Question

If \(\Delta=\begin{vmatrix}     1 & a & a^2 \\     1 & b & b^2 \\     1 & c & c^2 \end{vmatrix}\) and \(\Delta_1=\begin{vmatrix}     1 & 1 & 1 \\     bc & ca & ab \\     a & b & c \end{vmatrix}\) then

• A. Δ₁ ≠ Δ
• B. Δ₁ = Δ
• C. Δ₁ = -Δ
• D. Δ₁ = 3Δ

Hint:

When dealing with determinants, particularly in the case of Vandermonde determinants and symmetric forms, you can apply row operations to simplify the expression. In this case, expanding the determinant \( \Delta_1 \) along the first row and simplifying the resulting expression leads to the relation \( \Delta_1 = -\Delta \).

Answer 1

The Correct Option is C

The correct answer is (C) : Δ₁ = -Δ.

Answer 2

The correct answer is: (C) \( \Delta_1 = -\Delta \).

We are given two determinants:

\( \Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \)

\( \Delta_1 = \begin{vmatrix} 1 & 1 & 1 \\ bc & ca & ab \\ a & b & c \end{vmatrix} \)

We are asked to find the relationship between \( \Delta \) and \( \Delta_1 \). The determinant \( \Delta \) is a standard form known as the **Vandermonde determinant**, which has the following formula:

\( \Delta = (a - b)(b - c)(c - a) \)

Now, let's analyze the determinant \( \Delta_1 \): We can simplify \( \Delta_1 \) by applying row operations. In particular, notice that the first row is composed of all 1's, and the second and third rows have symmetric forms. By expanding along the first row and simplifying, we find that:

\( \Delta_1 = -\Delta \)

Therefore, the correct answer is (C) \( \Delta_1 = -\Delta \).