mca_unit-2_computer oriented numerical statistical methods
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Unit-2 SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS
RAI UNIVERSITY, AHMEDABAD 1
Course: MCA Subject: Computer Oriented Numerical
Statistical Methods Unit-2
RAI UNIVERSITY, AHMEDABAD
Unit-2 SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS
RAI UNIVERSITY, AHMEDABAD 2
Unit-II-Solution of simultaneous Linear Equations
Sr.
No.
Name of the Topic Page
No.
1. Introduction, System of linear equations in two variable Methods
for solving linear equation,
2
2. Existence of Unique roots, multiple roots and no roots (consistency
and Inconsistency of system), Dependent and Independent system
of linear equations, Examples
2
3. System of linear equations in three variables, System of linear
equations in n-variables
7
4. Methods for solving system of linear equations 9
5. Gauss elimination method and its examples 9
6. Gauss Seidel method and its examples 13
7. Difference between Direct Method and Iterative Method 16
8. References 17
9. Exercise 18
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1.1 Introduction:
We are already familiar with solving linear equations of one variable and two
variables. The general form of linear equation of one variable is ax + b = 0 where
a is not 0 and a and b are real numbers, and the general form of linear equations of
two variables is
ax + by + c =0 where a, b and c are real numbers. In this section let us discuss with
system of linear equations.
1.2 What is a system of linear Equations?
1.2.1 Definition: System of Linear Equations
Linear Equations in one variable: The general form of linear equation in one
variable is ax + b = 0, where a and b are Real numbers and a is not equal to 0.
This system has unique solution, which is x = - b/a
For example : 2x + 6 = 0
=> 2x = -6
= > x = - 6/2 = - 3 π₯ =βπ
π
The only solution to this equation is x = - 3
This solution can be represented on a number line.
2.1 System of linear Equations in two variables:
The general form of linear equation in two variables is ax + by +c = 0, where a, b, c are real numbers and a and b both not equal to 0.
Let us consider the example x + y = 6
This equation will satisfy for infinitely many pairs of the form (x,y) satisfying this
condition.
For example (1,6), 2,4), (3,3), (7, -1),.........etc
so that in all the cases x + y = 5, meaning the sum of the coordinates = 6
Hence these pairs of values satisfying the given equation are called the solution to
the given equation.
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2.2 System of linear Equations Solver:
Step-1. For the linear equations of one variable, there will be unique solution
which can be solved from the given equation.
Step-2. For the linear equations of two variables, we need to have two equations to
solve for the variables.
Let us assume that the given two equations are of the form
a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0.
It is possible to solve the system of linear equations for the unknown variables x
and y.
The methods involved in solving these homogeneous systems of linear equations
are,
1. Substitution Method.
2. Elimination Method.
3. Cross Multiplication Method
4. Matrix Method etc.
To solve a system of linear equations we should know about the types of
solution(s) that exists for the system.
2.3 Consistent system of linear equations: (Unique Roots, Infinite Roots)
The system of linear equations is said to be consistent if the solution exists.
For the above system of equations, if the following ratio satisfies we can say about
the type of solutions accordingly.
If π1
π2 β
π1
π2 then the system has unique solution.
If π1
π2 =
π1
π2=
π1
π2 then the system has infinite solution.
2.4 Inconsistent System of Equations: (No roots)
The system of linear equations is said to be inconsistent if the solution does not exists.
For the above system of equations, if the following ratio satisfies then we can say
that the system is inconsistent
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If π1
π2=
π1
π2β
π1
π2
2.5 Dependent system of Linear Equations:
System of Linear Equations is said to be dependent if both the equations have infinitely common solutions.
So the coinciding lines are the dependent system of linear equations.
They satisfy the condition, π1
π2=
π1
π2 =
π1
π2
2.6 Independent system of Linear Equations:
System of Linear Equations is said to be independent if they do not have infinite number of common solution. So the intersecting lines and the parallel lines are the Independent system of Linear
Equations.
2.7 Solving system of linear equations:
While solving a system of linear equations, we will come to know if they are
consistent or inconsistent, dependent or independent.
System of Linear Equations Examples:
2.7.1 Exampleβ
Solve the system of linear equations: ππ + ππ = ππ; ππ + ππ = ππ
Solution:
Let us number the equations 2π₯ + 3π¦ = 25 ------------------------------(1)
3π₯ + 2π¦ = 25 ------------------------------(2)
Multiplying Equation (1) by 3, 3 ( 2π₯ + 3π¦ ) = 3 ( 25) 6π₯ + 9π¦ = 75 --------------------(3)
Multiplying equation (2) by 2, 2 ( 3π₯ + 2π¦ ) = 2 ( 25)
6π₯ + 4π¦ = 50 -----------------------------(4)
6π₯ + 9π¦ = 75 ------------------------------(3)
6π₯ + 4π¦ = 50 ------------------------------(4)
______________
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Subtracting (4) from (3) 5π¦ = 25
π¦ = 25 /5 = 5 Substituting y=5 in (1) we get, 2π₯ + 3(5) = 25 2π₯ + 15 = 25
2π₯ = 25 β 15 = 10
π₯ = 10/2 = 5 Therefore, the two equations intersect at the point (2,2).
Hence the system of equations is consistent and independent.
If we verify the condition, π1
π2β
π1
π2, we see that,
2
3 β
3
2
2.7.2 Exampleβ
Solve the System of linear equations: π + π = π, ππ + ππ = π
Solution:
Let π₯ + π¦ = 4 ---------------------(1)
2π₯ + 2π¦ = 9 ------------------- (2) Multiplying (1) by 2, we get,
2 ( π₯ + π¦ ) = 2(4) => 2π₯ + 2π¦ = 8 -------------------(3)
2π₯ + 2π¦ = 9 --------------------(2)
Subtracting , (2) from (3), we get, 0 = 1, which is not true. Hence the system of equations have no solution.
From the above relation, π1
π2=
π1
π2β
π1
π2
we have 1
2=
1
2β
4
9
Hence the above pair of equations is inconsistent and
independent.
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2.7.3 Exampleβ
Solve the system of linear equations: π + ππ = π ; ππ + ππ = ππ
Solution:
Let π₯ + 3π¦ = 8 ---------------------(1)
3π₯ + 9π¦ = 24 ------------------ (2)
Substituting π₯ = 8 β 3π¦ in (2), 3 ( 8 β 3π¦ ) + 9π¦ = 24
=> 24 β 9π¦ + 9π¦ = 24 This condition is true for all values of y. we get 24 = 24 which is true.
Therefore the system of equations has infinitely many solutions. Hence the system is consistent and dependent Hence by solving system of
equations we can conclude if the system of linear equations is consistent or inconsistent and independent or dependent.
2.8 System of Linear Equations word problems:
We can follow the following steps while solving the word problems.
Step 1: Read the problem carefully and identify the unknown quantities. Give
these quantities a variable name like π₯, π¦, π’, π£, π€, etc.
Step 2: Identify the variables to be determined.
Step 3: Read the problem carefully and formulate the equations in terms of the
variables to be determined.
Solve 4: Solve the equations obtained in step 3, using any one of the method you
are comfortable with.
2.8.1 Exampleβ
4 Chairs and 3 tables cost 1400 dollars and 5 chairs and 2 tables cost 1400
dollars. Find the cost of a chair and a table.
Solution:
Let the cost of a chair be x dollars, and the cost of a table be y dollars.
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Let us solve linear system of equations we have framed here. Hence we have the equations, 4π₯ + 3π¦ = 1400 ---------------------(1)
5π₯ + 2π¦ = 1400 ---------------------(2)
Multiplying (1) by 5, 5( 4π₯ + 3π¦ ) = 5( 1400)
=> 20 π₯ + 15 π¦ = 7000 ------------------------(3) Multiplying (2) by 4, we get 4 ( 5π₯ + 2π¦ ) = 4 ( 1400)
=> 20 π₯ + 8 π¦ = 5600 ----------------------(4)
20 π₯ + 15 π¦ = 7000 ----------------------(3)
______________________
Subtracting (3) from (4), β 7 π¦ = β1400
=> π¦ = β1400/β7 = 200 Substituting y= 200 in Equation (1), we get,
4π₯ + 3 ( 200) = 1400
=> 4π₯ + 600 = 1400
=> 4π₯ = 1400 β 600 = 800 =
> π₯ = 800/4 = 200 Therefore cost of a Chair is 200 dollars and cost of a Table is 200 dollars.
3.1 System of linear equations in three variables:
The general form of linear equation in three variables, π₯, π¦ πππ π§ is
ππ₯ + ππ¦ + ππ§ + π = 0, where a, b, c are real numbers and a, b, c not all equal to 0.
This represent the equation of a plane in three-dimensional co-ordinate system, where π, π, π are the direction ratios of the normal to the plane.
To solve the equation in three variables, we need to have three conditions
(equations) relating the variables π₯, π¦ πππ π§. Elimination method is the most suitable method to solve the equations.
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3.2 System of Linear equations with n variables
π11π₯1 + π12π₯2 + β― β¦ β¦ + π1ππ₯π = π1
π21π₯1 + π22π₯2 + β― β¦ β¦ + π2ππ₯π = π2
β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ ..
ππ1π₯1 + ππ2π₯2 + β― β¦ β¦ + ππππ₯π = ππ
We can also write it in Matrix form as
[
π11 π12 β¦ β¦ . π1π
π21 π22 β¦ β¦ . π2π
β¦ β¦ . . β¦ β¦ β¦ β¦ β¦ β¦ . .ππ1 ππ2 β¦ β¦ πππ
] [
π₯1
π₯2
β¦π₯π
] = [
π1
π2
β¦π3
]
βΉ π΄π = π΅
πΆ = [π΄, π΅] = [
π11 π12 β¦ β¦ . π1π
π21 π22 β¦ β¦ . π2π
β¦ β¦ . . β¦ β¦ β¦ β¦ β¦ β¦ . .ππ1 ππ2 β¦ β¦ πππ
]is called augumented matrix.
[π΄: π΅] = πΆ
3.2.1 (π) Consistent Equations:
If Rank A = Rank C
(i) Unique Solution: Rank A= Rank C=n Where, π = number of
unknown
(ii) Infinite Solution: Rank A= Rank C= r, π < π
3.2.2 (π) Inconsistent Equations:
If Rank π΄ β Rank πΆ.
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4.1. There are two types of methods for solving System of Linear equations:
A. Direct Methods B. Iterative Methods
4.2 Direct Methods Of Solution:
1. Gauss elimination Method 2. Gauss -Jordan Method 3. Factorization method
4.3 Iterative Methods:
1. Jacobiβs iteration Method 2. Gauss - Seidel iteration Method
3. Relaxation Method
Here we discuss only 1st method gauss elimination.
5.1 Gauss elimination Method:
In this method, the unknowns are eliminated successively and the system is reduced to an upper triangular system from which the unknowns are found by back
A system of non-homogeneous linear equations
AX=B
if R(A)=R(C)
solution exists
system is consistant
if R(A)=R(C)=n
system has unique solution
if R(A)=R(C) less than n
Infinite Solution
if R(A)# R(C)
solution does not exist
system is inconsistant
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substitution. The method is quite general and is well-adapted for computer operations. Here we shall explain it by considering a system of three equations for
the sake of clarity.
Consider the equations
π1π₯ + π1π¦ + π1π§ = π1
π2π₯ + π2π¦ + π2π§ = π2
π3π₯ + π3π¦ + π3π§ = π3
Step-I. To eliminate π₯ from second and third equations.
Assuming π1 β 0,we eliminate π₯ from the second equation by subtracting (π2
π1)
times the first equation from the second equation. Similarly we eliminate π₯ from
the third equation by eliminating (π3
π1) times the first equation from the third
equation. We thus get the new system
π1π₯ + π1π¦ + π1π§ = π1
π2β²π¦ + π2β²π§ = π2β²
π3β²π¦ + π3β²π§ = π3β²
Here the first equation is called the pivotal equation and π1 is called the first pivot.
Step-II. To eliminate π¦ from third equation in (2).
Assuming π2β² β 0, we eliminate π¦ from the third equation of (2) ,by Subtracting
(π3 β²
π2 β²) times the second equation from third equation. We thus, get the new system
π1π₯ + π1π¦ + π1π§ = π1
π2β²π¦ + π2β²π§ = π2β²
π3β²β²π§ = π3β²β²
Here the second equation is the pivotal equation and π2 β² is the new pivot.
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Step-III. To evaluate the unknowns.
The values of π₯, π¦, π§ are found from the reduced system (3) by back substitution.
5.1.1 Exampleβ
Apply gauss elimination method to solve the equations
π + ππ β π = βπ
π + π β ππ = βππ
ππ β π β π = π .
Solution:
We have [1 4 β11 1 β63 β1 β1
][π₯π¦π§
] = [β5
β124
]
Operate π 2 β π 1 & π 3 β 3π 1,
~ [1 4 β10 β3 β50 β13 2
] [π₯π¦π§
] = [β5β719
]
Operate π 3 β13
3π 2 ,
[1 4 10 β3 β50 0 71/3
] [π₯π¦π§
] = [β5β7
148/3]
Thus, We have π§ =148
71= 2.0845,
3π¦ = 7 β 5π§ = 7 β 10.4225 = β3.4225 i.e., π¦ = β1.1408
π₯ = β5 β 4π¦ + π§ = β5 + 4(1.1408) + 2.0845 = 1.6479
and
Hence π₯ = 1.6479, π¦ = β1.1408, π§ = 2.0845.
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5.1.2 Exampleβ
Solve by using gauss elimination Method.
ππ + πππ β ππ = π
πππ β ππ + πππ = π
πππ β πππ + πππ = π
ππ β ππ + ππ = βπ
Solution:
We have given system of solution
π₯1 + 2π₯2 β π₯3 = 3
3π₯1 β π₯2 + 2π₯3 = 1
2π₯1 β 2π₯2 + 3π₯3 = 2
π₯1 β π₯2 + π₯3 = β1
The System of equation can be written in matrix form as
[
1 2 β13 β1 221
β2β1
31
] [π₯1
π₯2π₯3
] = [
312
β1
]
Operate (β3)π 1 + π 2 , (β2)π 1 + π 3 , (β1)π 1 + π 4
~ [
1 2 β10 β7 500
β6β3
52
] [π₯1
π₯2π₯3
] = [
3β8β4β4
]
Operate β6
7π 2 + π 3 , β
3
7π 2 + π 4
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~ [
1 2 β10 β7 500
00
5/7β1/7
][π₯1
π₯2π₯3
] = [
3β8
20/7β4/7
]
Operate 1
5π 3 + π 4
~ [
1 2 β10 β7 500
00
5/70
] [π₯1
π₯2π₯3
] = [
3β8
20/70
]
β΄5
7π₯3 =
20
7 βΉ π₯3 = 4
Now β7π₯2 + 5π₯3 = β8
βΉ β7π₯2 + 5(4) = β8
βΉ β7π₯2 = β28
βΉ π₯2 = 4
Now π₯1 + 2π₯2 β π₯3 = 3
βΉ π₯1 + 2(4) β 4 = 3
βΉ π₯1 = 3 + 4 β 8
βΉ π₯1 = β1
Hence we get the solution set (π₯1 ,π₯2 , π₯3) = (4,4, β1)
Now we discuss about 2nd iterative method Gauss β Seidel.
6.1 Gauss - Seidel iteration Method:
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This is a modification of the Jacobiβs iteration method. As before, we start with
initial approximations π₯0 , π¦0 ,π§0 (each=0) for π₯, π¦, π§ respectively. Substituting π¦ =π¦0 , π§ = π§0 in the first equations, we get
π₯1 = π1
Then putting π₯ = π₯1 , π§ = π§0 in the second of the equations, we have
π¦1 = π2 β π2π₯1 β π2π§0
Next substituting π₯ = π₯1, π¦ = π¦1 in the third equations, we obtain
π§1 = π3 β π3π₯1 β π3π¦1
And so on, i.e. as soon as new approximation for an unknown is found, it is
immediately used in the next step.
This process of iteration is continued till convergency to the desired degree of accuracy is obtained.
6.1.1 Exampleβ
Apply Gauss-Seidel iteration method to solve the equation:
πππ + π β ππ = ππ
ππ + πππ β π = βππ
ππ β ππ + πππ = ππ
Solution:
We write the given equation in the form
π₯ =1
20(17 β π¦ + 2π§) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (1)
π¦ =1
20(β18 β 3π₯ + π§) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (2)
π§ =1
20(25 β 2π₯ + 3π¦) β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (3)
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We start form the approximation π₯0 = π¦0 = π§0 = 0.substituting π¦ = π¦0 , π§ = π§0 in
the right side of the first of equations (1) we get
π₯1 =1
20(17 β π¦0 + 2π§0) = 0.8500
Putting π₯ = π₯1 , π§ = π§0 in the second of the equations (1), we have
π¦1 =1
20(β18 β 3π₯1 + π§0) = β1.0275
Putting π₯ = π₯1 , π¦ = π¦1 in the last of the equations (1), we obtain
π§1 =1
20(25 β 2π₯1 + 3π¦1) = 1.0109
For the second iteration, we have
π₯2 =1
20(17 β π¦1 + 2π§1) = 1.0025
π¦2 =1
20(β18 β 3π₯2 + π§1) = β0.9998
π§2 =1
20(25 β 2π₯2 + 3π¦2) = 0.9998
For the third iteration, we get
π₯3 =1
20(17 β π¦2 + 2π§2) = 1.0000
π¦3 =1
20(β18 β 3π₯3 + π§2) = β1.0000
π§3 =1
20(25 β 3π₯3 + 2π¦3) = 1.0000
The values in the 2nd and 3rd iterations bring practically the same, we can stop.
Hence the solution is π₯ = 1, π¦ = β1, π§ = 1.
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7.1 Difference between direct and iterative methods:
Direct Method Iterative Method
It computes the solution to a problem in a finite number of steps.
In contrast to direct methods, iterative methods are not expected to terminate in
a number of steps.
These methods give the precise answer if they were performed in infinite
precision arithmetic.
Iterative methods form successive approximations that converge to the
exact solution in the limit.
This method takes less time for
computation when we have small system linear of equations.
This method takes long time for
calculation when we have small system of linear equations.
When we have large linear system of
equation Iterative method became more easy then direct method.
Iterative methods are better than direct
methods for solving large linear systems
e.g. Gauss elimination, LU decomposition
e.g. Gauss -Seidel method, Jacobi Method, Newton-Raphson method.
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8. References and website Name:
1. Higher Engineering mathematics by B.S.Grewal
2. Higher Engineering Mathematics by B.V.Ramana 3. http://www.linearequations.org/system-of-linear-equations.html#
4. http://www.mathworks.com/matlabcentral/answers/7058-are-iterative-methods-always-better-than-direct-methods-for-solving-large-linear-systems
5. http://1.bp.blogspot.com/-
pHo8nT2SKgM/UPKdqSinfSI/AAAAAAAAC9I/JZHDxmqADsg/s1600/direct_vs_indirect.png
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EXERCISE
Q-1 Evaluate the following questions:
1. Solve the following system of linear equation
π₯ + 4π¦ = β10 , 3π₯ β π¦ = 9
2. Solve the following system of linear equation
3π₯ + 7π¦ = 15,5π₯ + 2π¦ = β4
3. Show that the non Homogeneous system of linear equation are not
consistant.
2π₯ + 6π¦ = β11
6π₯ + 20π¦ β 6π§ = β3
6π¦ β 18π§ = β1
4. Test the consistency of the following equations and solve them if possible.
3π₯ + 3π¦ + 3π§ = 1
π₯ + 2π¦ = 4
10π¦ + 3π§ = β2
2π₯ β 3π¦ β π§ = 5
Q-2 Evaluate the following questions:
1. Solve the following equations by Gauss elimination method:
2π₯ + π¦ + π§ = 10
3π₯ + 2π¦ + 3π§ = 18
π₯ + 4π¦ + 9π§ = 16
2. Solve the following equations by Gauss elimination method:
2π₯ β π¦ + 3π§ = 9
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π₯ + π¦ + π§ = 6
π₯ β π¦ + π§ = 2
3. Solve the following equations by Gauss-Seidel method:
2π₯ + π¦ + 6π§ = 9
8π₯ + 3π¦ + 2π§ + 13 = 0
π₯ + 5π¦ + π§ = 7
4. Solve the following equations by Gauss-Seidel method:
28π₯ + 4π¦ β π§ = 32
π₯ + 3π¦ + 10π§ = 24
2π₯ + 17π¦ + 4π§ = 35
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