Page 108 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes (g) Find a basis for V.
(h) For what 5 × 1 column matrices Y does the equation AX = Y have solutions X?
To solve these problems we form the augmented matrix A’ of the system AX = Y and apply an
appropriate sequence of row operations to A’.

 1 2 0 3 0 y 1   1 2 0 3 0 y 1 
 1 2  1  1 0 y   0 0  1  4 0  y  y 
 2   1 2 
 0 0 1 4 0 y   0 0 1 4 0 y 3  
3 
   
 2 4 1 10 1 y 4   0 0 1 4 1  2y  y 4 
1
 0 0 0 0 1 y   0 0 0 0 1 y 
 5  5 
 1 2 0 3 0 y 1   1 2 0 3 0 y 1 
 0 0 1 4 0 y  y   0 0 1 4 0 y  y 
 1 2   1 2 
 0 0 0 0 0  y  y  y    0 0 0 0 1 y 5 
1
3
2
   
 0 0 0 0 1  3y  y  y 4   0 0 0 0 0  y  y  y 3 
1
2
2
1
 0 0 0 0 1 y   0 0 0 0 0  3y  y  y  y 
 5   1 2 4 5
(a) If
 y 1 
 y  y 
 1 2 
PY =  y 5 
 
  y  y  y 3 
1
2
  3y  y  y  y 
 1 2 4 5
for all Y, then
 1 0 0 0 0 
 1  1 0 0 0 
 

P = 0 0 0 0 1 
 
  1 1 1 0 0 
  3 1 0 1  1 
 
hence PA is the row-reduced echelon matrix

 1 2 0 3 0
 0 0 1 4 0 
 

R = 0 0 0 0 1
 
 0 0 0 0 0 
 0 0 0 0 0 
 
It should be stressed that the matrix P is not unique. There are, in fact, many invertible
matrices P (which arise from different choices for the operations used to reduce A’) such
that PA = R.
(b) As a basis for W we may take the non-zero rows
 = (1 2 0 3 0)
1

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