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 of linear equations

A4)

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 .

Q5) Use Jacobi’s method to solve the system of equations:

A5)

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

Q6) Solve by Jacobi’s Method, the equations

A6)

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

Q7) Use Gauss –Seidel Iteration method to solve the system of equations

A7)

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

Q8) Solve the following system of equations

  By Gauss-Seidel method.

A8)

Rewrite the given system of equations as

Le t the initial approximation be

Thus the required solution is

Q9) Solve the system of equations

A9)

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

Q10) Find the inverse and determinant of

A10)

Here therefore exist.

On comparing the coefficient matrix with each column of RHS matrix.

From first term back substitution

We get 

From second and third we get