Module-5

Linear algebra

Question-1: Find the rank of a matrix M by echelon form.

M = 

Sol. First we will convert the matrix M into echelon form,

M =

Apply, , we get

M =

Apply  , we get

M =

Apply 

M =

We can see that, in this echelon form of matrix, the number of non – zero rows is 3.

So that the rank of matrix X will be 3.

Question-2: Find the rank of the following matrices by echelon form?

Let A =

Applying

A 

Applying

A 

Applying

A 

      Applying

A 

It is  clear  that  minor  of order 3  vanishes but  minor  of order  2  exists  as 

Hence rank of a given matrix A is 2 denoted by 

Question-3: Find the solution of the following homogeneous system of linear equations,

Sol. The given system of linear equations can be written in the form of matrix as follows,

Apply the elementary row transformation,

  , we get,

    , we get

  Here r(A) = 4, so that it has trivial solution,

Question-4: check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

Sol. Write the above system of linear equations in augmented matrix form,

Apply , we get

Apply

Here the rank of C is 3 and the rank of A is 2

Therefore both ranks are not equal. So that the given system of linear equations is not consistent.

Question-5: Check the consistency and find the values of x , y and z of the following system of linear equations.

2x + 3y + 4z = 11

X + 5y + 7z = 15

3x + 11y + 13z = 25

Sol. Re-write the system of equations in augmented matrix form.

                                  C = [A,B]

That will be,

Apply 

Now apply ,

We get,

~~

Here rank of A = 3

And rank of C = 3, so that the system of equations is consistent,

So that we can can solve the equations as below,

That gives,

x + 5y + 7z = 15  ……………..(1)

y + 10z/7 = 19/7 ………………(2)

4z/7 = 16/7 ………………….(3)

From eq. (3)

z = 4,

From 2,

From eq.(1), we get

x + 5(-3) + 7(4) = 15

That gives,

x = 2

Therefore the values of x , y , z are    2 , -3 , 4  respectively.

Question-6: Solve by the Gauss-Jordan elimination method-

Sol. Here

The augmented matrix is-

Therefore, the matrix A is in normal form-

Question-7: solve the following system of linear equations by using Guass seidel method-

6x + y + z = 105

4x + 8y + 3z = 155

5x + 4y - 10z = 65

Sol. The above equations can be written as,

   ………………(1)

………………………(2)

   ………………………..(3)

Now put z = y = 0 in first eq.

We get

x = 35/2

Put x = 35/2 and z = 0 in eq. (2)

We have,

Put the values of x and y in eq. 3

Again start from eq.(1)

By putting the values of y and z

y = 85/8 and z = 13/2

We get

The process can be showed in the table format as below

At the fourth iteration , we get the values of x = 14.98 , y = 9.98  , z = 4.98

Which are approximately equal to the actual values,

As x = 15  , y = 10 and  y = 5 ( which are the actual values)

Question-8: Find out the Eigen values and Eigen vectors of ?

Sol.  The Characteristics equation is given by

                                          Or

Hence the Eigen values are 0,0 and 3.

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Here number of unknowns are 3 and number of equation is 1.

Hence we have (3-1) = 2 linearly independent solutions.

Let

Thus the Eigen vectors corresponding to the Eigen value are (-1,1,0) and (-2,1,1).

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Taking last two equations we get

Or

Thus the Eigen vectors corresponding to the Eigen value are (3,3,3).

Hence the three Eigen vectors obtained are  (-1,1,0), (-2,1,1) and (3,3,3).

Question-9: Find the Eigen values of Eigen vector for the matrix.

Solution:

Consider the characteristic equation as

i.e.

i.e.

are the required eigen values.

Now consider the equation

… (1)

Case I:

Equation (1) becomes,

Thus and n = 3

3 – 2 = 1 independent variables.

Now rewrite the equations as,

Put

,

I.e.

The Eigen vector for

Case II:

If equation (1) becomes,

Thus

Independent variables.

Now rewrite the equations as,

Put

Is the Eigen vector for

Now

Case III:-

If equation (1) gives,

R1 – R2

Thus

Independent variables

Now

Put

Thus

Is the Eigen vector for

Question-10: Diagonalise the matrix

Let A =

The Eigen vectors are (4,1),(1,-1) corresponding to Eigen values .

Then    and also 

Also we know that    

Module-5

Linear algebra

Question-1: Find the rank of a matrix M by echelon form.

M = 

Sol. First we will convert the matrix M into echelon form,

M =

Apply, , we get

M =

Apply  , we get

M =

Apply 

M =

We can see that, in this echelon form of matrix, the number of non – zero rows is 3.

So that the rank of matrix X will be 3.

Question-2: Find the rank of the following matrices by echelon form?

Let A =

Applying

A 

Applying

A 

Applying

A 

      Applying

A 

It is  clear  that  minor  of order 3  vanishes but  minor  of order  2  exists  as 

Hence rank of a given matrix A is 2 denoted by 

Question-3: Find the solution of the following homogeneous system of linear equations,

Sol. The given system of linear equations can be written in the form of matrix as follows,

Apply the elementary row transformation,

  , we get,

    , we get

  Here r(A) = 4, so that it has trivial solution,

Question-4: check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

Sol. Write the above system of linear equations in augmented matrix form,

Apply , we get

Apply

Here the rank of C is 3 and the rank of A is 2

Therefore both ranks are not equal. So that the given system of linear equations is not consistent.

Question-5: Check the consistency and find the values of x , y and z of the following system of linear equations.

2x + 3y + 4z = 11

X + 5y + 7z = 15

3x + 11y + 13z = 25

Sol. Re-write the system of equations in augmented matrix form.

                                  C = [A,B]

That will be,

Apply 

Now apply ,

We get,

~~

Here rank of A = 3

And rank of C = 3, so that the system of equations is consistent,

So that we can can solve the equations as below,

That gives,

x + 5y + 7z = 15  ……………..(1)

y + 10z/7 = 19/7 ………………(2)

4z/7 = 16/7 ………………….(3)

From eq. (3)

z = 4,

From 2,

From eq.(1), we get

x + 5(-3) + 7(4) = 15

That gives,

x = 2

Therefore the values of x , y , z are    2 , -3 , 4  respectively.

Question-6: Solve by the Gauss-Jordan elimination method-

Sol. Here

The augmented matrix is-

Therefore, the matrix A is in normal form-

Question-7: solve the following system of linear equations by using Guass seidel method-

6x + y + z = 105

4x + 8y + 3z = 155

5x + 4y - 10z = 65

Sol. The above equations can be written as,

   ………………(1)

………………………(2)

   ………………………..(3)

Now put z = y = 0 in first eq.

We get

x = 35/2

Put x = 35/2 and z = 0 in eq. (2)

We have,

Put the values of x and y in eq. 3

Again start from eq.(1)

By putting the values of y and z

y = 85/8 and z = 13/2

We get

The process can be showed in the table format as below

At the fourth iteration , we get the values of x = 14.98 , y = 9.98  , z = 4.98

Which are approximately equal to the actual values,

As x = 15  , y = 10 and  y = 5 ( which are the actual values)

Question-8: Find out the Eigen values and Eigen vectors of ?

Sol.  The Characteristics equation is given by

                                          Or

Hence the Eigen values are 0,0 and 3.

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Here number of unknowns are 3 and number of equation is 1.

Hence we have (3-1) = 2 linearly independent solutions.

Let

Thus the Eigen vectors corresponding to the Eigen value are (-1,1,0) and (-2,1,1).

The Eigen vector corresponding to Eigen value is

Where X is the column matrix of order 3 i.e.

This implies that

Taking last two equations we get

Or

Thus the Eigen vectors corresponding to the Eigen value are (3,3,3).

Hence the three Eigen vectors obtained are  (-1,1,0), (-2,1,1) and (3,3,3).

Question-9: Find the Eigen values of Eigen vector for the matrix.

Solution:

Consider the characteristic equation as

i.e.

i.e.

are the required eigen values.

Now consider the equation

… (1)

Case I:

Equation (1) becomes,

Thus and n = 3

3 – 2 = 1 independent variables.

Now rewrite the equations as,

Put

,

I.e.

The Eigen vector for

Case II:

If equation (1) becomes,

Thus

Independent variables.

Now rewrite the equations as,

Put

Is the Eigen vector for

Now

Case III:-

If equation (1) gives,

R1 – R2

Thus

Independent variables

Now

Put

Thus

Is the Eigen vector for

Question-10: Diagonalise the matrix

Let A =

The Eigen vectors are (4,1),(1,-1) corresponding to Eigen values .

Then    and also 

Also we know that