Which one of the following does NOT equal | 1 x x^2 | | 1 y y^2 | | 1 z z^2 | A. | 1 x(x+1) x+1 | | 1 y(y+1) y+1 | | 1 z(z+1) z+1 | B. | 1 x+1 x^2+1 | | 1 y+1 y^2+1 | | 1 z+1 z^2+1 | C. | 1 x x^2 | | 0 y-x y^2-x^2 | | 0 z-x z^2-x^2 | D. | 2 x+y x^2+y^2 | | 2 y+z y^2+z^2 | | 2 z+x z^2+x^2 |
GATE 2013 · Engineering Mathematics · Determinant · medium
Answer: Option A does NOT equal the original Vandermonde determinant.
- Check option C (row subtraction): In option C: row 2 becomes [0, y-x, y^2-x^2] and row 3 becomes [0, z-x, z^2-x^2]. This is R2 - R1 and R3 - R1, which preserves the determinant. So C equals V.
- Check option D (row addition): Option D has rows [2, x+y, x^2+y^2], [2, y+z, y^2+z^2], [2, z+x, z^2+x^2]. These are pairwise sums of rows of the original: each new row i = row_i + row_j. Adding a multiple of one row to another preserves det, and here it can be verified by cyclic row addition keeping det unchanged. So D equals V.
- Check option A (column swap analysis): Option A: column 2 = [x(x+1), y(y+1), z(z+1)] = [x^2+x, y^2+y, z^2+z] and column 3 = [x+1, y+1, z+1]. This is C2' = C3_original + C2_original and C3' = C2_original + 1. Comparing with original (C2=x, C3=x^2): the columns have been rearranged so that x^2 terms appear in column 2 and x terms in column 3 — effectively a column swap (plus adding column 1 to each), which gives -V, not V.