<p>Let \(B^2 = I\) and \(AB = \begin{bmatrix} p & x & a \\ b & q & y \\ z & c & r \end{bmatrix}\). Given that \(\text{tr}(AB + AB^3 + \cdots + AB^{19}) = 210\), find the number of ordered triplets \((p, q, r)\) where \(p, q, r \in \mathbb{N}\) and \(p + q + r = 21\).</p>
Step-by-Step Solution
Key Concept: Since B² = I, we have B⁻¹ = B, so AB³ = AB, AB⁵ = AB, etc. The sum AB + AB³ + ··· + AB¹⁹ contains 10 identical terms, each equal to AB. Therefore tr(10·AB) = 10(p+q+r) = 210.
<p><strong>Step 1: Use the property B² = I</strong></p><p>Since B² = I, we have B⁻¹ = B. This means:</p><p>• AB¹ = AB</p><p>• AB³ = AB·B² = AB·I = AB</p><p>• AB⁵ = AB·B⁴ = AB·(B²)² = AB·I = AB</p><p>• In general, AB^(2k+1) = AB for all k ≥ 0</p><p><strong>Step 2: Identify all terms in the sum</strong></p><p>The sum AB + AB³ + AB⁵ + ··· + AB¹⁹ contains the terms with odd exponents from 1 to 19.</p><p>Odd exponents: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 → 10 terms total</p><p><strong>Step 3: Simplify using B² = I</strong></p><p>Since each term equals AB:</p><p>AB + AB³ + ··· + AB¹⁹ = 10·AB</p><p><strong>Step 4: Apply trace property</strong></p><p>tr(10·AB) = 10·tr(AB) = 10(p + q + r) = 210</p><p>Therefore: p + q + r = 21</p><p><strong>Step 5: Count solutions in natural numbers</strong></p><p>We need ordered triplets (p, q, r) where p, q, r ∈ ℕ (positive integers) and p + q + r = 21.</p><p>This is equivalent to distributing 21 among three variables with each ≥ 1.</p><p>Using stars and bars: Let p' = p - 1, q' = q - 1, r' = r - 1 (all ≥ 0)</p><p>Then p' + q' + r' = 18</p><p>Number of solutions = C(18+3-1, 3-1) = C(20, 2) = (20·19)/2 = 190</p><p>∴ Answer: <strong>190</strong></p>
Correct Answer: 190