Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>Let \(\{D_1, D_2, D_3, \ldots, D_n\}\) be the set of third-order determinants that can be made with the distinct non-zero real numbers \(a_1, a_2, \ldots, a_9\). Then</p>
<p>(1) \(\displaystyle\sum_{i=1}^{n} D_i = 1\)</p>
<p>(2) \(\displaystyle\sum_{i=1}^{n} D_i = 0\)</p>
<p>(3) \(D_i = D_j,\ \forall\ i, j\)</p>
<p>(4) None of these</p>
Step-by-Step Solution
Key Concept: Each of the 9 distinct numbers appears in exactly 4! positions across all possible 3×3 determinant arrangements, and the sum of all determinants equals the sum of products of rows weighted by their contribution to the determinant expansion formula.
<p><strong>Step 1:</strong> Recognize that we're summing determinants formed by all possible arrangements of 9 distinct numbers into 3×3 matrices.</p><p><strong>Step 2:</strong> In the determinant expansion, each number appears in different positions across all possible arrangements. By the principle of symmetry in determinant expansion (with positive and negative terms), each number contributes equally in positions that add and positions that subtract.</p><p><strong>Step 3:</strong> For any fixed element aᵢ, it appears in 8! arrangements in the top-left position (contributing +aᵢ × minor), and by symmetry of the determinant formula, its net contribution across all 9 positions (3 rows × 3 columns) and all sign patterns cancels out to zero.</p><p><strong>Step 4:</strong> Alternatively, consider that in the complete set of all determinants formed, each distinct number appears with equal frequency in positions that contribute positively and negatively to the sum, resulting in complete cancellation.</p><p>∴ D₁ + D₂ + D₃ + ... + Dₙ = <strong>0</strong></p>
Correct Answer: B