<p>Let \(a, b, c \in \mathbb{R}\), not all equal, and \[\Delta_1 = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}\], \[\Delta_2 = \begin{vmatrix} a+2b & b+3c & c+4a \\ b+2c & c+3a & a+4b \\ c+2a & a+3b & b+4c \end{vmatrix}\] then \(\dfrac{\Delta_2}{\Delta_1} =\) ________.</p>
Step-by-Step Solution
Key Concept: Express Δ₂ as a linear combination of rows using matrix operations: each row of Δ₂ can be written as a linear combination of rows of Δ₁. The determinant scales by the determinant of the transformation matrix.
<p><strong>Step 1:</strong> Express rows of Δ₂ as linear combinations of rows of Δ₁.</p><p>Let R₁, R₂, R₃ denote rows of Δ₁. Notice that:</p><p>Row 1 of Δ₂ = (a+2b, b+3c, c+4a) = 1·(a, b, c) + 2·(b, c, a) + 1·(c, a, b)</p><p>Row 2 of Δ₂ = (b+2c, c+3a, a+4b) = 1·(b, c, a) + 2·(c, a, b) + 1·(a, b, c)</p><p>Row 3 of Δ₂ = (c+2a, a+3b, b+4c) = 1·(c, a, b) + 2·(a, b, c) + 1·(b, c, a)</p><p><strong>Step 2:</strong> Write Δ₂ = M·Δ₁ where M is the transformation matrix:</p><p>$$M = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{vmatrix}$$</p><p><strong>Step 3:</strong> Calculate det(M). Since all three rows are identical: det(M) = 0</p><p><strong>Step 4:</strong> Therefore: Δ₂ = det(M)·Δ₁ = 0·Δ₁ = 0</p><p><strong>Step 5:</strong> Check Δ₁ directly. For the circulant matrix:</p><p>$$\Delta_1 = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$</p><p>Since a, b, c are not all equal, Δ₁ ≠ 0 (assuming a+b+c ≠ 0 in generic case).</p><p>∴ <strong>Answer: 0</strong></p>
Correct Answer: 0