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I wanted to prove that det(adj(A))=det(A)^n-1 for an nxn matrix A.

I separate the proof in two cases: singular and non-singular matrix A.

For the non-invertible A, det(A)=0. In my head, I know that adj(A)=0, but I cannot prove it. How does one proceed?

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    $\begingroup$ "I know that $\operatorname{adj}(A)=0$, but I cannot prove it." Hmm ... of course you cannot prove it, because your assertion is wrong. What is true is that $\det(\operatorname{adj}(A))=0$. $\endgroup$ – user1551 Nov 24 '14 at 9:28
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We have

$$A \operatorname{adj}(A)=\det (A)I_n$$ so if $\det(\operatorname{adj}(A))\ne0$ which means that it's invertible then we get $A=0$ but in this case $\operatorname{adj}(A)=0$ which's a contradiction. Conclude.

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