<p>Elements of a matrix <em>A</em> of order \(10 \times 10\) are defined as \(a_{ij} = \omega^{i+j}\) (where \(\omega\) is imaginary cube root of unity), then trace (<em>A</em>) of the matrix is</p>
Step-by-Step Solution
Key Concept: The trace is the sum of diagonal elements where i=j, so trace(A) = Σa_ii = Σω^(2i). Since ω³=1 and ω≠1, we need to find the sum of ω^2, ω^4, ω^6, ..., ω^20 using properties of cube roots of unity.
<p><strong>Step 1:</strong> Identify trace formula. trace(A) = Σ(i=1 to 10) a_ii = Σ(i=1 to 10) ω^(2i)</p><p><strong>Step 2:</strong> Use the property ω³ = 1. The exponents are 2,4,6,8,10,12,14,16,18,20. Reducing modulo 3: {2,1,0,2,1,0,2,1,0,2}</p><p><strong>Step 3:</strong> Group by residue class modulo 3:<br>• Exponent ≡ 0 (mod 3): positions 6,12,18 → ω⁰ = 1 appears 3 times → contribution = 3<br>• Exponent ≡ 1 (mod 3): positions 4,10,16 → ω¹ = ω appears 3 times → contribution = 3ω<br>• Exponent ≡ 2 (mod 3): positions 2,8,14,20 → ω² appears 4 times → contribution = 4ω²</p><p><strong>Step 4:</strong> Sum: trace(A) = 3 + 3ω + 4ω² = 3(1 + ω + ω²) + ω² = 3(0) + ω² = ω²</p><p>∴ Answer: A</p>
Correct Answer: A