The value of \( \begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix} \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (a-b)(b-c)(c-a) \)
The operations \( R_2\rightarrow R_2-R_1 \) and \( R_3\rightarrow R_3-R_1 \) create zeros in the first column:
\( \displaystyle \Delta=\begin{vmatrix}1 & a & a^2\\0 & b-a & b^2-a^2\\0 & c-a & c^2-a^2\end{vmatrix} \)
Taking out (b − a) from \( R_2 \) and (c − a) from \( R_3 \), using \( b^2-a^2=(b-a)(b+a) \):
\( \displaystyle \Delta=(b-a)(c-a)\begin{vmatrix}1 & a & a^2\\0 & 1 & b+a\\0 & 1 & c+a\end{vmatrix} \)
Expanding along the first column:
\( \displaystyle \Delta=(b-a)(c-a)\left[(c+a)-(b+a)\right] \)
\( \displaystyle =(b-a)(c-a)(c-b) \)
Since \( (b-a)(c-b)=(a-b)(b-c) \), this is
\( \displaystyle \Delta=(a-b)(b-c)(c-a) \)
Shortcut: Test simple values, put a = 0, b = 1, c = 2:
\( \displaystyle \begin{vmatrix}1 & 0 & 0\\1 & 1 & 1\\1 & 2 & 4\end{vmatrix}=1(4-2)=2 \)
Options (1), (2) and (4) all give 0 because a = 0, while Option (3) gives (0 − 1)(1 − 2)(2 − 0) = 2.
Tip: This is the standard Vandermonde determinant: if two of a, b, c are equal, two rows become identical and the determinant is 0, so (a − b), (b − c) and (c − a) are all factors.