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					                         Assignment No. 6
                        MTH 501 (Spring 2008)
                                                            Maximum Marks: 20
                                                    Due Date:- Tue: 28th July 2008

DON’T MISS THESE Important instructions:
    To solve this assignment, you should have good command over
     Lecture 33 to 37.
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   Question # 1:
                                                                        Marks=10
Use power method to find the eigenvalue and eigenvector for the given matrix starting
with x0 .(Do all necessary steps)
                   2       8     10         1
             A   8       3     4  , x  1
                                          o    
                  10
                           4      7        1
                                               
Perform 4 iterations without using Matlab.

Solution:
Iteration 1:
       2      8    10  1 2+8+10     20 
Ax 0 = 8
              3        1 = 8+3+4  = 15  ,
                    4                             u0 =21
       10
            4      7  1 10+4+7 
                                     21 
                                           
            20  0.95 
     Ax 0 1   
x1 =     = 15  = 0.71 
                         
     u0 21
            21  1
                      
                         

Iteration 2:

      2       8     10  0.95 
Ax1  8
              3      4  0.71 
                              
      10
              4      7  1
                              
                                
      1.9046  5.7136  10  17.61 
     7.6184  2.1426  4   13.76  , u1  19.3798
                                   
      9.523  2.8568  7  19.37 
                                   
                    17.6142  0.9088 
x2 
     Ax1
          
                1   13.761   0.7100 
      u1    19.3798                   
                               
                    19.3798  1
                                        
                                         


Iteration 3:
       2      8     10  0.9088 
Ax 2  8
             3       4  0.7100 
                                
       10
             4          1
                      7         
                                  
      1.8176  5.68  10  17.4976 
     7.2704  2.13  4   13.4004  , u2  18.928
                                   
      9.088  2.84  7  18.928 
                                   
            17.4976  0.9244 
        1                     
x3         13.4004   0.7079 
     18.928
            18.928   1
                              
                                 

Iteration 4:
       2      8     10  0.9244 
Ax 3  8
              3     4  0.7079 
                                
       10
              4     7  1
                                
                                  
      1.8488  5.6632  10  17.512 
     7.3952  2.1237  4   13.5189  ,
                                            u3  19.0756
      9.244  2.8316  7  19.0756 
                                     
              17.512  .9180 
x4 
         1    13.5189   .7087 
     19.0756                   
              19.0756   1 
                               

Iteration 5:
       2      8     10  .9180 
       8
Ax 4         3     4  .7087 
                               
       10
              4     7  1 
                               
      2.918  5.6696  10  18.5876 
     7.344  2.1261  4   13.4701  , u4  19.0148
                                    
      9.18  2.8348  7  19.0148 
                                    
              18.5876  .9775 
x5 
         1    13.4701   .7084 
     19.0148                    
                        
              19.0148   1 
                                  

Iteration 6:
       2      8     10  .9775 
Ax 5  8
              3     4  .7084 
                               
       10
              4         1 
                     7         
      1.955  5.6672  10  17.6222 
     7.82  2.1252  4   13.9452  , u5  19.6088
                                   
      9.775  2.8336  7  19.6088 
                                   
             17.6222  .8986 
x6 
        1    13.9452   .7111 
     19.6088                  
                       
             19.6088   1 
                                
Iteration 7:
       2      8     10  .8986 
       8
Ax 6         3     4  .7111 
                               
       10
              4     7  1 
                               
      1.7972  5.6888  10  17.486 
     7.1888  2.1333  4   13.3221  ,
                                                  u6  18.8304
      8.986  2.8444  7  18.8304 
                                     
              17.486  .9286 
x7 
         1    13.3221   .7074 
     18.8304                    
                        
              18.8304   1 
                                  




    Question # 2:
                                                                           Marks=10

Construct the general solution of x  Ax involving complex eigenfunctions and then
                                              3 1
obtain the general real solution. Where A          
                                              2 1
Solution:
Start with the trial function y(x)=μexp(-λx), μ a complex vector and λ a complex number.
Then after differentiation and rewriting the differential equation gets the form 0=(A-
λI)μexp(-λt). This is the well known equation for eigenvalues and eigenvectors of a
matrix. The general solution is the sum over the solution to the roots of the characteristic
equation and the eigenvectors thereof.

Characteristic equation for A: (3-λ)(1-λ)+2=0. λ1/2=2±i. The corresponding eigenvectors
are μ1=(-1-i,2) and μ2=(-1+i,2). So the complex solution is:


                                                                  .

Other way: x'=Ax has the solution y(x)=Exp(Ax) with the matrix exponential function.
The matrix exponential function is specially design to solve such linear differential
equations with constant coefficients.


ymatrix(t)=C                                                                      .
C a vector with complex entries. As is clearly seen the exponentials have the same
arguments, the solutions are linear depend of each other.

				
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mudassar samar mudassar samar mr http://www.niftye.com
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