Question Bank

UNIT–4

Matrices

Question-1: check whether the following matrix A is symmetric or not?

                                 A =      

Sol. As we know that if the transpose of the given matrix is same as the matrix itself then the matrix is called symmetric matrix.

So that, first we will find its transpose,

Transpose of matrix A ,

Here,

                                   A =

So that, the matrix A is symmetric.

Question-2: Find the inverse of matrix ‘A’ if-

Sol.

Here we have-

And the matrix formed by its co-factors of |A| is-

And

Therefore-

We know that-

Question-3: Find the inverse of matrix ‘A’ by using elementary transformation-

A =

Sol.  Write the matrix ‘A’ as-  

A = IA

Apply

Apply

Apply

Apply

So that

  = 

Question-4: 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-5: reduce the matrix A to its normal form and find rank as well.

Sol.  We have,

We will apply elementary row operation,

We get,

Now apply column transformation,

We get,

Apply

  , we get,

Apply   and  

Apply   

Apply   and  

Apply   and  

As we can see this is required normal form of matrix A.

Therefore the rank of matrix A is 3.

Question-6: 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-7: 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-8: 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-9: Show that any square matrix can be expressed as the sum of symmetric matrix and anti- symmetric matrix.

Sol.  Suppose A is any square matrix .

Then,

A =

Now,

(A + A’)’ = A’ + A

A+A’ is a symmetric matrix.

Also,

(A - A’)’ = A’ – A

Here A’ – A is an anti – symmetric matrix

So that,

Square matrix = symmetric matrix + anti-symmetric matrix

Question-10: expand the determinant

Sol. As we know, expansion of the determinant is given by,

Then we get,

Question-11: prove that (without expanding) the determinant given below is equal zero.

Sol.  Applying the operation,

        We get,

Now applying the operation,

We get,

Apply,

We get,

Here first and second row are identical then we know that by property it becomes zero.             

Question-12: Are the vectors , , linearly dependent. If so, express x1 as a linear combination of the others.

Solution:

Consider a vector equation,

i.e.

Which can be written in matrix form as,

Here & no. Of unknown 3. Hence the system has infinite solutions. Now rewrite the questions as,

Put

and

Thus

i.e.

i.e.

Since F11 k2, k3 not all zero. Hence are linearly dependent.

Question-13: At what value of P the following vectors are linearly independent.

Solution:

Consider the vector equation.

i.e.

This is a homogeneous system of three equations in 3 unknowns and has a unique trivial solution.

If and only if Determinant of coefficient matrix is non zero.

consider .

.

i.e.

Thus for the system has only trivial solution and Hence the vectors are linearly independent.

Qustion-14: 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

Questtion-15: 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-16: Find the characteristic equation of the matrix A =   and Verify cayley-Hamlton theorem.

Sol. Characteristic equation of the matrix, we can be find as follows-

Which is,

( 2 -  , which gives

According to cayley-Hamilton theorem,

  …………(1)

Now we will verify equation (1),

Put the required values in equation (1) , we get

Hence the cayley-Hamilton theorem is verified.

Question-17: Using Cayley-Hamilton theorem, find , if A = ?

Sol. Let A =

The characteristics equation of A is

Or

Or

By Cayley-Hamilton theorem   

L.H.S.

=

By Cayley-Hamilton theorem we have   

            Multiply both side by

.

Or 

=

=

Question-18: Find the inverse of matrix A by using Cayley-Hamilton theorem.

                                            A =

Sol.  The characteristic equation will be,

|A - | = 0

= 

= (2-

= (2 -

=

That is,

Or

We know that by Cayley-Hamilton theorem,

…………………….(1)t,

Multiply equation(1) by , we get

Or

Now we will find 

=

=

Hence the inverse of matrix A is,