Chapter 6 System of Linear Equation - Student

8/20/2019 Chapter 6 System of Linear Equation - Student

1/12

BIC 10303 ALGEBRA

CHAPTER 6: SYSTEM OF LINEAR EQUATION

LEARNING OUTCOME

At the end of this topic, students are able to:

i. understand the system of linear equation consisting two or more variables.

ii. use the suitable method in solving the system of linear equation.

iii. perform calculation either using substitution, elimination or matrices method.

INTRODUCTION:

• A system of linear equations is simply two or m ore linear equations using the same variables.

• If the system of linear equations is going to have a solution, then the solution will be an ordered pair ( ,) where and make both equations true at the same time.

NOTES:

INTRODUCTION OF SYSTEM LINEAR EQUATION:

• Linear equations are equations of the form + +⋯+ =, where ,…, are constants calledcoefficients, ,…, are variables , and b is the constant term.

• This equation is a linear equation in variables.• The variables are also often seen as

,

,

, etc.

The following equations are linear: The following equations are not linear

2 +3 = 7 3 − +4 =1 ( +) =

3 = + =1 =√ • The Variables

– Linear equations are functions which have two variables; independent and dependent variable.

– Independent variables are variables that you put into the equations

– Dependent variables are variables that you solve for.

– Example: = 2 – 4 • is independent variable• has to be the dependent variable

• A L inear Al gebraic Equati on (one equation in n unknowns) is defined as any equation of the form: + +⋯+ = Where ,,,…, and denote known constants whereas, ,,,…, denote unknowns

• A L inear Al gebraic Equati on ( equation in unknowns) is defined as any equation of the form: + + +⋯+ = + + +⋯+ = ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ + + +⋯+ = • A SOLUTION for × is any set of values for x1, x2, … , xn that simultaneously satisfies all of the

equations.

SYSTEM OF LINEAR EQUATIONS OF TWO VARIABLES:

Deciding whether an ordered pair is a solution of a linear system. The solution set of a linear system of equations

contains all ordered pairs that satisfy all the equations at the same time. + = + =

System of Linear Equations in two variables can be solved by using 3 method:

1) Graphing Method:

To solve the set of a linear system of equations with two variables by graphical method is by identifying where isthe graphs in tersect.

There are 4 steps to a solving a linear system using a graph


2/12

BIC 10303 ALGEBRA

Step 1: Put both equations in slope intercept form

Solve both equation for and Step 2: Graph both equations on the same coordinate plane

Step 3: Estimate where the graphs intercept

Label the solution

Step 4: Check to make sure your solution makes both equations true

Substitute and into both equations to verify the points is a solution toboth equations

Example: Solve the system of equations by graphical method + =5 and 2 − =4

Check: = 3, = 2; 3+2=5 (true), 2(3)− 2 =4 (true). Therefore, solution for the system of linear equation is = 3, = 2. There are 3 possible solutions to a system of linear equations in two variables that have been graphed:

1. One solution

When the two lines intercept at a single point,there will be one solution.

In this case, the solution is “consistent” and theequations are “independent”.

2. No solution

When the lines are parallel, there will be nosolutions.

The graphs are parallel lines. (Slopes areequal) In this case the system is “inconsistent”and the solution set is 0 or null.

3. Infinite number of solutions

If the lines are the same, there will be an infinite number of solutions.

The graphs are the same line. (Slopes and y-intercepts are the same) In this case, the equations are“dependent” and the solution set is an infinite set of ordered pairs.


3/12

BIC 10303 ALGEBRA

2) Method of Elimination:

There are 6 steps to a solving a linear system using a elimination method

The Step Example:Solve the system: 2 +3 = 19 and 3 − 7 =−6

Step 1: Write both equations in standard form 2 + 3 = 193 − 7 = −6

Step 2: Make the coefficients of one pair of variable

terms opposite

(Multiply one or both equations by appropriate

numbers so that the sum of the coefficients of

either x or y will be zero.)

Multiply the top equation by 3, the bottom equation

by -2

3[2 + 3 = 19] → 6 + 9 = 57−2[3 − 7 = −6] → −6 +14 = 12Step 3: Add the new equations to eliminate a variable Add the new equations to eliminate a variable

0 + 23 = 69Step 4: Solve the equation formed in Step 3 =69

23

= 3 Step 5: Substitute the result of Step 4 into either of the

original equations and solve for the othervalue.

Substitute into top equation and solve for 2 + 3(3) = 19

2 = 10 = 5

Solution Set: {(5,3) Step 6: Check the solution and write the solution set. Check the solution and write the solution set.

2(5) + 3(3) = 19 (True)

3(5) − 7(3) = −6 (True) 3) Method of Substituti on:

There are 5 steps to a solving a linear system using a substitute method

The Step Example:Solve the system: 4 + =5 and 2 − 3 =13

Step 1: Solve one of the equations for either variable Choose the variable to solve for in the topequation:

=

−4

+ 5

Step 2: Substitute for that variable in the other

equation

(The result should be an equation with just

one variable)

Substitute this variable into the bottom equation

2 − 3(−4+5) = 132 +12 − 15 =13→ 14 = 28

Step 3: Solve the equation from Step 2 14 =28 = 2 Step 4: Substitute the result of Step 3 into either of the

original equations and solve for the othervalue.

Substitute into original equations and solve for theother value

4(2) + =5 = −3 Solution Set: {(2,−3)

Step 5: Check the solution and write the solution set. Check the solution and write the solution set.

4(2)− 3 = 5 (True)2(2) − 3(−3) = 13 (True)


4/12

BIC 10303 ALGEBRA

SYSTEM OF LINEAR EQUATIONS OF MULTIPLE VARIABLES:

System of linear equations of multiple variable involving more than 2 variables. In this module we were focus on

three equations containing three variables. + + =

+

+

=

+ + =

System of linear equations in three variables can be solved by using 2 method:

1) Method of Elimination:

There are 6 steps to a solving a linear systemusing a elimination method

Example:Solve the system by using elimination method

3 + +2 = 11 2+2 − =−1 +3+2 =5

Step 1: Rewrite the system as two smaller

systems, each containing two of thethree equations.

First and second equation:

3++2 =11 2 +2 − =−1 First and third equation:

3 ++2 = 11 +3+2 =5 Step 2: Eliminate THE SAME variable in each

of the two smaller systems.

(Multiply one or both equations byappropriate numbers so that the sum ofthe coefficients of either , or will bezero.)

Let say we want to eliminate :3 ++2 = 11 (Multiply to -2)2 +2− =−1 −6 − 2− 4 =−22

2 +2− =−1 Sum up−4 − 5 =−23

3 ++2 =11 (Multiply to -3) +3+2 =5 −9 − 3− 6 =−33 +3+2 = 5 −8 − 4 =−28

Step 3: Write the resulting equations in twovariables together as a system ofequations.Then, solve the system for the tworemaining variables.

−4 − 5 =−23 (Multiply to 4)−8 − 4 =−28 (Multiply to -5)−16 − 20 =−92

40+20 =140 Sum up:24 =48 = 2 Substitute =2 into any of equation with two variable−4(2)− 5 =−23 −8−5 =−23

−5

=

−23+8

−5 =−15 = 3 Step 4: Substitute the value of the variables

from the system of two equations in oneof the ORIGINAL equations with threevariables.

Substitute =2 and = 3 into any of equation with three variable3(2) + +2(3) =11 6+ +6=11 =11− 6− 6 =−1 Solution Set: (2,−1,3)

Step 5: Check the solution and write thesolution set.

3(2)− 1+ 2(3) =11 (True)2(2) +2(−1)− 3 =−1 (True)2+3(

−1) +2(3) =5 (True)


5/12

BIC 10303 ALGEBRA

2) Method of Substituti on

There are 5 steps to a solving a linear system using asubstitute method

Example:Solve the system by using substitution method +2 − 3 =−5

2 −3+ = 4 −3 +4 − 2 =−8 Step 1: Solve one of the equations for either variable We choose for the most simple equation to be solve

First equation:

+2 − 3 =−5 =−5− 2 +3 Step 2: Substitute for that variable in the other

equation

Substitute into second and third equationSecond equation:2(−5− 2+3) − 3+ =4 −10−4+6 − 3+ =4 −7 +7 =14 −+ =2 Third equation:

−3(

−5

−2

+3

) +4

−2

=

−8

15+6 − 9+4 − 2 =−8 10 − 11 =−23 Step 3: Solve by using elimination method −+ =2 (Multiply to 10)

10 − 11 =−23 −10 +10 =20 10 − 11 =−23 Sum up:− =−3 = 3

Step 4: Substitute the result of Step 3 into equationwith two variable.

Then, substitute the variable into ORIGINALequation with three variable.

Substitute

=3 into any of equation with two variable

−+3=2 =1 Substitute = 3 and =1 into any of equation withthree variable +2(1)− 3(3) =−5 =2 Solution Set: {(2,1,3)

Step 5: Check the solution and write the solution set. 2+2(1)− 3(3) =−5 (True)2(2)− 3(1) +3 =4 (True)−

3(2) +4(1)

−2(3) =

−8 (True)


6/12

BIC 10303 ALGEBRA

6.4 MATRICES METHOD: INVERSE MATRIX

Consider the following system of equations:

+ + = + + = + + =

The system can be represented in matrix form as :

=

A X = B

The equations have a unique solution provided that A is a non-singular matrix, i.e. provided that det A ≠ 0. The equation

can be solved by :

X = A-1 B

The Step

Example:

By using Inverse Matrix, solve the following system:

4 +8 − 4 =4 3+8+5 =−11 −2++12 =−17

Step 1: Write the three equations in the matrix form, = 4 8 −43 8 5

−2 1 12

= 4−11

−17

A X = B

Step 2: Compute the determinant of matrix A, | | = 4 8 −43 8 5−2 1 12

| | = 4 8 51 12

− 8 3 5−2 12 +(−4) 3 8−2 1 | | =−80

Step 3: Find the cofactor of matrix A.

=+

8 51 12 −

3 58 12 +

3 88 1− 8 −4

1 12 +4 −4

8 12 −4 8

8 1

+8 −48 5

−4 −43 5

+4 83 8

= 91 −46 19−100 40 −2072 −32 8

Step 4: Determine the adjoint of matrix A, ( ).

(

)

=

= 91 −100 72−46 40 −3219

−20 8


7/12

BIC 10303 ALGEBRA

Step 5: Calculate the inverse of matrix A, . =

|| ( ) =1−80

91 −100 72−46 40 −3219 −20 8

Step 6: Solve the system and determine the value , and .

= = 1−80

91 −100 72−46 40 −3219

−20 8

4−11−

17

= −31−2

Solution Set: {(-3,1,-2)}


4(−3) + 8(1)−4(−2) = 4 (TRUE)3(−3) +8(1) +5(−2) =−11 (TRUE)−2(−3) +(1) +12(−2) =−17 (TRUE)


8/12

BIC 10303 ALGEBRA

6.5 MATRICES METHOD : ELIMINATION

Elimination with matrices is the technique most commonly used by computer software to solve systems of linear

equations. It finds a solution X to AX= B whenever the matrix A is invertible (det A ≠ 0). Two types of elimination

method with matrices:

a. Gaussian Elimination Method

b. Gauss-Jordan Elimination Method.

a. GAUSSIAN ELIMINATION METHOD.

The basic concept in solving a system using Gaussian Elimination Method:

Augmented matri x - Write the augmented matrix of the system.

Row-echelon form - Use elementary row operations to change the augmented matrix to row-echelon

form.

Back-substitution method - Write the new system of equations corresponds to the row-echelon form

of the augmented matrix and solve by back-substitution method.

The Step

Example: Solve the system by using Gaussian

Elimination Method:

4 +8 − 4 =4 3+8+5 =−11 −2++12 =−17

Step 1: Write into augmented matrix form. 4 8 −43 8 5−2 1 12 4−11−17

Step 2: Reduce the augmented matrix to reduced

row echelon form.

Note: Use elementary row operations to

change the augmented matrix to row-

echelon form.

4 8 −43 8 5−

2 1 12 4−11

−17

1

4 →

~ 1 2 −13 8 5−2 1 12 1−11−17

−3 → +2 →

~ 1 2 −10 2 80 5 10

18−110 1

2 →

~ 1 2 −10 1 40 5 10

1−7−15 − 5 →

~

1 2 −1 0 1 4

0 0 −10

1

−7

20

−1

10 →

~ 1 2 −10 1 40 0 1

1−7−2 + → − 4 →

Step 3: Write the new system of equations that

corresponds to the row-echelon form of the

augmented matrix.~ 1 2 −10 1 4

0 0 1 1−7−2

+2 − = 1+4 =−7 =−2


9/12

BIC 10303 ALGEBRA

Step 4: Solve the new system by using back-substitution method. +2 − =1 (1)

+4 =−7…(2) =−2…(3)

Substitute =−2 into equation (2),+4 =−7 +4(−2) =−7 = 1 Substitute =−2 and =1 into equation (1) +2 − =1 +2(1)− (−2) =1 =−3


Step 5: Check the solution and write the solution set. 4(−3) + 8(1)− 4(−2) = 4 (TRUE)3(−3) +8(1) +5(−2) =−11 (TRUE)−2(−3) +(1) +12(−2) =−17 (TRUE

b. GAUSS-JORDAN ELIMINATION METHOD.

The basic concept in solving a system using Gauss-Jordan Elimination Method:

Augmented matri x - Write the augmented matrix of the system.

Reduced Row-echelon form - Use elementary row operations to change the augmented matrix to row-

echelon form and further reduce to reduced row echelon form.

No need back-substitution method - Write the new system of equations corresponds to the reduced

row-echelon form of the augmented matrix and solve directly.

The Step

Example:

Solve the system by using Gauss-JordanElimination Method:

4 +8 − 4 =4 3+8+5 =−11 −2++12 =−17

Step 1: Convert into augmented matrix form.

4 8 −43 8 5−2 1 12 4−11−17

Step 2: Reduce the augmented matrix to reduced row

echelon form.

Note : Use elementary row operations to

change the augmented matrix to row-

echelon form (Gaussian Elimination) and

further reduce to reduced row echelon form.

4 8 −43 8 5−2 1 12 4−11−17

14 →

~ 1 2 −13 8 5−2 1 12 1−11−17

−3 → +2 → ~ 1 2 −10 2 8

0 5 10 18−110

1

2 →

~ 1 2 −10 1 40 5 10

1−7−

15 −

5

→


10/12

BIC 10303 ALGEBRA

~ 1 2 −1 0 1 40 0 −10

1−720

− 110

→

~ 1 2 −10 1 40 0 1

1−7−2 + → − 4 →

~

1 2 00 1 00 0 1

−11

−2

− 2 →

~ 1 0 00 1 00 0 1

−3 1−2 Step 3: Write the new system of equations that

corresponds to the reduced row-echelon formof the augmented matrix. Directly, solve thesystem.

1 0 00 1 00 0 1

−31−2 =−3 =1 =−2


Step 4: Check the solution and write the solution set. 4(−3) + 8(1)− 4(−2) = 4 (TRUE)3(

−3) +8(1) +5(

−2) =

−11 (TRUE)

−2(−3) +(1) +12(−2) =−17 (TRUE)


11/12

BIC 10303 ALGEBRA

6.6 MATRICES METHOD: CRAMER’S RULE

Cramer’s Rule for solving a system of three linear equations with three unknowns:

+ + = + + =

+ + = The system above be represented in the matrix form as:

=

A X = B

By substituting matrix B into each column in matrix A:

=

,

=

,

=

The solution for this system is given by:

= | | =

, = | | =

, = | | =

The equations have a unique solution provided that A is a non-singular matrix, i.e. provided that | | ≠ 0.

The Step

Example:

By using Cramer’s Rule, solve the followingsystem:

4 +8 − 4 =4

3+8+5 =−11 −2++12 =−17

Step 1: Write the three equations in the matrix form, = 4 8 −43 8 58 1 12

= 4−11−17

A X = B

Step 2: Substitute matrix B into 1st column of matrix A.Calculate .

= 8 −4− 8 5− 1 12 4 8 −43 8 5

8 1 12

=240−80 = −3

Step 3: Substitute matrix B into 2nd column of matrix A. Calculate . =

4 −43 − 58 − 12 4 8 −43 8 5

8 1 12

=−80−80 = 1


12/12

BIC 10303 ALGEBRA

Step 4: Substitute matrix B into 3rd column of matrix A. Calculate z.

= 4 8 3 8 −

8 1 − 4 8 −43 8 5

8 1 12

=160−80 = −2



4(−3) + 8(1)− 4(−2) = 4 (TRUE)3(−3) +8(1) +5(−2) =−11 (TRUE)−2(−3) +(1) +12(−2) =−17 (TRUE)

Chapter 6 System of Linear Equation - Student

Documents

Transcript of Chapter 6 System of Linear Equation - Student