<p>Evaluate \[\begin{vmatrix} {}^{x}C_1 & {}^{x}C_2 & {}^{x}C_3 \\ {}^{y}C_1 & {}^{y}C_2 & {}^{y}C_3 \\ {}^{z}C_1 & {}^{z}C_2 & {}^{z}C_3 \end{vmatrix}.\]</p>
Step-by-Step Solution
Key Concept: Express each binomial coefficient as a polynomial in its base (e.g., ₓC₁ = x, ₓC₂ = x(x-1)/2, ₓC₃ = x(x-1)(x-2)/6), then factor out common terms from rows to reveal a Vandermonde-like structure that yields a difference-of-roots product.
<p><strong>Step 1:</strong> Express binomial coefficients as polynomials:</p><p>ₓC₁ = x, ₓC₂ = x(x-1)/2, ₓC₃ = x(x-1)(x-2)/6</p><p>₍C₁ = y, ₍C₂ = y(y-1)/2, ₍C₃ = y(y-1)(y-2)/6</p><p>₍C₁ = z, ₍C₂ = z(z-1)/2, ₍C₃ = z(z-1)(z-2)/6</p><p><strong>Step 2:</strong> Factor out 1/2 from column 2 and 1/6 from column 3:</p><p>Determinant = (1/12) × \begin{vmatrix} x & x(x-1) & x(x-1)(x-2) \\ y & y(y-1) & y(y-1)(y-2) \\ z & z(z-1) & z(z-1)(z-2) \end{vmatrix}</p><p><strong>Step 3:</strong> Factor x, y, z from rows 1, 2, 3 respectively:</p><p>Determinant = (1/12)xyz × \begin{vmatrix} 1 & (x-1) & (x-1)(x-2) \\ 1 & (y-1) & (y-1)(y-2) \\ 1 & (z-1) & (z-1)(z-2) \end{vmatrix}</p><p><strong>Step 4:</strong> The remaining matrix is Vandermonde-type with parameters (x-1), (y-1), (z-1). Expanding gives:</p><p>(x-1-y+1)(y-1-z+1)(z-1-x+1) = (x-y)(y-z)(z-x)</p><p><strong>Step 5:</strong> Combine all factors:</p><p>∴ Answer: $\boxed{\frac{1}{12}xyz(x-y)(y-z)(z-x)}$</p>
Correct Answer: \(\frac{1}{12}xyz(x-y)(y-z)(z-x)\)