Page 98 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes RX = Z effectively express x ,...,x in terms of the (n – r) remaining x and the scalars z ,...,z . The
k1 kr j 1 r
last (m – r) equations are
0  z
r 1
 
0  z m
and accordingly the condition for the system to have a solution is z = 0 for i > r. If this condition
i
is satisfied, all solutions to the system are found just as in the homogeneous case, by assigning
arbitrary values the (n – r) of the x and then computing x from the ith equation.
j ki

Example 5: Let F be the field of rational numbers and

 1  2 1 
 
A  2 1 1
 
 0  5  1 
and suppose that we wish to solve the system AX = Y for some y , y and y . Let us perform a
1 2 3
sequence of row operations on the augmented matrix A’ which row-reduces A:

 1  2 1 y 1   1  2 1 y 1 
 2 1 1 y    0 5  1 (y  2 ) 

(2)
(2)
y
 2   2 1 
3
 0  5  1 y   0  5  1 y 3  
 1  2 1 y 1 
 1  2 1 y 1   
 0 5  1 (y  2 )   0  1  1 1 (y  2 )  
(1)
(2)
y
y
 2 1   5 5 2 1 
y
 0  0 0 (y  y  2 )  0 0 0 (y  y  2 ) 
1 
y
2
3
 3 2 1 
 1 0 3 1 (y  2y ) 
 5 5 1 2 
 
 0 1  1 1 (y  2 ) 
y
 5 5 2 1 
 0 0 0 (y  y  2 ) 
y
 3 2 1 
   
The condition that the system AX = Y have a solution is thus
2y – y + y = 0
1 2 3
and if the given scalars y satisfy this condition, all solutions are obtained by assigning a value
i
c to x and then computing
3
3 1
x =  c  (y  2y )
1 1 2
5 5
1 1
x = c  (y  2y )
2 2 1
5 5
Let us observe one final thing about the system AX = Y. Suppose the entries of the matrix A and
the scalars y ,...y happen to lie in a subfield F of the field F. If the system of equations AX = Y
1 m 1
has a solution with x ,...,x in F, it has a solution with x ,...,x in F . For, over either field, the
1 n 1 n 1
condition for the system to have a solution is that certain relations hold between y ,...,y in F
1 m 1
92 LOVELY PROFESSIONAL UNIVERSITY