Proof of an identity in determinants

  • Thread starter zorro
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  • #1
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How to prove that ||An||=|A|n2?

This property is used in my book but they did not give any explanation/proof of it.
Can someone help?

Edit: n2=n2
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Hi Abdul! :smile:

(you can do sup within sup: An2 :wink:)

It's just a special case of the general rule detAB = detA*detB. :smile:
 
  • #3
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It's just a special case of the general rule detAB = detA*detB. :smile:
here A=|An| and B=1
How does it get squared?
 
  • #4
tiny-tim
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Perhaps I'm misunderstanding the question :redface:

what did you mean by ||An|| ?
 
  • #5
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double determinant of An......... Is there any other meaning?
 
  • #6
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Any idea?
 
  • #7
Fredrik
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I'm not familiar with the term "double determinant". Can you define it?
 
  • #8
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ehh..there is no such identity :redface:
The step is actually |adj(adjA)|=||A|n-2A|=(|A|(n-2)n)|A|.
I thought it is goes like ||A|n-2|=|A|(n-2)n, but thats wrong.
I figured out that the above property is |kA|=kn|A|, where n is the order of A.
 

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