Unit - 2

Rate of convergence of the above methods

Q1) What is the rate of convergence for bisection and Regula-falsi method?

A1)

Bisection Mehtod: The percentage error    defined by

The rate  of convergence for  bisection method is  

i.e.  it  converges linearly.

Regula-Falsi Method: The rate of convergence is 

Here p=1.618  and  is linear.

Q2) Apply Gauss Elimination method to solve the equations)

A2)

Given                                                    Check Sum (sum of coefficient and constant)

                        -1                            …. (I)

                      -16                          …. (ii)

                          5 …. (iii)

(I)We eliminate x from (ii) and (iii)

Apply eq(ii)-eq(i) and eq(iii)-3eq(i) we get

                        -1                            ….(i)

                      -15                          ….(iv)

                          8            ….(v)

(II) We eliminate  y from eq(v)

Apply

                        -1                            ….(i)

                      -15                          ….(iv)

                      73            ….(vi)

(III) Back Substitution we get

From (vi) we get

From (iv) we get

From (i) we get 

Hence the solution of the given equation is

Q3) Solve the equation by Gauss Elimination Method)

A3)

Given

Rewrite the given equation as

                 … (i)

                             ….(ii)

                             ….(iii)

                                 …(iv)

(I)                We eliminate x from (ii),(iii) and (iv) we get

Apply eq(ii) + 6eq(i), eq(iii) -3eq(i), eq(iv)-5eq(i) we  get

                 …(i)

                             ….(v)

                       ….(vi)

                             …(vii)

(II)             We eliminate y   from (vi) and (vii) we get

Apply 3.8 eq(vi)-3.1eq(v) and 3.8eq(vii)+5.5eq(v) we get

                 …(i)

                             ….(v)

                         …(viii)

                             …(ix)

(III)           We eliminate z from eq (ix) we get

Apply 9.3eq (ix) + 8.3eq (viii), we get

                 … (i)

                             ….(v)

                         …(viii)

350.74u=350.74

Or u = 1

(IV)          Back Substitution

From eq(viii)

Form eq(v), we get

From eq(i) ,

Hence the solution of the given equation is x=5, y=4, z=-7 and u=1.

Q4) Apply Gauss Elimination Method to solve the following system of equation)

A4)

Given                    … (i)

              … (ii)

              … (iii)

(I)                We eliminate x from (ii) and (iii)

Apply we get

                … (i)

              … (iv)

             … (v)

(II)             We eliminate y from  (v)

Apply we get

                … (i)

              … (vi)

                      … (vii)

(III)           Back substitution 

From (vii)  

From (vi)

From (i)

Hence the solution of the equation is

Q5) Solve the system of linear equations

A5) Using partial pivoting  by Gauss elimination method we rewrite the given equations as

     (1)

        (2)

     (3)

Apply    and

     (1)

   (4)                       

              (5)

Apply )

     (1)

   (4)

Or     .

Putting value of z in we get .

Putting values of y and z in we get .

Hence the solution of the equation is .

Q6) By using Gauss-Jordan method, solve the system of equations-

A6)

Here the augmented matrix will be-

By performing elementary row transformation, and eliminations, we get-

Now making the pivots as 1, ((

We obtain,

Hence the solution of the system is-

Q7) What do you understand by matrix inversion using Gauss Jordan method?

A7)

We know that X will be the inverse of a matrix A if

Where I is an identity matrix of order same as A.

We write, using elementary operation we covert the matrix A in to a upper triangular  matrix. Then compare this matrix with each corresponding column and using back substitution. s

The determinant of this coefficient matrix is  the product of the diagonal elements  of the upper triangular matrix.

If this |A|=0 then inverse does not exist.

Q8) Solve the system of equations

A8)

Here A =       

The augmented system is 

Apply

Apply

Here 

Which is non zero so the inverse  exists.

Comparing the diagonal matrix with corresponding columns.

By back Substitution we get

Form first part of above           

Similarly we solve second and third part and obtain a matrix whose columns are

Which is the required inverse

Q9) Use Jacobi’s method to solve the system of equations)

A9)

Since

So, we express the unknown with large coefficient in terms of other coefficients.

Let the initial approximation be

2.35606

0.91666

1.932936

0.831912

3.016873

1.969654

3.010217

1.986010

1.988631

0.915055

1.986532

0.911609

1.985792

0.911547

1.98576

0.911698

Since the approximation in ninth and tenth iteration is same up to three decimal places, hence the solution of the given equations is

Q10) Solve by Jacobi’s Method, the equations

A10)

Given equation can be rewrite in the form

     … (i)

    ..(ii)

    ..(iii)

Let the initial approximation be

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

0.90025

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Putting these values on the right of the equation (i), (ii) and (iii) and so we get

Hence   solution approximately is

Q11) Use Jacobi’s method to solve the system of the equations

A11)

Rewrite the given equations

Let the initial approximation be

1.2

1.3

0.9

1.03

0.9946

0.9934

1.0015

Hence the solution of the above equation correct to two decimal places is

Q12) Use Gauss –Seidel Iteration method to solve the system of equations

A12)

Since

So, we express the unknown of larger coefficient in terms of the unknowns with smaller coefficients.

Rewrite the above system of equations

Let the initial approximation be

3.14814

                                                                   2.43217

2.42571

2.4260

Hence the solution correct to three decimal places is

Q13) Solve the following system of equations

  By Gauss-Seidel method.

A13) Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Q14) Solve the following equations by Gauss-Seidel Method

A14)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Q15) Explain power method.

A15)

Consider the eigen value problem

The Eigen values of a matrix A are given by the roots of the characteristic equation

If the matrix A is of order n, then expanding the determinant, we obtain the characteristic equation as

For the given matrix we write the characteristic equation, by expanding we find the roots . These are called the Eigen values.

Now suppose be the solution of the system of homogeneous equations, corresponding to the Eigen value

These vectors are called the Eigen vectors.

To find the approximate values of all the Eigen values and Eigen vectors, iteration method or power method is used.

Power method is used when only the largest and/or the smallest Eigen values of a matrix are desired.

Q16) Find the largest Eigen value and the corresponding Eigen vector of the matrix

Also find the error in the value of the largest Eigen value.

A16)

Let us choose the initial vector

Then

and

Now put , then-

Hence the largest Eigen value is-

And the corresponding Eigen vector is-

The error can be calculated as-

Q17) Give the steps used in power method.

A17)

Step-1: First we choose an arbitrary real vector , basically is chosen as-

Step-2: Compute , , , , ………… Put

Step-3: Compute , ,

Step-4: The largest Eigen value is

The error in can be find as-

The Eigen vector corresponding to is