Page 95 - DMTH502_LINEAR_ALGEBRA

Unit 5: Summary of Row-Equivalence

Notes
 11  11
 0 0 1  3   0 0 1  3 
   
(2)
(2)
  1 0 0 17    1 0 0 17 
 3   3 
 1 7    5 
 0 1    0 1 0 
  2 2     3  
The row-equivalence of A with the final matrix in the above sequence tells us in particular that
the solutions of
AX = 0

i.e., 2x – x + 3x + 2x = 0
1 2 3 4
x + 4x – x = 0
1 2 4
2x + 6x – x + 5x = 0
1 2 3 4
and

11
x  x = 0
4
3
3
17
x  x = 0
1
4
3
5
x  x = 0
4
2
3
are exactly the same. In the second system it is apparent that
11
x  x
3 4
3
17
x   x 4
1
3
5
x  x
2 4
3
 17 5 11 
Thus if x = C then we obtain a solution  C , C , C ,C  and also that every solution is of

4  3 3 3 
this form.
Self Assessment

 3  1 2
 
1. If A  2 1 1 , find all solutions of AX = 0 by row-reducing A.
 
 1   3 0 


i   (1 i ) 0 
 
2. Find a row-reduced matrix which is row-equivalent to A  1  2 1
 
 1  2i  1 

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