Page 1

Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

4-1 Review : Systems of linear equations in two variables Examples: 2x+y= 8 "is called linear Equation in two variables or first degree Equation in two variable" 

Systems of linear equations in two variables : ax+by=h cx+dy=k

" The

solution set : is the set of all ordered pairs which satisfy both equations"

 Choose the correct choice :

Which of the following belongs to the solution set of the system : 2x+ y = 8 x+3y=9 (A) ( 4 ,0)

(B ) ( 6 , 1)

(C) (3,2)

the solution ( 4 , 0 ) Satisfies the first equation only so that it is not a solution . ( 6 , 1 ) Satisfies the second equation only so that it is not a solution .

But ( 3 , 2 ) Satisfies both equations so that it is a solution .

The methods of solving such system  GRAPHING  SUBSTITUTION  ELIMINATION BY ADDITION

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

(1) GRAPHING : Geometrically the solutions is all common points in the two lines There are three cases Consistent

Inconsistent

If it has one or more solution

independent

If no solution exist

dependent

If it has one exactly solution "unique solution"

If it has more than one solution "an infinite number of solutions"

Example 1 : Solve by graphing

2x+ y = 8

(3,2)

x+3y=9 the solution  by using x and y intercepts 2x+ y = 8

x+3 y = 9

x y

x y

o 8

4 o

o 3

9 o The solution set = { ( 3 , 2 ) }

Matched Problem : Solve by graphing

x+ y = 4 2x - y = 2

x y

x y

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

( 2 )Substitution ( Algebraic method )  Choose one of two equations in a system and solve for one variable in terms of the other.  Substitute the result into other equation .  Solve the linear equation in one variable .  Find the other variable by substitute .

Example 2 : Solve by substitution :

2x+ y = 8 x+3y=9 Solution

2x+ y = 8

y = 8 - 2x

solve the first equation for y in terms of x Substitution in the second equation

x+3( 8- 2x) = 9 x + 24 – 6x = 9 -5 x= 9 - 24 -5 x= -15

divide both sides by – 5

x= 3 by substitute x = 3 in

y = 8 - 2x y = 8 – 2(3) = 8 - 6 = 2

The solution is x=3 , y=2 or (3 , 2) Check :

2x+ y

?

? x+3y = 9

?

? 3+3(2) = 9

= 8

2(3)+2 = 8 6+2

Υ

= 8

3+6

Υ =

9

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched Problem : Solve by substitution :

3x+2 y = -4 x + 3 y = 15 [ July 2009]

[ Answer : (-6,7)]

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched Problem : Depending on your answer in the previous problem solve the following

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

( 3 )Elimination by addition ( Algebraic method ) If you want to eliminate the variable( x) :  Make the coefficients of ( x) in each equation equal and have the opposite sign.

y ) to know the value of ( x ) .

 Substitute with (

If you want to eliminate the variable (

y)

:

 Make the coefficient of ( y) in each equation equal and has the opposite sign.  Substitute with ( x) to know the value of ( y) . Example 3 : Solve by using elimination by addition:

2x+ y = 8 x+3y=9 Solution We can start by eliminate ( y )

2x+ y = 8

Multiply by -3

-3( 2x+ y ) = -3(8) - 6 x – 3 y = - 24 So

- 6 x – 3 y = - 24 x+3y=9 -5x

= - 15

Divide both sides by -5

x = 3 substitute in the first equation ( or the second ) 2( 3 ) + y = 8 6+y=8 y= 8 – 6 = 2 The solution is x=3 , y=2 or (3 , 2) Check :

2x+ y

?

? x+3y = 9

?

? 3+3(2) = 9

= 8

2(3)+2 = 8 6+2

Υ

= 8

3+6

Υ =

9

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 4 : Solve by using elimination by addition:

3x - 2 y = 8 2x + 5 y = -1 Solution

We can start by eliminate ( x )

3x - 2 y = 8

Multiply by -2

- 6x +4 y = -16

2x + 5 y = -1 Multiply by 3

6x + 15 y = -3

- 6x +4 y = -16 6x + 15 y = -3 19 y = -19 Divide both sides by 19 y = -1 substitute in the second equation 2x + 5 (-1) = -1 2x + 5 (-1) = -1 2x - 5 = -1 2x =-1 + 5 2x = 4 x=2 The solution is x=2 , y=-1 or (2 , -1) Check : ? 3x-2 y = 8

?

2x+5 y = -1 ?

3(2) - 2(-1) = 8 ÎĽ

6+2 = 8

?

2(2) + 5(-1) = -1 ÎĽ

4 -5 = -1

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched Problem : Solve the following system using elimination by addition :

5 x - 2 y = 12 2x+3y=1

Matched Problem :

[ 3/11/2007] [answer ( 3,2)]

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 5 : Solve by using elimination by addition: 2x + 6 y = - 3 x+3y =2 Solution Multiply the second equation by -2 and adding we obtain

2x +6 y = -3 -2x -6 y = -4 0 = -7 Not possible ( so we have obtained a contraction ) The system has no solution The solution set is the empty set

Example 6 : Solve by using elimination by addition: x -

y = 4

-2x +

y = -8 Solution

Multiply the top equation by 2 and adding we obtain

2x - y = 8 -2x + y = - 8 0 =0

(Equations Are Equivalent )

Their graphs coincide and the system is dependent So : We solve any of the 2 equations for ( x) or ( y ) Solve second equation for y

y = -8+2x

Let x = K

( where t is a real number )

y = -8+2K So : The solution = ( K , -8+2K ) Where The Variable K is called parameter Note : Replacing K with a real number produces A Particular Solution K=2 K = -1 K=5 ( -1 , -10) (2 , -4) ( 5 , 2) 31 Ahmedmaths2010@yahoo.com

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched Problem : Solve the following system using elimination by addition :

6x - 5 y = 10 -12 x + 10 y = -20

Matched Problem : : Solve the following system using elimination by addition :

6x - 5 y = 5 -12 x + 10 y = -20

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

4-2 Systems of Linear Equations and Augmented Matrices Matrices: A matrix is a rectangular array of numbers written within brackets in M rows and N columns . [ The number of rows always given first ] . An element

For example : [

]

[

]

[

]

Matrix A

Matrix B

Matrix C

Arranged in2 rows and 3 columns

Arranged in4 rows and 3 columns

Arranged in2 rows and 2 columns

 Each number in a matrix is called an element . by  If matrix has M rows and N columns , it is called an M Ă— N matrix . M and N are called the dimensions of the matrix .  A matrix with N rows and N columns is called square matrix of order N. For example :Matrix C is square matrix of order 2 .  The Column matrix :is a matrix with only one column . đ?&#x;? For example : đ?&#x;? with 3 rows and 1 column . đ?&#x;Ž  The Row matrix : is a matrix with only one row . For Example : đ?&#x;? đ?&#x;? đ?&#x;‘ with 1 row and 3 columns . Double subscript notation a i j The position of an element If A= [

]=[

a 11 ( a sub one one ) = 5 a 21 = 1

] ,

,

a 12 = 3

,

a 12 = 0

a 22 = -2

,

a 23= 3

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

The following system of linear equations

3x - 2 y = 8

3x1 - 2 x2

The subscript form

2x + 5 y = -1

= 8

2x1 + 5 x2 = -1

Related Matrices :

[

đ?&#x;‘ đ?&#x;?

đ?&#x;? ] đ?&#x;“

[

Coefficient matrix

đ?&#x;– ] đ?&#x;?

[

Constant matrix

đ?&#x;‘ đ?&#x;?

đ?&#x;? đ?&#x;“

đ?&#x;– ] đ?&#x;?

Augmented coefficient matrix

 We say that two augmented matrices row equivalent denoted by the symbol placed between the two matrices  Operation That Produce Row – Equivalent Matrices : ( A) Two rows are interchanged ( R i Rj ) đ?&#x;? đ?&#x;“ đ?&#x;? R1 R2 đ?&#x;‘ đ?&#x;? đ?&#x;– [ ] ] đ?&#x;‘ đ?&#x;? đ?&#x;– đ?&#x;? đ?&#x;“ đ?&#x;? ( B) A row is multiplied by a nonzero constant ( k R i [

Ri )

đ?&#x;’ đ?&#x;’ đ?&#x;? đ?&#x;– đ?&#x;?đ?&#x;? R 1 → R 1 [ đ?&#x;? đ?&#x;? ] [ ] đ?&#x;? đ?&#x;“ đ?&#x;? đ?&#x;? đ?&#x;“ đ?&#x;? ( C) A constant multiple of one row is added to another row ( k R i+ R j

[

đ?&#x;‘ đ?&#x;?

đ?&#x;” đ?&#x;Ž

đ?&#x;— ] đ?&#x;?

đ?&#x;? đ?&#x;‘

R1+ R 2

→R2

[

1

2

3

đ?&#x;‘ đ?&#x;?

đ?&#x;” đ?&#x;?

đ?&#x;— ] đ?&#x;“

Solving Linear Systems Using Augmented Matrices The Possible Final Matrix Forms For a System of 2 Linear Equations in 2 variables

Exactly one solution

Consistent and independent

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;?

đ?’Ž ] đ?’?

Infinitely many solutions Consistent and dependent

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž

đ?’Ž ] đ?&#x;Ž

No Solutions Inconsistent

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž

đ?’Ž đ?’‘

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Rj )


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 1 : Solve using augmented matrix method

3x1+4 x 2= 1

x1 – 2 x2 = 7 the solution Step 1 :Need a 1 here

Step 2 :Need a 0 here

Step 3 :Need a 1 here

Step 4 :Need a 0 here

[

đ?&#x;’ đ?&#x;?

đ?&#x;? ] đ?&#x;•

đ?&#x;? đ?&#x;‘

đ?&#x;? đ?&#x;’

đ?&#x;• ] đ?&#x;?

-3

6

-21

đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;?đ?&#x;Ž

[

[

đ?&#x;‘ đ?&#x;?

đ?&#x;• ] đ?&#x;?đ?&#x;Ž

0

2

[

đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;?

đ?&#x;• ] đ?&#x;?

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;?

đ?&#x;‘ ] đ?&#x;?

Therefore , x1 = 3

R1

R2

(-3) R 1 + R 2 → R 2

(

)R2→R2

-4

( 2 ) R 2+ R 1 → R 1

and x2 = -2

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 2 : Solve using augmented matrix method

x1 -2 x 2= 1 2x1 –x2 = 5 the solution

Step 2 :Need a 0 here

Step 3 :Need a 1 here

Step 4 :Need a 0 here

đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;?

đ?&#x;? ] đ?&#x;“

-2

4

-2

đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;‘

đ?&#x;? ] đ?&#x;‘

0

2

2

[

đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;?

đ?&#x;? ] đ?&#x;?

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;?

đ?&#x;‘ ] đ?&#x;?

[

[

Therefore , x1 = 3

(-2) R 1 + R 2 → R 2

( )R2→R2

( 2 ) R 2+ R 1 → R 1

and x2 = 1

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Finite Mathematics

Matched Problem : Solve

Chapter ( 4 ) systems of linear equations ,matrices

using augmented matrix method 2x1 + x 2 = 0 x1 - 2 x2 = -5 Answer : x1 = -1 , x2 = 2

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 3 : Solve using augmented matrix method

2x1 - x 2= 4 -6 x1 +3x2 = -12 the solution Step 1 :Need a 1 here

[

Step 2 :Need a 0 here

đ?&#x;? đ?&#x;”

đ?&#x;? đ?&#x;‘

đ?&#x;’ ] đ?&#x;?đ?&#x;?

đ?&#x;?

đ?&#x;? đ?&#x;?

đ?&#x;?

đ?&#x;‘

đ?&#x;?đ?&#x;?

đ?&#x;” 6

-3

)R1→R1

( 6 ) R 1 +R 2 → R 2

24

đ?&#x;?

đ?&#x;? đ?&#x;?

đ?&#x;?

đ?&#x;Ž

đ?&#x;Ž

đ?&#x;Ž

Infinitely many solutions ( consistent and dependent ) The matrix corresponds to the system x1 -

x2=2

0 = 0 So : let x 2=k x 1=2 +

( where k any real number ) k

the solution for the system is ( 2 + So: the solution set = { ( 2 +

k ,k)

k ,k)

k

R}

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Finite Mathematics

Matched Problem : Solve

Chapter ( 4 ) systems of linear equations ,matrices

using augmented matrix method - 2x1 + 6 x 2 = 6

3 x1 - 9 x2 = -9

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example 4 : Solve using augmented matrix method

2x1 +6 x 2 = -3 x1 +3x2 = 2 the solution Step 1 :Need a 1 here

[

đ?&#x;? đ?&#x;?

Step 2 :Need a 0 here

đ?&#x;” đ?&#x;‘

đ?&#x;?

đ?&#x;‘

đ?&#x;?

đ?&#x;”

-2

-6

đ?&#x;?

đ?&#x;‘

đ?&#x;Ž

đ?&#x;Ž

đ?&#x;‘ ] đ?&#x;? đ?&#x;?

R1

R2

( -2 ) R 1 +R 2 → R 2

đ?&#x;‘ -4

đ?&#x;? đ?&#x;•

This is an augmented matrix of the system x 1 +3 x 2 = 2 0 = -7 So the system is inconsistent and ther are no solution So: the solution set = Empty set

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Finite Mathematics

Matched Problem : Solve

Chapter ( 4 ) systems of linear equations ,matrices

using augmented matrix method 2x1 - x 2 = 3 4 x1 - 2 x2 = -1

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Finite Mathematics

Matched Problem : Solve

Chapter ( 4 ) systems of linear equations ,matrices

using augmented matrix method

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

4-3 Gauss-Jordan Elimination  We consider systems involving more than two variables . Reduced Matrices : The Possible Final Matrix Forms For a System of 2 Linear Equations in 2 variables

Exactly one solution Consistent and independent

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;?

đ?’Ž ] đ?’?

Infinitely many solutions Consistent and dependent

[

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž

No Solutions Inconsistent

đ?’Ž ] đ?&#x;Ž

đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž

đ?’Ž đ?’‘

Where m , n , and p are real numbers , p ≠0

For large linear systems :THE REDUCED FORM A matrix is said to be in Reduced Row Echelon Form (RR EF) IF: 1) Each row consisting entirely of zeros is below any row having at least one nonzero element .[ Each row consisting entirely of zeros should be at the bottom ]. Not in reduced form

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

reduced form

đ?&#x;‘ đ?&#x;Ž đ?&#x;’

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;‘ đ?&#x;’ đ?&#x;Ž

2) The leftmost nonzero element in each row is 1 . Not in reduced form

reduced form

đ?&#x;? đ?&#x;Ž đ?&#x;Ž đ?&#x;? đ?&#x;? đ?&#x;Ž đ?&#x;Ž đ?&#x;? đ?&#x;Ž đ?&#x;? đ?&#x;Ž đ?&#x;‘ đ?&#x;Ž đ?&#x;? đ?&#x;Ž đ?&#x;‘ đ?&#x;Ž đ?&#x;Ž đ?&#x;? đ?&#x;“ đ?&#x;Ž đ?&#x;Ž đ?&#x;? đ?&#x;“ 3) All other elements in the column containing the leftmost 1 of a given row are zeros . Not in reduced form

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;‘ đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;? đ?&#x;‘ đ?&#x;“

reduced form

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;? đ?&#x;‘ đ?&#x;“

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

4) The leftmost 1 in any row is to the right of the leftmost 1 in the row above . Not in reduced form

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;? đ?&#x;‘ đ?&#x;“

reduced form

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;? đ?&#x;‘ đ?&#x;“

Examples from previous exam

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Solving Systems By Gauss-Jordan Elimination Example : ( 1 ) Solve by gauss-Jordan elimination : 2x1 -2x2+x3 = 3 3x1 + x2 - x3 = 7 x1 -3x2+2 x3 = 0

Need a 0 here

Solution đ?&#x;? đ?&#x;‘ đ?&#x;?

Need a 1 here

đ?&#x;? đ?&#x;? đ?&#x;‘

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;‘ đ?&#x;• đ?&#x;Ž

R1

đ?&#x;? đ?&#x;‘ đ?&#x;?

R3

đ?&#x;‘ đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;Ž đ?&#x;• đ?&#x;‘

(-3) R 1 + R 2 → R 2 (-2) R 1 + R 3 → R 3

Need a 0 here

Need a 1 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;‘ đ?&#x;?đ?&#x;Ž đ?&#x;’

đ?&#x;? đ?&#x;• đ?&#x;‘

đ?&#x;Ž đ?&#x;• đ?&#x;‘

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;?

( đ?&#x;?đ?&#x;Ž ) R 2 → R 2

đ?&#x;‘ đ?&#x;? đ?&#x;’

đ?&#x;? đ?&#x;Ž đ?&#x;Ž. đ?&#x;• đ?&#x;Ž. đ?&#x;• đ?&#x;‘ đ?&#x;‘

(3)R2+R1→R1 (- 4) R 2 + R 3 → R 3

Need a 0 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;Ž. đ?&#x;? đ?&#x;Ž. đ?&#x;• đ?&#x;Ž. đ?&#x;?

đ?&#x;?. đ?&#x;? đ?&#x;Ž. đ?&#x;• đ?&#x;Ž. đ?&#x;?

(-5)R3→R3

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;Ž. đ?&#x;? đ?&#x;Ž. đ?&#x;• đ?&#x;?

đ?&#x;?. đ?&#x;? đ?&#x;Ž. đ?&#x;• đ?&#x;?

( 0.1 ) R 3 + R 1 → R 1 (0.7) R 3 + R 2 → R 2

Need a 1 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;? đ?&#x;Ž đ?&#x;?

So : The solution for the system is x1 = 2 ,

x2 = 0 , x3 = -1

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched problem : Solve by gauss-Jordan elimination : 3x1 +x2 -2x3 = 2 x1 - 2 x2 + x3 = 3 2x1 - x2 - 3 x3 = 3

Answer : x1 = 1 , x2= -1 , x3 = 0

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Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example : ( 2 ) Solve by gauss-Jordan elimination : 2x1 - 4x2+ x3 = - 4 4x1 -8 x2 +7 x3 = 2 -2 x1 +4x2 - 3 x3 = 5 Need a 0 here

Solution đ?&#x;? đ?&#x;’ đ?&#x;?

Need a 1 here

đ?&#x;’ đ?&#x;– đ?&#x;’

đ?&#x;? đ?&#x;• đ?&#x;‘

đ?&#x;’ đ?&#x;? đ?&#x;“

đ?&#x;? đ?&#x;?

R1

đ?&#x;? đ?&#x;’ đ?&#x;?

R1

Need a 1 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž. đ?&#x;“ đ?&#x;• đ?&#x;‘

đ?&#x;? đ?&#x;? đ?&#x;“

(- 4) R 1 + R 2 → R 2 ( 2) R 1 + R 3 → R 3

Need a 0 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;– đ?&#x;’

đ?&#x;Ž. đ?&#x;“ đ?&#x;“ đ?&#x;?

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;?đ?&#x;Ž đ?&#x;?

đ?&#x;‘ đ?&#x;? đ?&#x;“

(

đ?&#x;? ) đ?&#x;“

R2→R2

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž. đ?&#x;“ đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;? đ?&#x;?

(

−đ?&#x;? đ?&#x;?

)R2+R1→R1

( 2 ) R 2+ R 3 → R 3

The last row produces a contradiction

So : the system has no solution ( inconsistent)

47 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched problem : Solve by gauss-Jordan elimination : 2x1 -4x2 - x3 = - 8 4 x1 - 8 x2 +3 x3 = 4 -2x1 +4 x2 + x3 = 11

Answer : The system has no solution ( inconsistent)

48 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example : ( 3 ) Solve by gauss-Jordan elimination : 3x1 +6x2-9 x3 = 15 2x1+4 x2 -6 x3 = 10 -2 x1 -3x2 +4 x3 = -6 Need a 0 here

Solution đ?&#x;‘ đ?&#x;? đ?&#x;?

Need a 1 here

đ?&#x;” đ?&#x;’ đ?&#x;‘

đ?&#x;— đ?&#x;” đ?&#x;’

đ?&#x;?đ?&#x;“ đ?&#x;?đ?&#x;Ž đ?&#x;”

đ?&#x;? đ?&#x;‘

R1

R1

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;’ đ?&#x;‘

đ?&#x;‘ đ?&#x;“ đ?&#x;” đ?&#x;?đ?&#x;Ž đ?&#x;’ đ?&#x;”

(- 2) R 1 + R 2 → R 2 ( 2) R 1 + R 3 → R 3

Need a 0 here

đ?&#x;? đ?&#x;Ž đ?&#x;Ž đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;?

đ?&#x;‘ đ?&#x;Ž đ?&#x;? đ?&#x;? đ?&#x;? đ?&#x;Ž

đ?&#x;“ đ?&#x;Ž đ?&#x;’ đ?&#x;‘ đ?&#x;’ đ?&#x;Ž

R2

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

R3

đ?&#x;? đ?&#x;? đ?&#x;Ž

đ?&#x;‘ đ?&#x;? đ?&#x;Ž

đ?&#x;“ đ?&#x;’ đ?&#x;Ž

(2) R 2 + R 1→ R 1

So : the system has infinitely many solutions (consistent and dependent)

The corresponding system is :

x1+x3 = -3 x2 - 2x3 = 4

Let So

x3=k

( where k is any real number )

x1=-x3 -3= -k-3 x2 = 2x3 +4=2k +4

The solution is ( x1 , x2 ,x3 ) =( -k-3 , 2k +4 , k ) Notice : for any particular solution if t = 0 the solution become ( -3,4,0) 49 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched problem : Solve by gauss-Jordan elimination : 2x1 - 2x2 - 4x3 = - 2 3 x1 - 3 x2 -6 x3 = -3

Answer : (5k+4 , 3k+ 5 , k)

-2x1 +3 x2 + x3 = 7

50 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Example : ( 4 ) Solve by gauss-Jordan elimination : x1 +2x2 + 4 x3 + x 4 – x5 = 1 2 x1 +4x2 + 8 x3 +3 x 4 – 4 x5 = 2 x1 +3x2 + 7 x3

+3 x5 = -2

Solution

Need a 0 here

(- 2) R 1 + R 2 → R 2

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;’ đ?&#x;‘

đ?&#x;’ đ?&#x;? đ?&#x;– đ?&#x;‘ đ?&#x;• đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;?

đ?&#x;’ đ?&#x;Ž đ?&#x;‘

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;? đ?&#x;? đ?&#x;Ž

đ?&#x;’ đ?&#x;‘ đ?&#x;Ž

đ?&#x;? đ?&#x;? đ?&#x;?

đ?&#x;? đ?&#x;? đ?&#x;’ đ?&#x;? đ?&#x;‘ đ?&#x;? đ?&#x;? đ?&#x;? đ?&#x;’

đ?&#x;? đ?&#x;Ž đ?&#x;‘

đ?&#x;? đ?&#x;’ đ?&#x;?

đ?&#x;? đ?&#x;‘ đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;‘ đ?&#x;Ž

đ?&#x;‘ đ?&#x;? đ?&#x;?

đ?&#x;— đ?&#x;’ đ?&#x;?

đ?&#x;• đ?&#x;‘ đ?&#x;Ž

đ?&#x;? đ?&#x;Ž đ?&#x;Ž

đ?&#x;Ž đ?&#x;? đ?&#x;Ž

đ?&#x;? đ?&#x;‘ đ?&#x;Ž

đ?&#x;Ž đ?&#x;Ž đ?&#x;?

đ?&#x;‘ đ?&#x;? đ?&#x;?

đ?&#x;• đ?&#x;‘ đ?&#x;Ž

( -1) R 1 + R 3 → R 3

R2

R3

(- 2) R 2 + R 1 → R 1

(-3) R 3+ R 1 → R 1 R 3+ R 2 → R 2

Matrix in reduced form

The corresponding system is : x1-2x3 -3x5= 7 x2 +3x3 +2x5=-3 x4 - 2x5 =0

SO

The repeated variables are x5 and x3 so Let x3 = K and x5 = N for any real numbers K , N

x1= 2x3 +3x5 + 7

x1= 2K +3N + 7

x2 =-3x3 -2x5 -3 x4 = 2x5

The solution ď Œ (2K +3N +

x2 =-3K -2N -3 x4 = 2N 7 , -3K -2N -3 , K , 2N , N )

51 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Matched problem : Solve by gauss-Jordan elimination : x1 - x2 +2x3

-2x5 = 3

-2 x1 +2 x2 -4 x3 窶度4 + x5= -5 3x1 -3 x2 +7 x3+x4- 4x5 = 6

Answer : (k+7 , k , n-2 , -3n-1 , n) Where : k , s any real numbers

52 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

53 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

THE MATCHED PROBLEM

If

k Z= P there are 3 cases :

( 1 ) EXACTLY ONE SOLUTION :

( Where k , p are real numbers ) k≠ 0

( 2 ) INFINITLY MANY SOLUTIONS : ( 3 ) NO SOLUTION :

k= 0

AND( INTERSECTION)

P =0

k = 0 AND P ≠ 0

54 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

55 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

If

Chapter ( 4 ) systems of linear equations ,matrices

k Z= P there are 3 cases :

( 1 ) EXACTLY ONE SOLUTION :

( Where k , p are real numbers ) k≠ 0

( 2 ) INFINITLY MANY SOLUTIONS : ( 3 ) NO SOLUTION :

k= 0

AND( INTERSECTION)

P =0

k = 0 AND P ≠ 0

56 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Application: A company that rents small moving trucks wants to purchase 25 Trucks with a combined capacity of 28,000 cubic feet . three different types of Trucks are available : a 10-foot truck with a capacity of 350 cubic feet , a 14-foot truck with a capacity of 700 cubic feet , and a 24-foot truck with a capacity of 1,1400 cubic feet . how many of each type of truck should the company purchase ?

Solution The number of each type of truck : x1= number of 10 – foot trucks x2= number of 14 – foot trucks x3= number of 24 – foot trucks ( Total number of trucks ) : ( Total capacity ) :

x1 + x2

+

x3 = 25

350 x1 + 700 x2

+

1,400 x3 = 28,000

By form the augmented coefficient matrix of the system đ?&#x;?

đ?&#x;?

đ?&#x;?

đ?&#x;‘đ?&#x;“đ?&#x;Ž đ?&#x;•đ?&#x;Žđ?&#x;Ž đ?&#x;?, đ?&#x;’đ?&#x;Žđ?&#x;Ž

đ?&#x;?

đ?&#x;?

đ?&#x;?đ?&#x;“

đ?&#x;?

đ?&#x;?

đ?&#x;’

đ?&#x;–đ?&#x;Ž

đ?&#x;?

đ?&#x;?

đ?&#x;?

đ?&#x;?đ?&#x;“

đ?&#x;Ž

đ?&#x;?

đ?&#x;‘

đ?&#x;“đ?&#x;“

đ?&#x;?

đ?&#x;Ž

đ?&#x;?

đ?&#x;Ž

đ?&#x;?

đ?&#x;‘

x2 + 3 x 3 = 55 Let x 3 = k

R2 → R

đ?&#x;?đ?&#x;–đ?&#x;Žđ?&#x;Žđ?&#x;Ž

đ?&#x;?

x1 – 2 x 3 = - 30

đ?&#x;? đ?&#x;‘đ?&#x;“đ?&#x;Ž

đ?&#x;?đ?&#x;“

đ?&#x;‘đ?&#x;Ž

-R1 +R 2 → R 2

-R2 +R 1 → R 1

Matrix is in reduced

đ?&#x;“đ?&#x;“

or or

x1 = 2 x 3 - 30 x2 = - 3 x 3 + 55

Then for ( k )any real number the solution

x1 = 2 k - 30 x2 = - 3 k + 55 x3= k 57 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Interpretation : If we want the possible values of k that will produce meaningful solutions x 3 = k [ must be non-negative whole number : 1 , 2 , 3 , ‌‌ ] x1 = 2 k - 30 ≼ 0

k ≼ 15

x2 = - 3 k + 55 ≼ 0

k≤

15

The only possible values of k are : 15 , 16 , 17 , 18 .

=

16

17

đ?&#x;?đ?&#x;–

đ?&#x;? đ?&#x;‘

18

đ?&#x;?đ?&#x;–

đ?&#x;? đ?&#x;‘

The next table display these solutions : k 15 16 17 18

X1 0 2 4 6

X2 10 7 4 1

X3 15 16 17 18

58 Ahmedmaths2010@yahoo.com

Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

Problems from The previous exams

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Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

MATCHED PROBLEM

The solution

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Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

MATCHED PROBLEM

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Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

MATCHED PROBLEM

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Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)


Finite Mathematics

Chapter ( 4 ) systems of linear equations ,matrices

MATCHED PROBLEM

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Prof: Ahmed Ahmed ( For any inquiry mob : 94982102)

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the first chapter : linear equations and graphs finite mathematics for business,economics ,life science and social science

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