Vandermonde determinant from 3-by-3

Question

I am a novice at math and need help calculating the determinant of the Vandermonde matrix, which is:

$V=\begin{bmatrix} 1&a_1&a_1^2&\dots&a_1^{n-1}\\ 1&a_2&a_2^2&\dots&a_2^{n-1}\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ 1&a_n&a_n^2&\dots&a_n^{n-1} \end{bmatrix}$

$det(V)={\displaystyle \prod_{1\le i\lt j\le n} (a_j-a_i)}$

I want to start with the following 3-by-3 and generalize to the n-by-n.

$A=\begin{bmatrix} 1&a_1&a_1^2\\ 1&a_2&a_2^2\\ 1&a_3&a_3^2 \end{bmatrix}$

So far I have:

$det(A) = a_2a_3(a_3-a_2)-a_1a_3(a_3-a_1)+a_1a_2(a_2-a_1)$

But I cannot figure out how to simplify this into a form like the determinant for $V$ above. What do I need to do?

Answer 1

I am a novice as well. I tried figuring this out.

$$\begin{bmatrix}1 & a& a^2 \\1&b&b^2 \\1 &c& c^2 \end{bmatrix}$$ $R_2 \rightarrow R_2 -R_1 \\ R_3 \rightarrow R_3 -R_1$ $$\begin{bmatrix}1 & a& a^2 \\0&b-a&b^2-a^2 \\0 &c-a& c^2-a^2 \end{bmatrix} $$ Taking out $(b-a)(c-a)$ from rows $R_2,~ R_3$ $$(b-a)(c-a) \cdot \begin{bmatrix}1 & a& a^2 \\0&1&b+a \\0 &1& c+a \end{bmatrix}$$ Taking Determinant gets us. $$|A| = (b-a)(c-a)(c-b)$$