Question Bank (unit-6)

Question-1: :   Prove that the following matrix is orthogonal:

Let A  =   then  A’= 

Now,   AA’=

   Thus AA’=   = I

Hence the matrix is orthogonal.

Question-2: 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-3: Find the rank of a matrix A by echelon form.

                       A =

Sol. Convert the matrix A into echelon form,

                                 A =

Apply 

                                 A =

Apply    , we get

                                     A =

Apply , we get

                                   A =

Apply ,

                                   A =

Apply  ,

                                 A =

Therefore the rank of the matrix will be 2.

Question-4: Find the rank of a matrix A by echelon form.

                                 A =

Sol. Transform the matrix A into echelon form, then find the rank,

We have,

                        A =

Apply, 

                     A =

Apply ,

                       A =

Apply

                      A =

Apply

                      A =

Hence the rank of the matrix will be 2.

Question-5: Find the inverse of the matrix   by row transformation?

       Let  A=

     By Gauss-Jordan method

       We have

=

       Apply

           Apply we get

            Apply we get

             Apply

Apply

Hence

Question-6: : Find the inverse of the matrix   by row transformation?

Let A=

By Gauss-Jordan method

We have

              =

Apply we get

Apply

Apply

Apply

Apply

Apply

Hence

Question-7: Find the inverse of 

Let A=

By Gauss Jordan method

Apply

Apply

Apply 

Apply

Hence the inverse of matrix A is

Question-8: Find the inverse of

Let A=

By Gauss-Jordan Method [A:I]

                                        =

Apply

Apply 

Apply

Apply

Apply

Hence the inverse of matrix A is

Question-9: find the solution of the following linear equations.

+ - = 1

- + = 2

- 2 + = 5

Sol. These equations can be converted into the form of matrix as below-

Apply the operation, and     , we get

We get the following set of equations from the above matrix,

      + - = 1  …………………..(1)

          + 5 = -1  …………………(2)

                         = k

Put x = k  in eq. (1)

We get,

                   + 5k = -1

Put these values in eq. (1), we get

                                                                           and  

These equations have infinitely many solutions.

Question-10: solve the following system of equations by gauss-jordan method

+ - 2 = 0

+ + = 0

- 7 + = 0

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

By applying operation  -     we get,

Now apply,   

We get,

Apply,   

The set of the equations we get from above matrices,

+ + = 0 …………………..(1)

21 + = 0  …………………………..(2)

Suppose

From equation-2, we get

21

Now from equation-1:

- ()  + 4b + 9a = 0

We get,

Question-11: solve the following system of equations by gauss-jordan method

2x + y + 2z = 10

X + 2y + z = 8

3x + y – z = 2

Sol.  We will write the equations in matrix form as follows,

Now apply operation, 

Apply  , we get,

Apply,    , we get,

Apply,  / -3 , we get

Apply

Apply / -4, we get

Finally apply                   

Therefore the solution of the set of linear equations will be,

x = 1 , y = 2 , z = 3

Question-12: solve the following system of equations by gauss-jordan method.

6a + 8b +6c + 3d = -3

6a – 8b + 6c – 3d = 3

        8b           - 6d = 6

Sol. Change the system of linear equations into matrix form,

The augmented matrix format will be,

    Apply 

Apply

Apply 

  Apply    and

Apply 

Apply   

This is the reduced row echelon form,

The system of linear equations becomes,

a + c = 0

       b = 0

        d = -1

Suppose c = t be a free variable,

Then the solution will be, 

a = -t

b = 0

c = t

d = -1

For any number ‘t’.

Question-13: Find out the Eigen values and Eigen vectors of ?

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-14: Find out the Eigen values and Eigen vectors of

Let A =

The characteristics equation of A is .

                                           Or 

                                                Or 

                                                      Or

                                                       Or

The Eigen vector corresponding to Eigen value is

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

         Or

On solving we get 

Thus the Eigen vectors corresponding to the Eigen value is (1,1,1).

The Eigen vector corresponding to Eigen value is

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

Or    

On solving or .

Thus the Eigen vectors corresponding to the Eigen value is (0,0,2).

The Eigen vector corresponding to Eigen value is

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

   Or    

On solving we get    or .

Thus the Eigen vectors corresponding to the Eigen value is (2,2,2).

Hence three Eigen vectors are (1,1,1), (0,0,2) and (2,2,2).

Question-15: Verify the Cayley-Hamilton theorem and find the inverse.

The characteristics equation of A is

                                            Or

                                           Or

                                           Or

                                           Or

                                            Or

             By Cayley-Hamilton theorem

                L.H.S.

          =

=

=

Multiply both side with in

                                              Or

        Or

                     =

Question-16: :  Diagonalise the matrix

Let A=

The three Eigen vectors obtained are  (-1,1,0), (-1,0,1) and (3,3,3) corresponding to Eigen values .

Then    and

Also we know that    

                              Question Bank (unit-6)

                              Question Bank (unit-6)

Question-1: :   Prove that the following matrix is orthogonal:

Let A  =   then  A’= 

Now,   AA’=

   Thus AA’=   = I

Hence the matrix is orthogonal.

Question-2: 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-3: Find the rank of a matrix A by echelon form.

                       A =

Sol. Convert the matrix A into echelon form,

                                 A =

Apply 

                                 A =

Apply    , we get

                                     A =

Apply , we get

                                   A =

Apply ,

                                   A =

Apply  ,

                                 A =

Therefore the rank of the matrix will be 2.

Question-4: Find the rank of a matrix A by echelon form.

                                 A =

Sol. Transform the matrix A into echelon form, then find the rank,

We have,

                        A =

Apply, 

                     A =

Apply ,

                       A =

Apply

                      A =

Apply

                      A =

Hence the rank of the matrix will be 2.

Question-5: Find the inverse of the matrix   by row transformation?

       Let  A=

     By Gauss-Jordan method

       We have

=

       Apply

           Apply we get

            Apply we get

             Apply

Apply

Hence

Question-6: : Find the inverse of the matrix   by row transformation?

Let A=

By Gauss-Jordan method

We have

              =

Apply we get

Apply

Apply

Apply

Apply

Apply

Hence

Question-7: Find the inverse of 

Let A=

By Gauss Jordan method

Apply

Apply

Apply 

Apply

Hence the inverse of matrix A is

Question-8: Find the inverse of

Let A=

By Gauss-Jordan Method [A:I]

                                        =

Apply

Apply 

Apply

Apply

Apply

Hence the inverse of matrix A is

Question-9: find the solution of the following linear equations.

+ - = 1

- + = 2

- 2 + = 5

Sol. These equations can be converted into the form of matrix as below-

Apply the operation, and     , we get

We get the following set of equations from the above matrix,

      + - = 1  …………………..(1)

          + 5 = -1  …………………(2)

                         = k

Put x = k  in eq. (1)

We get,

                   + 5k = -1

Put these values in eq. (1), we get

                                                                           and  

These equations have infinitely many solutions.

Question-10: solve the following system of equations by gauss-jordan method

+ - 2 = 0

+ + = 0

- 7 + = 0

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

By applying operation  -     we get,

Now apply,   

We get,

Apply,   

The set of the equations we get from above matrices,

+ + = 0 …………………..(1)

21 + = 0  …………………………..(2)

Suppose

From equation-2, we get

21

Now from equation-1:

- ()  + 4b + 9a = 0

We get,

Question-11: solve the following system of equations by gauss-jordan method

2x + y + 2z = 10

X + 2y + z = 8

3x + y – z = 2

Sol.  We will write the equations in matrix form as follows,

Now apply operation, 

Apply  , we get,

Apply,    , we get,

Apply,  / -3 , we get

Apply

Apply / -4, we get

Finally apply                   

Therefore the solution of the set of linear equations will be,

x = 1 , y = 2 , z = 3

Question-12: solve the following system of equations by gauss-jordan method.

6a + 8b +6c + 3d = -3

6a – 8b + 6c – 3d = 3

        8b           - 6d = 6

Sol. Change the system of linear equations into matrix form,

The augmented matrix format will be,

    Apply 

Apply

Apply 

  Apply    and

Apply 

Apply   

This is the reduced row echelon form,

The system of linear equations becomes,

a + c = 0

       b = 0

        d = -1

Suppose c = t be a free variable,

Then the solution will be, 

a = -t

b = 0

c = t

d = -1

For any number ‘t’.

Question-13: Find out the Eigen values and Eigen vectors of ?

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-14: Find out the Eigen values and Eigen vectors of

Let A =

The characteristics equation of A is .

                                           Or 

                                                Or 

                                                      Or

                                                       Or

The Eigen vector corresponding to Eigen value is

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

         Or

On solving we get 

Thus the Eigen vectors corresponding to the Eigen value is (1,1,1).

The Eigen vector corresponding to Eigen value is

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

Or    

On solving or .

Thus the Eigen vectors corresponding to the Eigen value is (0,0,2).

The Eigen vector corresponding to Eigen value is

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

   Or    

On solving we get    or .

Thus the Eigen vectors corresponding to the Eigen value is (2,2,2).

Hence three Eigen vectors are (1,1,1), (0,0,2) and (2,2,2).

Question-15: Verify the Cayley-Hamilton theorem and find the inverse.

The characteristics equation of A is

                                            Or

                                           Or

                                           Or

                                           Or

                                            Or

             By Cayley-Hamilton theorem

                L.H.S.

          =

=

=

Multiply both side with in

                                              Or

        Or

                     =

Question-16: :  Diagonalise the matrix

Let A=

The three Eigen vectors obtained are  (-1,1,0), (-1,0,1) and (3,3,3) corresponding to Eigen values .

Then    and

Also we know that