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I would like to understand how the following fact : detA⋅I n ⋅adj(adj(A))=(detA) n−1 ⋅A

allows us to determine/assert that if A isn't invertible then (adj(adj(A)) = 0.

If A isn't invertible I can't devide by detA.....

Thanks , Guy

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    What does your question refer to? Do you have a link?2012-10-20
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    http://math.stackexchange.com/questions/92837/proof-mathrmadj-mathrmadja-mathrmdetan-2-cdot-a-for-a/92842#928422012-10-20
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    I could understand all the proof but I don't understand how it helps to assert that if A isn't invertible then adj(adj(A)) = 0.2012-10-20

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As you already noted: This is not possible but it is not claimed in the link you provided.

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    It is claimed in the comments. Do you mean It's not the right way to prove it?2012-10-20
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    I might confuse you. I want to proof that if A isn't invertible then adj(adj(A)) is zero. Isn't the link I provided is the right way to do so>?2012-10-20
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    No it is not, and as I said no one claims it there. Just read the comments, can't find it.2012-10-20
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    It is so weird because the following fact was written by my teacher : If A has rank n-1 then adj(adj(A)) = 0.2012-10-20
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    The claim is right, but what you thought of as a proof is not.2012-10-20
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    Oh...... Can you please give me a direction to prove it ?2012-10-20
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    There is one attempt in the components of the link you provided.2012-10-20
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6191/discussion-between-julian-kuelshammer-and-guy)2012-10-20