# If $\det(\mathrm{adj}\,A) \ne 0$, then $\det(A) \ne 0$.

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...

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Is $A$ a $5\times 5$ matrix? –  Michael Albanese Oct 13 '12 at 14:22
What is $\mathrm{adj} A$? –  userNaN Oct 13 '12 at 14:31
–  anon Oct 13 '12 at 14:38
@anon thanks!${}$ –  userNaN Oct 13 '12 at 14:40

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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I don't understand why did you set |B| = 0. My question may not be clear. I mean if we know that |adjA| isn't zero how can we assert that |A| isn't zero. Now in your proof you set |B|=|adjA|=0 and you show that |A|=0. how does it help me? –  SyndicatorBBB Oct 13 '12 at 15:13
@user44471: You should change "equal" to "not equal" in the title then, so that the title matches the actual question. –  Hans Lundmark Oct 13 '12 at 15:29
Yes my mistake... sorry. –  SyndicatorBBB Oct 13 '12 at 15:48

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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