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#1 2006-04-05 00:07:24

RauLiTo
Full Member

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a question in the DETERMINANTS

| X   X   1 +X |
| Y   Y   1+Y   | = 0             
| Z   Z   1+Z  |

that was given ...

prove that X Y Z = -1

help please guys i have an exam tomorrow neutral

Last edited by RauLiTo (2006-04-05 00:08:19)


ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

#2 2006-04-05 00:56:56

George,Y
Super Member

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Re: a question in the DETERMINANTS

decompenent the last colume, and reform
| X   X   1 +X |                |1  X   X |      |1  X   X|
| Y   Y   1+Y   | = -1(- 1)| 1  Y   Y | +XYZ| 1  Y   Y| = (1+XYZ)VD 
| Z   Z   1+Z  |                |1  Z   Z |         | 1  Z   Z |

VD -Vander monde Determinant
if x≠ y≠ z
xyz=-1

Last edited by George,Y (2006-04-05 00:57:31)


X'(y-Xβ)=0

#3 2006-04-05 02:47:02

RauLiTo
Full Member

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Re: a question in the DETERMINANTS

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 ...

#4 2006-04-05 16:05:07

George,Y
Super Member

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Re: a question in the DETERMINANTS

Reread the determinant properties, and then you will understand
|column1 column2 column3a+column3b|
=|column1 column2 column3a|+|column1 column2 column3b|


X'(y-Xβ)=0

#5 2006-04-07 10:44:25

George,Y
Super Member

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Re: a question in the DETERMINANTS

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
|

| = k|
|
|
|= - |
| = |
|

Vandermonde Determinant
 
this link

Last edited by George,Y (2006-04-07 10:52:48)


X'(y-Xβ)=0

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