A = 
Write the matrix ‘A’ as A = IA Apply , we get Apply Apply Apply Apply So that, = 
M = 
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. 
A = 
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. 
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. 
2x + y + 2z = 0 x + y + 3z = 0 4x + 3y + bz = 0 Which has (1) Trivial solution (2) Nontrivial solution 
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 nontrivial solution (2) Convert the system of equations into matrix form AX = O Apply Respectively , we get the following resultant matrices
For nontrivial solutions, r(A) = 2 < n b – 8 = 0 b = 8 
Rewrite 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. 
Here we have And here Now by using cramer’s rule 
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) 
Applying We get,
Example: Solve Sol: given Apply We get 
Are the vectors , , linearly dependent. If so, express x1 as a linear combination of the others. 
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. 
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

A = 
Characteristic equation will be = 0 ( 7  (7 (7 Which gives, Or According to cayleyHamilton theorem, …………………….(1) In order to verify cayleyHamilton theorem , we will find the values of So that, Now
Put these values in equation(1), we get = 0 Hence the cayleyhamilton theorem is verified. 
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Let A = The characteristics equation of A is Or Or Or By CayleyHamilton theorem L.H.S: = =0=R.H.S Multiply both side by on Or Or [ Or 