Your Federal Quarterly Tax Payments are due April 15th

# mth501 assignment Question and solutions by mudassar9

VIEWS: 2 PAGES: 5

vu mth501 assignment question and solution

• pg 1
```									                         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.
 Upload assignments properly through LMS, (No Assignment will be
accepted through email).
 All students are directed to use the font and style of text as is used in
this document.
 Don’t use colorful back grounds in your solution files.
 Use Math Type or Equation Editor etc for mathematical symbols.
 This is not a group assignment, it is an individual assignment so be
careful and avoid copying others’ work. If some assignment is found
to be copy of some other, both will be awarded ZERO MARKS. It
also suggests you to keep your assignment safe from others. No
excuse will be accepted by anyone if found to be copying or letting
others copy.

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.

```
To top