Prove that the determinant of the Vandermonde matrix \[ \begin{bmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix} \] is \((b - a)(c - a)(c - b)\).
Solution
Subtract the first row from the second and third rows. This does not change the determinant: \[ \det\begin{bmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{bmatrix} = \det\begin{bmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{bmatrix}. \] Factor \(b-a\) from the second row and \(c-a\) from the third, using \(b^2-a^2=(b-a)(b+a)\) and \(c^2-a^2=(c-a)(c+a)\): \[ =(b-a)(c-a) \det\begin{bmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{bmatrix}. \] Expanding along the first column leaves \[ (b-a)(c-a)\det\begin{bmatrix}1&b+a\\1&c+a\end{bmatrix} =(b-a)(c-a)(c-b). \]