A1)

Write the matrix ‘A’ as-  

A = IA

Apply , we get

Apply

Apply

Apply

Apply

So that,

  =

A2)

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.

A3)

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.

A4)

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.

A5)

For trivial solution, we already know that the values of x , y and z will be zerp, so that ‘b’ can have any value.

Now for non-trivial solution-

(2)

Convert the system of equations into matrix form-

AX = O

Apply Respectively , we get the following resultant matrices

For non-trivial solutions, r(A) = 2 < n

b – 8 = 0

b = 8

A6)

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.

A7)

Here we have-

And here-

Now by using cramer’s rule-

A8)

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)

A9)

Applying

          We get,

Example: Solve-

Sol:

given-

Apply-

We get-

A10)

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 F11k2, k3 not all zero. Hence are linearly dependent.

A11)

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    

A12)

Characteristic equation will be-

= 0

( 7 -

(7-

(7-

Which gives,

Or

According to cayley-Hamilton theorem,

   …………………….(1)

In order to verify cayley-Hamilton theorem , we will find the values of 

So that,

Now

Put these values in equation(1), we get

= 0

Hence the cayley-hamilton theorem is verified.

A13)

Let A =

The characteristics equation of A is

Or

Or 

Or 

By Cayley-Hamilton theorem 

                                   L.H.S:

                                                 =   =0=R.H.S

Multiply both side by on 

                                               Or

                                              Or [

                                               Or