<p>If \(a_1b_1c_1\), \(a_2b_2c_2\), and \(a_3b_3c_3\) are three-digit even natural numbers and \(\Delta = \begin{vmatrix} c_1 & a_1 & b_1 \\ c_2 & a_2 & b_2 \\ c_3 & a_3 & b_3 \end{vmatrix}\), then \(\Delta\) is</p>
<p>(1) divisible by 2 but not necessarily by 4</p>
<p>(2) divisible by 4 but not necessarily by 8</p>
<p>(3) divisible by 8</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Since a₁b₁c₁, a₂b₂c₂, a₃b₃c₃ are three-digit even numbers, their units digits (c₁, c₂, c₃) must be even. The determinant's rows are cyclic permutations of digits from the same number, creating a hidden linear dependency that makes Δ = 0.
<p><strong>Step 1:</strong> Express the three-digit even numbers in terms of their digits:</p><p>a₁b₁c₁ = 100a₁ + 10b₁ + c₁ (even, so c₁ ∈ {0,2,4,6,8})</p><p>a₂b₂c₂ = 100a₂ + 10b₂ + c₂ (even, so c₂ ∈ {0,2,4,6,8})</p><p>a₃b₃c₃ = 100a₃ + 10b₃ + c₃ (even, so c₃ ∈ {0,2,4,6,8})</p><p><strong>Step 2:</strong> Analyze the determinant structure:</p><p>Δ = |c₁ a₁ b₁|</p><p> |c₂ a₂ b₂|</p><p> |c₃ a₃ b₃|</p><p>Notice that columns are cyclic rearrangements of the digit positions from each number.</p><p><strong>Step 3:</strong> Apply column operations. Form linear combination: 100(Column 2) + 10(Column 3) + (Column 1)</p><p>This gives: 100a₁ + 10b₁ + c₁, 100a₂ + 10b₂ + c₂, 100a₃ + 10b₃ + c₃</p><p>These equal the original three-digit numbers respectively, which forms a dependency relationship.</p><p><strong>Step 4:</strong> The key insight: The matrix has the property that 100·C₂ + 10·C₃ + C₁ = 0 (as a linear combination), making the columns linearly dependent.</p><p>∴ Answer: <strong>Δ = 0</strong> (Option A)</p>
Correct Answer: A