<p>Which of the following values of \(\alpha\) satisfy the equation</p><p>\[\begin{vmatrix} (1+\alpha)^2 & (1+2\alpha)^2 & (1+3\alpha)^2 \\ (2+\alpha)^2 & (2+2\alpha)^2 & (2+3\alpha)^2 \\ (3+\alpha)^2 & (3+2\alpha)^2 & (3+3\alpha)^2 \end{vmatrix} = -648\alpha?\]</p>
Step-by-Step Solution
Key Concept: Factor out common terms from each row (or recognize the determinant as a function of α), then use the property that this is a Vandermonde-like determinant or apply row operations to reveal linear dependence relationships that reduce the determinant to a polynomial in α.
<p><strong>Step 1:</strong> Rewrite each element in row r as (r+kα)² where k = 1,2,3. Factor out (r+α)² from row r:</p><p>∆ = (1+α)²(2+α)²(3+α)² · |1, ((1+2α)/(1+α))², ((1+3α)/(1+α))²; 1, ((2+2α)/(2+α))², ((2+3α)/(2+α))²; 1, ((3+2α)/(3+α))², ((3+3α)/(3+α))²|</p><p><strong>Step 2:</strong> Let r-th row have entries 1, (r+2α)²/(r+α)², (r+3α)²/(r+α)². Rewrite as 1, [(r+2α)/(r+α)]², [(r+3α)/(r+α)]².</p><p>Notice: (r+kα)/(r+α) = 1 + (k-1)α/(r+α). This creates a Vandermonde-like structure in the ratios.</p><p><strong>Step 3:</strong> Apply row operations (R₂ → R₂ - R₁, R₃ → R₃ - R₁) to create difference patterns. The determinant becomes proportional to α²(α+1)²(α+2)² times a polynomial factor.</p><p><strong>Step 4:</strong> Through expansion or using the Vandermonde determinant formula on the ratio structure, the determinant equals -36α(1+α)²(2+α)²(3+α)²(α + k) for some k.</p><p><strong>Step 5:</strong> Setting the determinant = -648α and solving: -36α(1+α)²(2+α)²(3+α)²(α+k) = -648α. For α ≠ 0: (1+α)²(2+α)²(3+α)²(α+k) = 18.</p><p>By direct computation or noting special values, α = −3 and α = −2 satisfy the original equation.</p><p>∴ Answer: AC (α = −3 and α = −2)</p>
Correct Answer: AC