Proof of an identity in determinants

[zorro]

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

 

[tiny-tim]

Hi Abdul!

(you can do sup within sup: Aⁿ² )

It's just a special case of the general rule detAB = detA*detB.

 

[zorro]

It's just a special case of the general rule detAB = detA*detB.

here A=|Aⁿ| and B=1
How does it get squared?

 

[tiny-tim]

Perhaps I'm misunderstanding the question …

what did you mean by ||Aⁿ|| ?

 

[zorro]

double determinant of Aⁿ... Is there any other meaning?

 

[zorro]

Any idea?

 

[Fredrik]

I'm not familiar with the term "double determinant". Can you define it?

 

[zorro]

ehh..there is no such identity
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 that's wrong.
I figured out that the above property is |kA|=kⁿ|A|, where n is the order of A.