Matrices & Determinants
Determinants
Grade 12
Question:
<p>Consider a set S = {2, 3, 5, 7, 11, 13, 17, 19, 23}, a collection of 1<sup>st</sup> 9 prime numbers. Let {D₁, D₂, D₃, ....... D_n} be the set of third order determinantes that can be made with all the 9 elements of set 'S'. Then which one of the following is true about K = D₁ + D₂ + ....... + D_n</p>
<p>(a) K is a 3 digit number</p>
<p>(b) K → 0</p>
<p>(c) K is a 4 digit number not divisible by any element of set 'S'</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Each element of S appears in the same position (row, column) across all n = 9! third-order determinants equally often. By symmetry, the sum of all determinants equals zero because determinant is a multilinear alternating function, and the contributions cancel when each element occupies each position equally.
<p><strong>Step 1:</strong> Recognize that {D₁, D₂, ..., D_n} contains all possible 3×3 determinants formed using distinct elements from S.</p><p><strong>Step 2:</strong> There are 9!/(9-9)! = 9! total arrangements, but for 3×3 matrices we use 9P9 = 9! ways to arrange all 9 elements (each determinant uses 3 elements in order). Actually, we form C(9,9) × 3! × 3! × 3! = 9!/(1) distinct determinants = 9!÷1 ordered selections.</p><p><strong>Step 3:</strong> By the multilinearity property of determinants: when we sum over ALL possible 3×3 matrices with entries from S, each element of S occupies each of the 9 positions (3 rows × 3 columns) in exactly (9-1)!/(3-1)! of the determinants.</p><p><strong>Step 4:</strong> For any fixed arrangement in rows 1,2 with all 6 elements, the sum over all possible row 3 arrangements: ∑(using remaining 3 elements) det = 0 by the property that summing a multilinear alternating function over complementary sets yields zero.</p><p><strong>Step 5:</strong> By repeated application, K = D₁ + D₂ + ... + D_n = <strong>0</strong></p><p>∴ Answer: B (K = 0)</p>
Correct Answer: B