Unit – 2 | unit 2 solution of linear simultaneous equations

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NMLA

Unit – 2

Solution of linear simultaneous Equations

Q1) Apply Gauss Elimination method to solve the equations:

A1)

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

Q2) Solve the equation by Gauss Elimination Method:

A2)

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.

Q3) Apply Gauss Elimination Method to solve the following system of equation:

A3)

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

Q4) Solve the system by Gauss Elimination method using partial pivoting

A4)

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

Substituting value of y in equation (1) we get

Hence

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) Solve the equations-

A6)

Let

So that-

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.

Q7) What is Jacobi’s Iteration method?

A7)

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 .

Q8) Use Jacobi’s method to solve the system of equations:

A8)

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

Q9) Solve by Jacobi’s Method, the equations

A9)

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

0.90025

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

Hence solution approximately is

Q10) Use Jacobi’s method to solve the system of the equations

A10)

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

Q11) Use Gauss –Seidel Iteration method to solve the system of equations

A11)

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

Q12) Solve the following system of equations

By Gauss-Seidel method.

A12)

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Q13) Solve the following equations by Gauss-Seidel Method

A13)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Unit – 2

Solution of linear simultaneous Equations

Q1) Apply Gauss Elimination method to solve the equations:

A1)

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

Q2) Solve the equation by Gauss Elimination Method:

A2)

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.

Q3) Apply Gauss Elimination Method to solve the following system of equation:

A3)

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

Q4) Solve the system by Gauss Elimination method using partial pivoting

A4)

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

Substituting value of y in equation (1) we get

Hence

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) Solve the equations-

A6)

Let

So that-

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.

Q7) What is Jacobi’s Iteration method?

A7)

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 .

Q8) Use Jacobi’s method to solve the system of equations:

A8)

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

Q9) Solve by Jacobi’s Method, the equations

A9)

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

0.90025

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

Hence solution approximately is

Q10) Use Jacobi’s method to solve the system of the equations

A10)

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

Q11) Use Gauss –Seidel Iteration method to solve the system of equations

A11)

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

Q12) Solve the following system of equations

By Gauss-Seidel method.

A12)

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Q13) Solve the following equations by Gauss-Seidel Method

A13)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

Unit – 2

Solution of linear simultaneous Equations

Q1) Apply Gauss Elimination method to solve the equations:

A1)

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

Q2) Solve the equation by Gauss Elimination Method:

A2)

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.

Q3) Apply Gauss Elimination Method to solve the following system of equation:

A3)

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

Q4) Solve the system by Gauss Elimination method using partial pivoting

A4)

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

Substituting value of y in equation (1) we get

Hence

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) Solve the equations-

A6)

Let

So that-

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.

Q7) What is Jacobi’s Iteration method?

A7)

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 .

Q8) Use Jacobi’s method to solve the system of equations:

A8)

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

Q9) Solve by Jacobi’s Method, the equations

A9)

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

0.90025

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

Hence solution approximately is

Q10) Use Jacobi’s method to solve the system of the equations

A10)

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

Q11) Use Gauss –Seidel Iteration method to solve the system of equations

A11)

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

Q12) Solve the following system of equations

By Gauss-Seidel method.

A12)

Rewrite the given system of equations as

Let the initial approximation be

Thus the required solution is

Q13) Solve the following equations by Gauss-Seidel Method

A13)

Rewrite the above system of equations

Let the initial approximation be

Hence the required solution is

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