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| X   X²   1 +X³ |
| Y   Y²   1+Y³   | = 0             
| Z   Z ²  1+Z³  |
that was given ...
prove that X Y Z = -1
help please guys i have an exam tomorrow ![]()
Last edited by RauLiTo (2006-04-04 02:08:19)
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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decompenent the last colume, and reform 
| X   X²   1 +X³ |                |1  X   X² |      |1  X   X²|
| Y   Y²   1+Y³   | = -1(- 1)| 1  Y   Y² | +XYZ| 1  Y   Y²| = (1+XYZ)VD  
| Z   Z ²  1+Z³  |                |1  Z   Z ²|         | 1  Z   Z ²|
VD -Vander monde Determinant
if x≠ y≠ z
xyz=-1
Last edited by George,Y (2006-04-04 02:57:31)
X'(y-Xβ)=0
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thank you very much ... but unfortunately i dind't understand it well ... can someone explain it for me ? !
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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Reread the determinant properties, and then you will understand
|column1 column2 column3a+column3b|
=|column1 column2 column3a|+|column1 column2 column3b|
X'(y-Xβ)=0
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Sorry you may not find these properties in your text book
|a  b  c|  |a  b' c|    |a  b+b'  c|
|d  e  f|+|d  e'  f| = |d  e+e'  f|
|g  h  i|   |g  h'  i|     |g  h+h'  i|
see the bottom of this link
other properties used
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Vandermonde Determinant 
  
this link
Last edited by George,Y (2006-04-06 12:52:48)
X'(y-Xβ)=0
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