Explain what it means for the determinant of the matrix, representing an operator $F$, to be independent of the basis of the vector space. Prove this property of the determinant.
I'm not exactly sure what the first bit is asking, but for the second bit, I though that if we let the matric corresponding to that determinant equal $B$, then by conjugation we would get $B^{-1}AB$ and so we get
$$\det(B^{-1}AB)= \det(B^{-1})\det(A) \det(B) = \det(A)$$
which shows that the determinant of $A$ is independent of the determinant of the matrix representing an operator.
Is this correct?
Let $F:V\rightarrow V$ be a linear operator on the $n$-dimensional vector space $V$. If $\mathcal A = \{a_1,\ldots,a_n\}, \mathcal B = \{b_1,\ldots,b_n\}, \mathcal C = \{c_1,\ldots,c_n\}$ and $\mathcal D = \{d_1,\ldots,d_n\}$ are 4 different bases for $V$, then we get two (most likely) different matrices representing $F$, $$[F]_{\mathcal A }^{\mathcal B}~~~\text{and}~~~[F]_{\mathcal C}^{\mathcal D}.$$
Here the $ij$-th entry of $[F]_{\mathcal A }^{\mathcal B}$ is the $j$-th coefficient of $F(a_i)$ when $F(a_i)$ is written as a linear combination of the basis $\mathcal B$. Similarly for the matrix $[F]_{\mathcal C}^{\mathcal D}$.
The question is wanting you to prove that $$\det[F]_{\mathcal A }^{\mathcal B}=\det[F]_{\mathcal C}^{\mathcal D}.$$