Given a $3\times 3$ matrix $A$, with $\det(A) = \frac{1}{8}$, find $\det(3A)$, $\det((6A)^{-1})$

Question

How do I go about solving this? The $\det(3A)$ is not simply $3\cdot\frac{1}{8}$ correct?

Answer 1

As matrix $A$ is $3\times3$, we have $\det{(kA)}=k^3\times\det{(A)}$ for $k \in \mathbb{C}$ $$\therefore\det{(3A)}=3^3\times\det{(A)}=3^3\times\frac{1}{8}=\frac{27}{8}$$ $$\det{(6A)^{-1}}=\bigl(\det{(6A)}\bigr)^{\!-1}=\bigl(6^3\times\det{(A)}\bigr)^{\!-1}=\Bigl(6^3\times\frac{1}{8}\Bigr)^{\mkern-5mu-1}=\frac{1}{27}$$