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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
|
Vandermonde Determinant
this link
Last edited by George,Y (2006-04-06 12:52:48)
X'(y-Xβ)=0
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