Linear System

8/2/2019 Lecture 2 Linear System

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Lecture 2

LINEAR SYSTEM OF EQUATIONS

Learning outcomes: by the end of this lecture

1. You should know,

a)What is a linear system of equationsb)What is a homogeneous systemc)How to represent a linear system in matrix formd)What is a coefficient matrixe)What is an augmented matrix

2. You should be able tosolve a linear system of

equations using:

a)Inverse matrix methodb) Row operations:

i.Gauss-elimination method (REF)ii.Gauss-Jordan method (RREF)


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Definition of a Linear Equation inn Variables:

A linear equation inn variable nxxx ,,, 21 has the

formbxaxaxa nn 2211 ,

where the coefficients baaa n ,,,, 21 are real numbers

(usually known). The number of 1a is the leading

coefficient and 1x is the leading variable.

The collection ofseverallinear equations isreferred to asthe system of linear equations.

Definition of System ofm Linear Equation in

n Variables:

A system ofm linear equations inn variables

is a set of m equations, each of which is linear

in the same n variables:

mnmnmm

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

where ,,1,,,2,1,, njmiba iij are constants.

Example:1Consider the following system of linear

equations:


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3

67

832

3,42523

43

31

321

321

321

xx

xxx

nmxxx

xxx

Example: 2 Which of the following are linear

equations?2

1 2 3

1 2 3

( ) 3 2 7 ( ) (sin ) 4 (log5) ( ) 23

1

( ) 2 4 ( ) sin 2 3 0 ( ) 4

x

a x y b x x x e c x y

d e y e x x x f x y

( ) and ( ) are linear equations.a b ( ), ( ), ( ), and ( ) are not linear.c d e f

Number of Solutions of a System of LinearEquations

Consider the following systems of linear equations

(a)1

3

x y

x y

(b)4

2

x y

x y

(c)6 2 8

3 4

x y

x y

For a system of linear equations, precisely one of

the following is true:

(a) The system has exactly one solution.(b)The system has no solution.(c) The system has infinitely many solutions. Consistent and InconsistentA system of linear equations is called consistentif

it has at least one solution and inconsistentif it has

no solution.

)sol

1x y

3x y 2x y

4x y


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EquivalentTwo systems of linear equations are said to be

equivalentif they have the same set of solutions.

BackSubstitutionWhich of the following systems is easier to solve?

2 3 9

( ) 3 7 6 22

2 5 5 17

x y z

a x y z

x y z

2 3 9

( ) 3 5

2

x y z

b y z

z

System (b) is said to be in row-echelon form. To

solve such a system, use a procedure called back

substitution.

Augmented Matrices and CoefficientMatrices

Consider the m n linear system11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

...

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

Let

11 12 1

21 22

1 1

2

11 12 1

21 22 2

2 1 2

22

1

, , |

n

n

n

n

m m m m mm m mn n

b b

b

a a a

a a aA

a a a

a a a

a a

b B A b

a ba

b

aba

A is called thecoefficient matrix of the system.

B is called theaugmentedmatrix of the system.

b is called the constant matrix of the system.

Itis possible to write the system


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11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

...

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

in the following matrix form

BXA

b

b

b

m

2

1

2

1

21

22221

11211

nmnmm

n

n

x

x

x

aaa

aaa

aaa

Example:65y-2

13

x

yx BXA

6

1

y

x

52

13

Row-EquivalentTwo m n matrices are said to be row-equivalent ifone can be obtained by the other by a series of

elementary row operations.

Now we are in the stage to tackle the question.

How to solve a linear system AX = B?

First: Row operations

The key to solve a system of linear equations is to

transform the original augmented matrix to some

matrix with some properties via a few elementary

row operations. As a matter of fact, we can solve

any system of linear equations by transforming


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the associate augmented matrix to a matrix in

some form.

The form is referred to as the reduced row

echelon form.

1.1 Gauss-elimination method (REF)

Step 1: Form augmented matrix bA :

Step 2: Transform bA : to row echelon form

matrix DC : using row operations

Step 3: Solve the system corresponding to DC : ,

using back substitutionExample: Solve the following system of

equations3-z-y2-x3

1yx

5z3y-2

x

using Gauss elimination.

Sol.

3:123

1:011

5:312

11 R21 R

3:123

1:0112

5:2

32

11

R33R

RR

13

212

R

R

2/21:2/112/10

23:

23

230

25:

23

211

33

22

R2

R3

2

R

R

21:1110

1:1102

5:2

32

11

323 RR R


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22:1200

1:1102

5:2

32

11

33 R121 R

611:100

1:1102

5:2

32

11

So, z = 11/6 , y = 5/6, x = 1/6

1.2 Gauss-Jordan Reduction Method (RREF)

Step 1: Form augmented matrix bA :

Step 2: Transform bA : to reduced row echelon

form matrix FH : using row operations

Step 3: for each nonzero row in FH : , solve the

corresponding equations

Example: Solve the following linear system of

equations3z-x3

8zy2x

9z32y

x

using Gauss-Jordan

reduction method

Sol.

3:103

8:112

9:321

313

212

R3R-

R2R

R

R

24:1060

10:550

9:321

22 R51 R


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24:1060

2:110

9:321

121

323

R2R

R6R

R

R

12:400

2:110

5:101

33 R41 R

3:100

2:110

5:101

131

232

RR

RR

R

R

3:100

1:010

2:001

So, x = 2, y = -1, z = 3

C.W Solve the following system by Gaussian-

Jordan elimination.2 8 10

2 2

7 17 7 1

y z

x y z

x y z

.

0 2 8 10 1 2 1 2 1 2 1 2

1 2 1 2 0 2 8 10 0 2 8 10

7 17 7 1 7 17 7 1 0 3 14 15

1 2 1 2 1 2 1 2 1 2 0 2 1 0 0 12

0 2 8 10 0 1 4 5 0 1 0 5 0 1 0 5

0 0 2 0 0 0 1 0 0 0 1 0 0 0 1 0

The solution is 12, 5, 0.x y z

)sol


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Examples

I. Exactly one solution:Solve for the following system:

33

82

932

31

321

321

xx

xxx

xxx

[Solution:]The Gauss-Jordan reduction is as

follows:

Step 1: The augmented matrix is

3

8

9

1

1

3

0

1

2

3

2

1

.

Step 2:The matrix in reduced row echelon form

is

3

1

2

1

0

0

0

1

0

0

0

1

Step 3: The solution is 3,1,2 321 xxx

II.Infinite number of solutions:Solve for the following system:


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10

153

0242

21

321

xx

xxx


follows:


1

0

0

2

5

4

3

2

Step 2: The matrix in reduced row echelon

form is

1

2

3

5

1

0

0

1

Step 3: The linear system corresponding to the

matrix in reduced row echelon form is

13

25

32

31

xx

xx

The solutions are 1 3 2 32 5 , 1 3x x x x

3x is free variable or parameter and let 3 ,x t t R therefore

Rttxtxtx ,,31,52 321

III.No solution:Solve for the following system:

62

1753

5422

431

4321

4321

xxx

xxxx

xxxx


follows:

Step 1:The augmented matrix is

6

11

5

2

7

4

1

5

3

0

3

2

1

1

1

Step 2: The matrix in reduced row echelon form


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is

1

0

0

0

3

2

0

2

1

0

1

0

0

0

1

Step 3: The linear system corresponding to the


10

032

02

432

431

xxx

xxx

Since ,10 there is no solution

Example: Solve for the following linear system:

5723

-1132

-3952

352

4321

4321

4321

4321

xxxx

xxxx

xxxx

xxxx

[Solution:] The Gauss-Jordan reduction is as

follows:


5

11

3

3

7

3

9

5

2

1

1

2

3

1

5

1

1

2

2

1

Step 2:After elementary row operations, the


0

3

2

5

0

2

3

2

0

1

0

0

0

0

1

0

0

0

0

1

.

Step 3:The linear system corresponding to the


32

23

52

43

42

41

xx

xx

xx

The solutions are 1 4 2 4 3 45 2 , 2 3 , 3 2x x x x x x


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Rttxtxtxtx ,,23,32,25 4321

Example: Find conditions on a such that the

following system has no solution, one solution, orinfinitely many solutions. 1( 2) 1

2 2 ( 2) 1

x ay z

x a y z

x y a z

1 1 1 1 1 1 1 1 1

1 2 1 1 0 2 2 0 0 0 2 2 0 0

2 2 2 1 0 2 2 1 0 0 1

1 1 1 1

Case1: 1 0 0 1 1

0 0 0 0

1 1 1

Case2: 1 0 1 0 0

0 0 1

1 0 1 1

(a) 0 0 1 0 0

0 0 0 1

a a a

a a a

a a a a

a

a

a

a

a

1 1 1

(b) 0 0 1 0 0

10 0 1

a

a

a

1 : has infinitely many solutions.

0 : has no solutions.

1 and 0 : has exactly one solution.

a

a

a a

)sol

Lecture 2 Linear System

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