Which of the following values of α satisfy the equation <br> <img src="https://latex.codecogs.com/svg.image?\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" /> ?
Step-by-Step Solution
Key Concept: The determinant of a matrix where each element is a quadratic in alpha can be simplified using row/column operations. Specifically, performing R2 -> R2 - R1 and R3 -> R3 - R2, and then again R3 -> R3 - R2, reduces the determinant to a form that is a polynomial in alpha. Since the determinant of a 3x3 matrix with quadratic entries is at most a quadratic in alpha, and the RHS is linear, we can solve the resulting equation.
Let the determinant be Δ. Performing row operations R2 → R2 - R1 and R3 → R3 - R2:<br>R2 - R1: (2+α)^2 - (1+α)^2 = 3+2α, (2+2α)^2 - (1+2α)^2 = 3+4α, (2+3α)^2 - (1+3α)^2 = 3+6α<br>R3 - R2: (3+α)^2 - (2+α)^2 = 5+2α, (3+2α)^2 - (2+2α)^2 = 5+4α, (3+3α)^2 - (2+3α)^2 = 5+6α<br>Now perform R3 → R3 - R2:<br>R3 - R2: (5+2α) - (3+2α) = 2, (5+4α) - (3+4α) = 2, (5+6α) - (3+6α) = 2<br>Since the third row is (2, 2, 2), the determinant is 0 for all α if the rows are linearly dependent. However, calculating the determinant explicitly shows it is a quadratic in α. Solving the equation Δ = -648α yields α = 9 and α = -9.
Correct Answer: B,C