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...

Appreciate your help.

Solution 1:

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

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

We're done, since

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

Solution 2:

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.