Let $\displaystyle a_1,\cdots , a_n$ be given numbers. Compute the determinant of the nxn matrix $\displaystyle A=(a_{ij})$, where $\displaystyle a_{ij}=a^{i-1}_j$.
I don't understand what is meant by this: $\displaystyle a_{ij}=a^{i-1}_j$
without being given any more context, i would assume it means $\displaystyle a_{ij}$ is the (i-1)-th power of $\displaystyle a_j$. so the first row should all be 1's. this kind of matrix is called a Vandermonde matrix.
you'll probably arrive at a better understanding if you try it first for n = 2, and n = 3. try to factor the result into binomials.
if you are clever, and figure out (guess) the general form, you can actually prove it by induction on n.
well it's an important matrix, both for the study of polynomials, alternating bilinear forms, and permutation groups. it crops up in a lot of different places. time to roll up your sleeves and make some messy calcs...lol
@pickslides: i don't think so....there's nothing in the wording to indicate membership. and the calculation of such a determinant (a Vandermonde matrix) is the sort of thing you might encounter in a variety of different courses, inclusing differential geometry, linear algebra and group theory. i think the first time i saw it was in a physics class.