In general, the rearrangement of equation is done even if pivot element is non-zero to improve the accuracy of solution by reducing the round off errors involved in elimination process,
Trang 1largest coefficient of y We continue this process till last equation This procedure is known as
partial pivoting In general, the rearrangement of equation is done even if pivot element is non-zero to improve the accuracy of solution by reducing the round off errors involved in elimination process, by getting a larger determinant, which is done by finding a largest element
of the row as the pivotal element
Complete Pivoting: If the order of elimination of x1 , x2, x3, is not important, then we may choose at each stage the largest coefficient of the whole matrix of coefficients We may search the largest value, not only in rows but also in columns After searching largest value, we bring at the diagonal position This method of elimination is known as complete pivoting The superiority of this method is that it gives the solution of a system, provided its determinant does not vanish in finite number of steps
A system of equations A X = B is said to be ill-conditioned or unstable if it is highly sensitive
to small changes in A and B i.e., small change in A or B causes a large change in the solution
of the system On the other hand if small changes in A and B give small changes in the solution, the system is said to be stable, or, well conditioned Thus in a ill-conditioned system, even the small round off errors effect the solutions very badly Unfortunately it is quite difficult to recognize an ill-conditioned system
For example, consider the following two almost identical systems
x1−x2=1 x1−x2=1
1 1.00001
x − x2 =0 and x1−0.99999 x2 =0
Respective solutions are:
(100001, 100000) and (–99999, –100000) obviously the two solutions differ very widely Therefore the system is ill conditioned
Example 3 Show that the following system of linear equations is ill-conditioned.
7x 10y = 1−
5x + 7y = 0.7 Sol On solving the given equations we get x= 0 and y= 0.1.
Now, we make slight changes in the given system of equations The new system becomes
7x+ 10y= 1.01
5x+7y=0.69
Here we get x= − 0.17 and y= 0.22.
Hence the given system is ill-conditioned
ELIMINATION METHOD
The solution of system of equations will have some rounding error, we will discuss a technique called as ‘iterative refinement’ which leads to reduced rounding errors and often a reasonable solution for some ill-conditioned problems is obtained
Trang 2Consider the system of equations:
a x b y c z d
a x b y c z d
a x b y c z d
Let x′, y′, z′ be an approximate solution, Substituting these values on the left-hand sides, we
get new values of d1, d2, d3 as d′1, d′2, d′3, so their new system becomes;
a x b y c z d
a x b y c z d
a x b y c z d
′+ ′+ ′= ′
′+ ′+ ′= ′
Subtracting each equation in ( )2 from the corresponding equations in ( )1 , we get
a x b y c z k
a x b y c z k
a x b y c z k
where, x e = −x x y, e = −y y z′, e = −z z′ and k i= −d i d i′
We now solve the system ( )3 for x y z e, e, e giving z= +x′ x e, y= +y′ y e, z= +z′ z e which will be better approximations for x y z, , We can repeat the process for improving the accuracy
Example 4 An approximate solution of the system 2x + 2y – z = 6, x – y + 2z = 8; – x + 3y + 2z = 4 is given by x = 2.8, y′ = 1, z = 1.8 Using the iterative method improve this solution Sol Substituting the approximate value x′ = 2.8, y′ = 1, z′ = 1.8 in the given equations,
We get
( ) ( )
2 2.8 +2 1 −1.8 5.8=
( )
( ) ( )
−2.8 3 1+ +2 1.8 =3.8
Subtracting each equation in ( )1 from the corresponding given equations, we get
2x e+2y e− =z e 0.2
2 0.6
where x e = x – 2.8, y e = y –1, z e = z –1.8
Solving the equations (2), we get x e = 0.2, y e = 0, z e = 0.2
This gives the better solution x=3, y=1, z=2, which incidentally is the exact solution
Ans.
LINEAR EQUATION
All the previous methods seen in solving the system of simultaneous algebraic linear equations are direct methods Now we will see some indirect methods or iterative methods
Trang 3This iterative methods is not always successful to all systems of equations If this method
is to succeed, each equation of the system must possess one large coefficient and the large coefficient must be attached to a different unknown in that equation This condition will be satisfied if the large coefficients are along the leading diagonal of the coefficient matrix When this condition is satisfied, the system will be solvable by the iterative method The system,
a11 1x +a x12 2+a x13 3 =b1
a21 1x +a x32 2+a x23 3 =b2
a31 1x +a x32 2+a x33 3=b3
will be solvable by this method if
a11 > a12 + a13
a22 > a21 + a23
a33 > a31 + a32
In other words, the solution will exist (iterating will converge) if the absolute values of the leading diagonal elements of the coefficient matrix A of the system AX = B are greater than the sum of absolute values of the other coefficients of that row The condition is sufficient but not necessary
Under the category of iterative method, we shall describe the following two methods:
(i) Jacobi’s method (ii) Gauss-Seidel method.
8.6.1 Jacobi’s Method or Gauss-Jacobi Method
Let us consider the system of simultaneous equations
a1x b y c z+ 1 + 1 =d1
a3x b y c z+ 3 + 3 =d3
such that a1, ,b2 and c3 are the largest coefficients of x,y,z, respectively So that convergence is
assured Rearranging the above system of equations and rewriting in terms of x y z, , , as:
( 1 1 1 )
1
1
x d b y c z a
( 2 2 2 )
2
1
y d a x c z b
( 3 3 3 )
3
1
z d a x b y c
let x0, y0, z0 be the initial approximations of the unknowns x, y and z Then, the first approximation
are given by
1 ( 1 1 0 1 0)
1
1
x d b y c z a
1 ( 2 2 0 2 0)
2
1
b
Trang 41 ( 3 3 0 3 0)
3
1
c
Similarly, the second approximations are given by
2 ( 1 1 1 1 1)
1
1
x d b y c z a
2 ( 2 2 1 2 1)
2
1
y d a x c z b
2 ( 3 3 1 3 1)
3
1
z d a x b y c
Proceeding in the same way, if x n,y z n, n are the nth iterates then
1 ( 1 1 1 )
1
1
x d b y c z a
1 2 2 2
2
1
y d a x c z b
1 3 3 3
3
1
z d a x b y c
The process is continued till convergency is secured
Note: In the absence of any better estimates, the initial approximations are taken as x0 =0, y0=0,
z = .
Example 5 Solve the following system of equation using Jacobi’s method
5x y + z = 10−
2x + 4y = 12
x + y + 5z = 1−
Start with the solution (2, 3, 0).
Sol Given system of equation can be written in the folliwng form, if we assume, x0 , y0,
z0 as initial approximation:
1=1{10+ 0+ 0}
5
1 =1{12 2− 0}
4
1= 1{− −1 0− 0}
5
Now if x0 =2, y0 =3, z0 =0, then
First approximation: 1=1{10 3 0+ − =} 2.6
5
x
Trang 51= 1{12 4− }=2.0
4
y
1=1{− − −1 2 3}= −1.2
5
z
Second approximation: 2 = 1{10 2 1.2+ + }=2.64
5
x
2 { }
1
12 5.2 1.70 4
2 { }
1
1 2.6 2 1.12 5
z = − − − = −
Third approximation: 3 = 1{10 1.7 1.12+ + }=2.564
5
x
3 1{ }
12 5.28 1.680 4
3 = 1{− −1 2.64 1.7− }= −1.068
5
z
Fourth approximation: 4 = 1{10 1.68 1.068+ + }=2.5496
5
x
4 =1{12 5.128} 1.7180− =
4
y
5
Fifth approximation: x5 = 1
5{10 + 1.718 + 1.0428} = 2.553
y4 = 1
4{12 – 5.0992} = 1.725,
z5 = 1
5{–1 – 2.5496 – 1.718} = – 1.054
Hence, approximating solution after having some other approximations is (up to 3 decimal places)
x = 2.556
y = 1.725
z = –1.055 Ans.
Example 6 Solve the following system of equations by Jacobi iteration method.
3x + 4y + 15z = 54.8, x + 12y + 3z = 39.66 and 10x + y – 2z = 7.74 Sol The coefficient matrix of the given system is not diagonally dominant Hence we rearrange
the equations, as follows, such that the elements in the coefficient matrix are diagonally dominant
10x + y – 2z = 7.74
x + 12y + 3z = 39.66 3x + 4y + 15z = 54.8
Trang 6Now, we write the equations in the form
1 7.74 2 10
1 39.66 3 12
1 54.8 3 4 15
(1)
We start from an approximation = x0 = y0 = z0 = 0
Substituting these on RHS of (1), we get
First approximation:
x1= 1
10[7.74 – 0 + 2(0)] = 0.774
y1= 1
12[39.66 – 0 – 3(0)] = 1.1383333
z1= 1
15[54.8 – 3(0) – 4(0)] = 3.6533333
Second approximation:
x2= 1
10[7.74 – 1.1383333 + 2(3.6533333)] = 1.3908333
y2= 1
12[39.66 – 0.744 – 3(3.6533333)] = 2.3271667
z2= 1
15[54.8 – 3(0.744) – 4(1.1383333)] = 3.1949778
Third approximation:
x3= 1
10[7.74 – 2.3271667 + 2(3.1949778)] = 1.1802789
y3= 1
12[39.66 – 1.3908333 – 3(3.1949778)] = 2.3903528
z3= 1
15[54.8 – 3(1.3908333) – 4(2.3271667)] = 2.7545889
Fourth approximation:
x4= 1
10[7.74 – 2.3903528 + 2(2.7545889)] = 1.0858825
y4= 1
12[39.66 – 1.1802789 – 3(2.7545889)] = 2.5179962
z4= 1
15[54.8 – 3(1.1802789) – 4(2.3903528)] = 2.7798501
Fifth approximation:
x5= 1 7.74 2.5179962 2 2.7798501( ) 1.0781704
y5= 1 39.66 1.0858825 3 2.7798501( ) 2.5195473
z5= 1 54.8 3 1.0858825( ) (4 2.5179962) 2.7646912
Trang 7Sixth approximation:
6 1 7.74 2.5195473 2 2.7646912( ) 1.0749835
10
x = − + =
1 39.66 1.0781704 3 2.7646912 2.5239797 12
y = − − =
6 ( ) ( )
1 54.8 3 1.0781704 4 2.5195473 2.76582 15
z = − − =
Seventh approximation:
7 1 7.74 2.5239797 2 2.76582( ) 1.074766
10
7 = − − ( )=
1 39.66 1.0749835 3 2.76582 2.523963 12
y
7 ( ) ( )
1 54.8 3 1.0749835 4 2.5239797 2.7652754 15
z = − − = From the sixth and seventh approximations:
x=1.075, y=2.524and z=2.765 correct to three decimals Ans
8.6.2 Guass-Seidel Method
This is a modification of Gauss-Jacobi method As before, the system of the linear equations
a x b y c z+ + =d
a x b y c z+ + =d
a x b y c z+ + =d
is written as
( 1 1 1 )
1
1
x d b y c z a
( 2 2 2 )
2
1
y d a x c z b
( 3 3 3 )
3
1
z d a x b y c
and we start with the intial approximation x0, y0, z0 Substituting y0 and z0 in Eqn ( )1 , we get
1 ( 1 1 0 1 0)
1
1
x d b y c z a
Now substituting x=x z1, =z0 in Eqn ( )2 , we get
1 ( 2 2 1 2 0)
2
1
y d a x c z b
Substituting x=x1, y=y1 in Eqn ( )3 , we get
1 ( 3 3 1 3 1)
3
1
z d a x b y c
Trang 8This process is continued till the value of x y z, , , are obtained to the desired degree of
accuracy In general, kth iteration can be written as
1 ( 1 1 1 )
1
1
x d b y c z a
1 ( 2 2 1 2 )
2
1
y d a x c z b
1 ( 3 3 1 3 1)
3
1
z d a x b y c
The rate of convergence of Gauss-Seidel method is roughly twice that of Gauss-Jacobi method
Example 7 Solve by Gauss-Seidel iteration method the system of equations
8x – 3y + 2z = 20; 6x + 3y + 12z = 35 and 4x + 11y – z = 33.
Sol From the given equations, we have
1( )
20 3 2 8
1(33 4 )
11
1 ( )
35 6 3 12
Putting y=0, z=0 in RHS of ( )1 , we get 20 2.5
8
x= = Putting x=2.5, z=0 in RHS of ( )2 , we get
1 33 4 2.5( ) 2.0909091
11
y= − = Putting x=2.5, y=2.0909091 in RHS of ( )3 , we get
1 ( ) ( )
1
35 6 2.5 3 2.0909091 1.1439394 12
z = − − = For the second approximation:
2 ( 1 1)
1
20 3 2 8
x = + y − z
1 20 3 2.0909091( ) (2 1.1439394) 2.9981061
8
2 [ 2 1]
1
33 4 11
y = − x +z
1 33 4 2.9981061( ) 1.1439394 2.0137741
11
2 [ 2 2]
1
35 6 3 12
z = − x − y
1 35 6 2.9981061( ) (3 2.0137741) 0.9141701
12
Trang 9Third approximation:
3
1
20 3 2.0137741 2 0.9141701 3.0266228 8
x = + − =
3
1
33 4 3.0266228 0.9141701 1.9825163 11
y = − + =
3
1
35 6 3.0266228 3 1.9825163 0.9077262 12
z = − − =
Fourth approximation:
4
1
8
4
1
33 4 3.0165121 0.90777262 1.9856071 11
y = − + =
4
1
35 6 3.0165121 3 1.9856071 0.9120088 12
z = − − =
Fifth approximation:
5
1
8
5
1
33 4 3.0166005 0.9120088 1.9859643 11
y = − + =
5
1
35 6 3.0166005 3 1.9859643 0.9118753 12
z = − − =
Sixth approximation:
6
1
20 3 1.9859643 2 0.9118753 3.0167568 8
x = + − =
6
1
33 4 3.0167568 0.9118753 1.9858913 11
y = − + =
6
1
35 6 3.0167568 3(1.9858913 0.9118099 12
z = − − =
Seventh approximation:
x7 = 1
18[20 + 3(1.9858913) –2(0.9118099)] = 3.0167568
y7 = 1
11[33 – 4(3.0167568) + 0.9118099] = 1.9858894
z7 = 1
12[35 – 6(3.0167568) –3(1.9858894)] = 0.9118159
Since at the sixth and seventh approximations, the values of x y z, , , are the same, correct
to four decimal places, we can stop the iteration process
∴ x=3.0167, y=1.9858, z=0.9118
We find that 12 iteration are necessary in Gauss-Jacobi Method to get the same accuracy
as achieved by 7 iterations in Gauss-Seidel method
Trang 10Example 8 Solve the following system of equations using Gauss-Seidel method:
10x + y + 2z = 44
2x +10y + z = 51
x + 2y + 10z = 61 Sol Given system of equations can be written as:
1( )
10
x= − −y z
1 (51 2 )
10
y= − x z−
1 ( )
10
z= − −x y
If we start by assuming y0 = =0 z0 then, we obtain
1 ( )
1
44 0 0 4.4 10
Now we susbtitute x=4.4 and z0 =0 for y1 and we obtain
1 1 (51 8.8 0) 4.22
10
y = − − = Similarly, we obtain 1 1 (61 4.4 2 4.22) 4.816
10
Now for second approximation, we obtain
x2 =4.0154
y2 =3.0148
z2 =5.0955
Third approximation is given by
x3 =3.0794
y3 =3.9746
z3=4.9971
Similarly, if we proceed up to eighth approximation, then, we obtain
x8 =3.00
y8 =4.00
z8=5.00
PROBLEM SET 8.1
1
1 Apply Gauss-Elimination method to solve the system of equations
x+ y z− = −
3x y z− − =4 [Ans 117,
71
x= 81,
71
y=− 148
71
z= ]