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I'm trying to understand the reason why $A$ is invertible only if $\mathrm{adj}\,A$ is invertible.

That's what I have right now: $A\, \mathrm{adj}\,A = |A|\cdot I$.

So if we take $\det$ of both sides we get: $|A\,\mathrm{adj}\,A| = ||A|\cdot I|$

and then: $|A| \cdot |\mathrm{adj}\,A| = |A|^n$

but now I'm stuck...

Appreciate your help.

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    @anon thanks!${}$2012-10-13

2 Answers 2

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Suppose $\det{(\operatorname{adj}{A})} \neq 0$ but $\det{A} = 0$. Since $A \operatorname{adj} A = 0$ and $\operatorname{adj}{A}$ is invertible, we have $A=0$, so $\operatorname{adj}{A} = 0$, giving $\det{(\operatorname{adj}{A})} = 0$ which is a contradiction.

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For simplicity put $\,B:=adj\, A\,$ , so:

$AB=|A|\cdot I\Longrightarrow |A||B|=|A|^n$

We're done, since

$|B|=0\Longrightarrow |A|^n=0\Longrightarrow |A|=0$

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    Yes my mistake... sorry.2012-10-13