How to find the determinant of a scaled adjacency matrix?

Question

If $A$ is a $3 \times 3$ matrix and $\det(A) = -1$, what would $\det(\frac{1}{3} \operatorname{adj}(3A))$ be?

I know that $\det(\operatorname{adj}(A)) = \det(A)^{n-1}$, so I guessed it would be $(27 \times -1)^2$, but I can't seem to figure out where that $1/3$ goes.

I apologize for the poor formatting. I am still learning how to do them correctly.

Answer 1

Let $B=\operatorname{adj}(3A) $ and $C= \frac{1}{3} \operatorname{adj}(3A)$ $$ \det(C) =\det\left(\frac{B}{3}\right) =\frac{1}{3^3}\det(B)$$

As you've mentioned already, $\det(\operatorname{adj}(A)) = \det(A)^{n-1}$ $$ \det(B) =\det(\operatorname{adj}(3A))=(\det(3A))^{3-1}=(-27)^2=3^6$$ Therefore, $$ \det\left(\frac{1}{3} \operatorname{adj}(3A)\right) =\det(C)=\frac{1}{3^3} \cdot3^6=3^3=\color{red}{27}$$

Answer 2

Since $\det(A)=-1\ne0$, $A$ is invertible. Hence \begin{aligned} \det\left(\frac13\operatorname{adj}(3A)\right) &=\det\left(\frac13\det(3A)(3A)^{-1}\right)\\ &=\det\left(\frac133^3\det(A)\frac13A^{-1}\right)\\ &=\det\left(-3A^{-1}\right)\\ &=(-3)^3\det(A)^{-1}\\ &=27. \end{aligned}