Question

If \(\begin{vmatrix} a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2 \end{vmatrix} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}, \lambda \ne 0\), then \(k\) is equal to:

• A. \(4\lambda abc\)
• B. \(-4\lambda abc\)
• C. \(4\lambda^2\)
• D. \(-4\lambda^2\)

Hint:

Apply \(R_2 \to R_2 - R_3\) to get \(4\lambda(a, b, c)\), then clean up \(R_3\) using \(R_1\) and \(R_2\).

Answer 1

The Correct Option is C

Let \(\Delta\) be the left determinant. Apply \(R_2 \to R_2 - R_3\): \((a+\lambda)^2 - (a-\lambda)^2 = 4a\lambda\), so the new second row is \(4\lambda(a, b, c)\).
Apply \(R_3 \to R_3 - R_1\): \((a-\lambda)^2 - a^2 = -2a\lambda + \lambda^2\), so the third row becomes \(\lambda^2(1, 1, 1) - 2\lambda(a, b, c)\).
Apply \(R_3 \to R_3 + \frac{1}{2}R_2\): the third row becomes \(\lambda^2(1, 1, 1)\).
Hence \(\Delta = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4\lambda a & 4\lambda b & 4\lambda c \\ \lambda^2 & \lambda^2 & \lambda^2 \end{vmatrix} = 4\lambda \cdot \lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = 4\lambda^3 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}\).
Comparing with \(k\lambda \cdot (\ldots)\): \(k\lambda = 4\lambda^3\), so \(k = 4\lambda^2\).
Answer: (3) \(4\lambda^2\)