Page 169 - Buku Aljabar Linear & Matriks

Solusi:

u ) 1 , 1 , 1 ( 1 1 1
1) v 1 = 1 = = ( , , )
u 1 3 3 3 3

2) u2 – proyw1 u2 = u2 –  u2, v1  v1
2 1 1 1 2 1 1
( 1 ,0 = ) 1 , − ( , , ) = ( − , , )
3 3 3 3 3 3 3

u − proy u 3 2 1 1 2 1 1
v = 2 w 2 = (− , , )= (− , , )
2
u 2 − proy w u 2 6 3 3 3 6 6 6

3) u3 – proyw2 u3 = u3 –  u3, v1  v1 –  u3, v2  v2

1
=  − , 1 
 ,0

 2 2 

u − u ,v v − u ,v v  1 1 
v 3 = 3 3 1 1 3 2 2 =  −,0 , 
u 3 − u 3 ,v 1 v 1 − u 3 ,v 2 v 2  2 2 

Jadi:

 1 1 1 
v 1 =  , , 


 3 3 3 
 2 1 1  Membentuk basis
v 2 = − , , 


 6 6 6  orthonormal untuk
3
 1 1  R
v 3 =  ,0 − , 
 2 2 

160 | R u a n g - r u a n g V e k t o r