5 Question in Linear Algebraplease see the attachment Note: i want (Norman Wagner) to solve the HW.thanks.

Sol: (1) The augmented matrix  ,C A B
1 1
3 3 1
2 2 1 2 2
0 0 0 1 1
2 1 1 1 1
1 1 0.5 1 1
1
0 0 0 1 1
2
2 1 1 1 1
1 1 0.5 1 1
0 0 0 1 1 2
0 1 0 1 1
1 1 0.5 1 1
0 1 0 1 1 Interchange the 2nd and 3rd
0 0 0 1 1
R R
R R R
 
 

 
  
 
 
 
 
  
 
 
  
 
    
 
 
   
 
  
2 2
2 2 3
1 1 3
1 1 2
row
1 1 0.5 1 1
0 1 0 1 1 2
0 0 0 1 1
1 1 0.5 1 1
0 1 0 0 0
0 0 0 1 1
1 1 0.5 0 0
0 1 0 0 0
0 0 0 1 1
1 0 0.5 0 0
0 1 0 0 0
0 0 0 1 1
R R
R R R
R R R
R R R
 
 
  
 
  
 
 
  
 
  
 
 
  
 
  
 
 
  
 
  

This is the reduced row echelon form.
The rank of C is equal 3 and the rank of A is also equal to 3.
     number of unknowns R A R C 
So its infinite numbers of solutions.
1 3
2
4
0.5 0
0
1
x x
x
x
 



3 Let , thenx k

1 2 40.5 , 0, 1x k x x   
So the solution set is
1 2 3 40.5 , 0, , 1,x k x x k x    
Sol: (3) (a) Vector equation,
1 2 3
1 2 4 3
3 1 5 9
5 0 10 15
x x x
       
       
    
       
               
(b) Matrix equation,
1 2 3
1 2 3
1 2 3
2 4 3
3 1 5 9
5 0 10 15
x x x
x x x
x x x
   
   
   
(c) First let
1 2 3
1 2 4
3 , 1 , 5
5 0 10
a a a
     
     
   
     
           

We want to find
1 2 3, , and x x x such that,

1 1 2 2 3 3b x a x a x a  
We will consider the augmented matrix,  1 2 3, , ,B a a a b
2 2
1 2 4 3
3 1 5 9
5 0 10 15
1 2 4 3
0 7 7 0 ...