Question

If $\Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = k(a-b)(b-c)(c-a)$, then $k$ is equal to

• A. $-2$
• B. $1$
• C. $2$
• D. $abc$

Hint:

Vandermonde determinant $\prod_{1 \le i<j \le n} (x_j - x_i)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The determinant is the Vandermonde determinant.
Step 2: Detailed Explanation:
The Vandermonde determinant $\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (b-a)(c-a)(c-b) = (a-b)(b-c)(c-a)$ (up to sign).
Compute: $(b-a)(c-a)(c-b) = (a-b)(c-a)(b-c) = -(a-b)(b-c)(c-a)$. Actually, $(b-a)(c-b)(c-a) = (a-b)(b-c)(c-a)$. Yes, they are equal. So $k=1$.
Step 3: Final Answer:
$k = 1$.