b2−a2=(b−a)(b+a) and c2−a2=(c−a)(c+a). Factor (b−a) from R2 and (c−a) from R3:
(b−a)(c−a)100a11a2b+ac+a
The first column has (1,0,0), so the determinant reduces to the 2×2 minor:
(b−a)(c−a)⋅11b+ac+a=(b−a)(c−a)⋅[(c+a)−(b+a)]
=(b−a)(c−a)(c−b)
Final answer: (b−a)(c−a)(c−b) or equivalently (a−b)(b−c)(c−a)⋅(−1).
Why This Works
The Vandermonde determinant has a beautiful product structure: the determinant is the product of all pairwise differences. This pattern generalises — for an n×n Vandermonde matrix, the determinant is ∏i<j(xj−xi).
Practical consequence: the determinant vanishes if and only if at least two of a,b,c are equal — that’s because the corresponding rows become identical.
Alternative Method
Direct expansion along the first row gives a(b2c−bc2)+(other terms) — a mess. Use the cofactor formula and then group cleverly. The row-reduction approach above is much cleaner.
Recognise Vandermonde patterns instantly: rows of the form (1,x,x2,...,xn−1). JEE often disguises them by permuting columns or transposing — look for the structure underneath.
Common Mistake
Students often expand directly without factoring, ending up with a polynomial in a,b,c that’s hard to simplify. The trick is row operations FIRST, then factor. Also: sign errors when stating the answer — (a−b)(b−c)(c−a)=−(b−a)(c−b)(a−c) etc. Pick one canonical form.
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