Unit - 2

Solution of linear simultaneous Equations

2.1 Direct Methods-Gauss Elimination Method

In this method we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to upper triangular system. At last with help of back substitution we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

To convert the above system into upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

                                         …          ……                …..         …

                                         …          ……                 ….         …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

Thus .

Then we calculate the values of .

Note: In (i) the coefficient is the pivot element and the equation is called the pivot equation. If then the above method fails and if it is close to zero the round off  error may occur.

If or very small compared   to other coefficient of the equation, then we find the largest available coefficient in the column given below the pivot equation and then interchange the two rows to obtain new pivot variable this is known as partial pivoting.

Example 1

Apply Gauss Elimination method to solve the equations:

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

Example 2:

Solve the equation by Gauss Elimination Method:

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.

Example 3:

Apply Gauss Elimination Method to solve the following system of equation:

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

Example 4: Solve the system by Gauss Elimination method using partial pivoting

Given exact solution is

Given equations are

Using partial pivoting we rewrite the given equations as

                                      (1)

                                   (2)

Using Gauss elimination method

   Multiplying (1) by (-0.0003120/0.5000) +  (2) we get

Or 

Substituting value of y in equation (1) we get

Hence

Example 5: Solve the system of linear equations

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 .

2.2 Method of Factorization

The method is based on the fact that every matrix A can be expressed as the product of a lower triangular matrix and an upper triangular matrix              , provided all the principal minors of A are non-singular.

Which means-

If , then-

Now consider the equations-

We can write it as-

Where-

Let

Where-

Equation (1) becomes-

Writing-

Equation (3) becomes-

which is equivalent to the equations-

Solving these for we know V. Then equation (4) becomes-

From which can be found by back substitution.

We write (2) as to find the matrix L and U-

Multiplying the matrix on the left and equating corresponding elements from both sides, we get-

3.    

4.    

5.    

We compute the elements of L and U in the following manner-

1. First row of U
2. First column of L
3. Second row of U
4. Second column of L
5. Third row of U

Example: Solve the equations-

Sol.

Let

So that-

3.    

4.    

5.    

So

Thus-

Writing UX = V,

The system of given equations become-

By solving this-

We get-

Therefore the given system becomes-

Which means-

By back substitution, we get the values of x, y and z.

2.3 Iterative Methods-Jacobi’s method, Gauss –Seidal Method

Jacobi’s Iteration method and Gauss-Seidal method:

Let us consider the system of simultaneous linear equation

The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get

Take the initial approximation we get the values of the first approximation of .

By the successive iteration we will get the desired the result.

Example 1 Use Jacobi’s method to solve the system of equations:

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

Example 2 Solve by Jacobi’s Method, the equations

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

Example 3 Use Jacobi’s method to solve the system of the equations

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

Gauss Seidel method:   

This is the modification of the Jacobi’s Iteration.  As above in Jacobi’s Iteration, we take first approximation as and put in the right hand side of the first equation of (2) and let the result be  . Now we put right hand side of second equation of (2) and suppose the result is now put in the RHS of third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to desired accuracy.

Example 1 Use Gauss –Seidel Iteration method to solve the system of equations

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

Example 2 Solve the following system of equations

    By Gauss-Seidel method.

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Example 3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

Unit - 2

Solution of linear simultaneous Equations

2.1 Direct Methods-Gauss Elimination Method

In this method we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to upper triangular system. At last with help of back substitution we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

To convert the above system into upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

                                         …          ……                …..         …

                                         …          ……                 ….         …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

Thus .

Then we calculate the values of .

Note: In (i) the coefficient is the pivot element and the equation is called the pivot equation. If then the above method fails and if it is close to zero the round off  error may occur.

If or very small compared   to other coefficient of the equation, then we find the largest available coefficient in the column given below the pivot equation and then interchange the two rows to obtain new pivot variable this is known as partial pivoting.

Example 1

Apply Gauss Elimination method to solve the equations:

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

Example 2:

Solve the equation by Gauss Elimination Method:

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.

Example 3:

Apply Gauss Elimination Method to solve the following system of equation:

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

Example 4: Solve the system by Gauss Elimination method using partial pivoting

Given exact solution is

Given equations are

Using partial pivoting we rewrite the given equations as

                                      (1)

                                   (2)

Using Gauss elimination method

   Multiplying (1) by (-0.0003120/0.5000) +  (2) we get

Or 

Substituting value of y in equation (1) we get

Hence

Example 5: Solve the system of linear equations

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 .

2.2 Method of Factorization

The method is based on the fact that every matrix A can be expressed as the product of a lower triangular matrix and an upper triangular matrix              , provided all the principal minors of A are non-singular.

Which means-

If , then-

Now consider the equations-

We can write it as-

Where-

Let

Where-

Equation (1) becomes-

Writing-

Equation (3) becomes-

which is equivalent to the equations-

Solving these for we know V. Then equation (4) becomes-

From which can be found by back substitution.

We write (2) as to find the matrix L and U-

Multiplying the matrix on the left and equating corresponding elements from both sides, we get-

3.    

4.    

5.    

We compute the elements of L and U in the following manner-

1. First row of U
2. First column of L
3. Second row of U
4. Second column of L
5. Third row of U

Example: Solve the equations-

Sol.

Let

So that-

3.    

4.    

5.    

So

Thus-

Writing UX = V,

The system of given equations become-

By solving this-

We get-

Therefore the given system becomes-

Which means-

By back substitution, we get the values of x, y and z.

2.3 Iterative Methods-Jacobi’s method, Gauss –Seidal Method

Jacobi’s Iteration method and Gauss-Seidal method:

Let us consider the system of simultaneous linear equation

The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get

Take the initial approximation we get the values of the first approximation of .

By the successive iteration we will get the desired the result.

Example 1 Use Jacobi’s method to solve the system of equations:

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

Example 2 Solve by Jacobi’s Method, the equations

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

Example 3 Use Jacobi’s method to solve the system of the equations

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

Gauss Seidel method:   

This is the modification of the Jacobi’s Iteration.  As above in Jacobi’s Iteration, we take first approximation as and put in the right hand side of the first equation of (2) and let the result be  . Now we put right hand side of second equation of (2) and suppose the result is now put in the RHS of third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to desired accuracy.

Example 1 Use Gauss –Seidel Iteration method to solve the system of equations

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

Example 2 Solve the following system of equations

    By Gauss-Seidel method.

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Example 3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

Unit - 2

Solution of linear simultaneous Equations

Unit - 2

Solution of linear simultaneous Equations

Unit - 2

Solution of linear simultaneous Equations

Unit - 2

Solution of linear simultaneous Equations

Unit - 2

Solution of linear simultaneous Equations

2.1 Direct Methods-Gauss Elimination Method

In this method we eliminate successively the unknown so that the equation (1) remain with the single unknown and reduce to upper triangular system. At last with help of back substitution we calculate the values of the remaining unknowns.

Consider a system of n linear equation in n unknown

To convert the above system into upper triangular matrix we eliminate from the second, third, fourth …., n equations above by multiplying the first equation by added them to the corresponding equations second, third, fourth,…., n equation. We get

                                         …          ……                …..         …

                                         …          ……                 ….         …

Repeating the above method for we get finally the upper triangular form.

Upper Triangular form of above

Thus .

Then we calculate the values of .

Note: In (i) the coefficient is the pivot element and the equation is called the pivot equation. If then the above method fails and if it is close to zero the round off  error may occur.

If or very small compared   to other coefficient of the equation, then we find the largest available coefficient in the column given below the pivot equation and then interchange the two rows to obtain new pivot variable this is known as partial pivoting.

Example 1

Apply Gauss Elimination method to solve the equations:

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

Example 2:

Solve the equation by Gauss Elimination Method:

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.

Example 3:

Apply Gauss Elimination Method to solve the following system of equation:

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

Example 4: Solve the system by Gauss Elimination method using partial pivoting

Given exact solution is

Given equations are

Using partial pivoting we rewrite the given equations as

                                      (1)

                                   (2)

Using Gauss elimination method

   Multiplying (1) by (-0.0003120/0.5000) +  (2) we get

Or 

Substituting value of y in equation (1) we get

Hence

Example 5: Solve the system of linear equations

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 .

2.2 Method of Factorization

The method is based on the fact that every matrix A can be expressed as the product of a lower triangular matrix and an upper triangular matrix              , provided all the principal minors of A are non-singular.

Which means-

If , then-

Now consider the equations-

We can write it as-

Where-

Let

Where-

Equation (1) becomes-

Writing-

Equation (3) becomes-

which is equivalent to the equations-

Solving these for we know V. Then equation (4) becomes-

From which can be found by back substitution.

We write (2) as to find the matrix L and U-

Multiplying the matrix on the left and equating corresponding elements from both sides, we get-

3.    

4.    

5.    

We compute the elements of L and U in the following manner-

1. First row of U
2. First column of L
3. Second row of U
4. Second column of L
5. Third row of U

Example: Solve the equations-

Sol.

Let

So that-

3.    

4.    

5.    

So

Thus-

Writing UX = V,

The system of given equations become-

By solving this-

We get-

Therefore the given system becomes-

Which means-

By back substitution, we get the values of x, y and z.

2.3 Iterative Methods-Jacobi’s method, Gauss –Seidal Method

Jacobi’s Iteration method and Gauss-Seidal method:

Let us consider the system of simultaneous linear equation

The coefficients of the diagonal elements are larger than the all other coefficients and are non zero. Rewrite the above equation we get

Take the initial approximation we get the values of the first approximation of .

By the successive iteration we will get the desired the result.

Example 1 Use Jacobi’s method to solve the system of equations:

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

Example 2 Solve by Jacobi’s Method, the equations

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

Example 3 Use Jacobi’s method to solve the system of the equations

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

Gauss Seidel method:   

This is the modification of the Jacobi’s Iteration.  As above in Jacobi’s Iteration, we take first approximation as and put in the right hand side of the first equation of (2) and let the result be  . Now we put right hand side of second equation of (2) and suppose the result is now put in the RHS of third equation of (2) and suppose the result be the above method is repeated till the values of all the unknown are found up to desired accuracy.

Example 1 Use Gauss –Seidel Iteration method to solve the system of equations

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

Example 2 Solve the following system of equations

    By Gauss-Seidel method.

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Example 3 Solve the following equations by Gauss-Seidel Method

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.