Matrices - One problem with two optional solutions, don't know which one is correct I have the following question:

Let $A$ be a $3\times 3$ matrix such that $|\text{adj}(3A)|=3$. Find $|A|$.

I solved the question in two different methods, but one method gave a solution which doesn't exist in the other one.
**First method:**$$|\text{adj}(3A)|=|3A|^2=|3A|\cdot{|3A|}=3^3|A|\cdot{3^3|A|}=3$$Hence, $|A|^2=3^5 \Rightarrow |A|=\pm{\frac{1}{3^{2.5}}}$.
Second method: It is known that $A\cdot\text{adj}(A)=|A|I$, hence $$3^3|A|\cdot3=|3A||\text{adj(3A)}|=|3A\cdot\text{adj}(3A)|=\Big||3A|I\Big|=\Big|3^3|A|I\Big|=3^9|A|^3$$hence, $3^4|A|\left(1-3^5|A|^2\right)=0$, i.e $|A|=0 \ \vee \ |A|=\pm\frac{1}{3^{2.5}}$.
Why doesn't the first method include the option $|A|=0$?
Edit: I think that A is invertible iff its adjugate is invertible. In this case, the option $|A|=0$ implies $|\text{adj} (3A)|=0$ which contradicts what is given. 
Is that correct? 
Where am I mistaken?
Please help, thanks.
 A: Both methods are fine, the first is simply more accurate.
Your first methods tells you either $|A|=  \frac{1}{3^{2.5}}$ or $|A|=-  \frac{1}{3^{2.5}}$.
The second method is weaker and only gives you
$|A|= 0$ or $|A|=  \frac{1}{3^{2.5}}$ or $|A|=-  \frac{1}{3^{2.5}}$.
However, by the first method you have shown that $|A| = 0$ is not possible. 
It would also be true to say for any matrix with  $det(arg(3A))=3$ you have either $|A|=12$ or   $|A|=  \frac{1}{3^{2.5}}$ or $|A|=-  \frac{1}{3^{2.5}}$.
Also one could argue that you have not solved the task completely with your first method. It remains to show if there exists a matrix with  $|A|=  \frac{1}{3^{2.5}}$ and $|arg(3A)|=3$; and if there exists a matrix with $|arg(3A)|=3$ and $|A|=-  \frac{1}{3^{2.5}}$ (Similarly, if you would have only used your second method, you should have checked if there exists a matrix with $det(arg(3A))=3$ and $|A|=0$).
Edit: To answer your question in the edit, it holds
$$ |adj(3A)| = |3^2 adj(A)| = 3^2|A|^2  .$$
Therefore $adj(A)$ is invertible if and only if $A$ is invertible. Therefire $|A|=0$ contradicts your assumption.
