Question

If \( a \neq b \neq c \), \( \Delta_1 = \begin{vmatrix} 1 & a^2 & bc 1 & b^2 & ca 1 & c^2 & ab \end{vmatrix} \), \( \Delta_2 = \begin{vmatrix} 1 & 1 & 1 a^2 & b^2 & c^2 a^3 & b^3 & c^3 \end{vmatrix} \) and \( \frac{\Delta_1}{\Delta_2} = \frac{6}{11} \), then \( 11(a + b + c) = \)

• A. \( 0 \)
• B. \( 1 \)
• C. \( ab + bc + ca \)
• D. \( 6(ab + bc + ca) \)

Hint:

For standard cyclic determinants containing powers of \(a, b, c\), you can skip the tedious row-reduction steps entirely by remembering the standard factor formulas:
• Power sequence \((1, 2, 3)\) yields \((a-b)(b-c)(c-a)(ab+bc+ca)\)
• Power sequence \((1, 1, 3)\) (with a linear middle row) yields \((a-b)(b-c)(c-a)(a+b+c)\) Memorizing these two patterns makes ratio problems an absolute breeze!

Answer 1

The Correct Option is D

Concept: Determinant expressions containing cyclic variables can often be transformed into matching structures by multiplying rows or columns by appropriate scaling variables (\(a, b, c\)). Additionally, the value of a determinant is unchanged if rows and columns are completely swapped (taking the transpose), i.e., \(|A| = |A|^T\). Interchanging any two individual rows or columns swaps the algebraic sign of the determinant.

Step 1: Transforming the first determinant \(\Delta_1\) to create a symmetric structure.
The first determinant is given as: \[ \Delta_1 = \begin{vmatrix} 1 & a^2 & bc 1 & b^2 & ca 1 & c^2 & ab \end{vmatrix} \] Multiply the 1st row (\(R_1\)) by \(a\), the 2nd row (\(R_2\)) by \(b\), and the 3rd row (\(R_3\)) by \(c\). To keep the value unchanged, we divide the entire determinant by \(abc\): \[ \Delta_1 = \frac{1}{abc} \begin{vmatrix} a & a^3 & abc b & b^3 & abc c & c^3 & abc \end{vmatrix} \] Now, factor out the common term \(abc\) from the third column (\(C_3\)): \[ \Delta_1 = \frac{abc}{abc} \begin{vmatrix} a & a^3 & 1 b & b^3 & 1 c & c^3 & 1 \end{vmatrix} = \begin{vmatrix} a & a^3 & 1 b & b^3 & 1 c & c^3 & 1 \end{vmatrix} \]

Step 2: Using column interchanges to align \(\Delta_1\) with the layout of \(\Delta_2\).
Let's rearrange the columns to put the column of ones first. First, interchange column 2 and column 3 (\(C_2 \leftrightarrow C_3\)), which introduces a negative sign: \[ \Delta_1 = -\begin{vmatrix} a & 1 & a^3 b & 1 & b^3 c & 1 & c^3 \end{vmatrix} \] Next, interchange column 1 and column 2 (\(C_1 \leftrightarrow C_2\)), which introduces another negative sign, making it positive: \[ \Delta_1 = \begin{vmatrix} 1 & a & a^3 1 & b & b^3 1 & c & c^3 \end{vmatrix} \] Taking the transpose of this matrix (converting rows to columns) leaves the determinant value unchanged: \[ \Delta_1 = \begin{vmatrix} 1 & 1 & 1 a & b & c a^3 & b^3 & c^3 \end{vmatrix} \quad \cdots (1) \]

Step 3: Relating \(\Delta_1\) and \(\Delta_2\) through factoring.
The second determinant is given as: \[ \Delta_2 = \begin{vmatrix} 1 & 1 & 1 a^2 & b^2 & c^2 a^3 & b^3 & c^3 \end{vmatrix} \] Both \(\Delta_1\) and \(\Delta_2\) share identical 1st and 3rd rows. They are standard Vandermonde-type cyclic variations. Let us use standard algebraic expansion differences for cyclic polynomials: \[ \Delta_1 = \begin{vmatrix} 1 & 1 & 1 a & b & c a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c) \] \[ \Delta_2 = \begin{vmatrix} 1 & 1 & 1 a^2 & b^2 & c^2 a^3 & b^3 & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(ab+bc+ca) \]

Step 4: Evaluating the required expression using the given ratio.
Taking the ratio of the two expanded determinants: \[ \frac{\Delta_1}{\Delta_2} = \frac{(a-b)(b-c)(c-a)(a+b+c)}{(a-b)(b-c)(c-a)(ab+bc+ca)} \] Since \(a \neq b \neq c\), the common differences are non-zero and can be cancelled out safely: \[ \frac{\Delta_1}{\Delta_2} = \frac{a+b+c}{ab+bc+ca} \] We are given that this ratio equals \(\frac{6}{11}\): \[ \frac{a+b+c}{ab+bc+ca} = \frac{6}{11} \] Cross-multiplying to match our target expression: \[ 11(a+b+c) = 6(ab+bc+ca) \] This directly corresponds to option (D).