Given $\operatorname{adj}A=\begin{bmatrix} -1 & -2 & 1\\ 3 & 0 & -3 \\ 1 & -4 & 1 \end{bmatrix}$ . Find $A$.
My Attempt
We know that $|\operatorname{adj}A|=|A|^{n-1}\Rightarrow |A|=\pm\sqrt{|\operatorname{adj}A|}=\pm2\sqrt3$.
Also it is well known that $A(\operatorname{adj}A)=|A| I\Rightarrow A=|A|(\operatorname{adj} A)^{-1}=\pm\frac{1}{2\sqrt3}\begin{bmatrix} -12 & -2 & 6\\ 6 & -2 & 0 \\ -12 & 6 & 6 \end{bmatrix}$
My doubt is that should there be a unique $A$ or there are two possibilities as shown in working above.