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.

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.
