<p><em>A</em> and <em>B</em> are different matrices of order <em>n</em> satisfying <em>A</em><sup>3</sup> = <em>B</em><sup>3</sup> and <em>A</em><sup>2</sup><em>B</em> = <em>B</em><sup>2</sup><em>A</em>. If det.(<em>A</em> − <em>B</em>) ≠ 0, then find the value of det.(<em>A</em><sup>2</sup> + <em>B</em><sup>2</sup>).</p>
Step-by-Step Solution
Key Concept: From A³ = B³ and A²B = B²A, we can derive that (A-B) and (A²+AB+B²) are related through factorization. The condition A²B = B²A combined with A³ = B³ forces A²+AB+B² to be singular.
<p><strong>Step 1:</strong> From A³ = B³, we have A³ - B³ = 0, which factors as (A-B)(A²+AB+B²) = 0.</p><p><strong>Step 2:</strong> Since det(A-B) ≠ 0, the matrix (A-B) is invertible, so we must have A²+AB+B² = 0 (the zero matrix).</p><p><strong>Step 3:</strong> From A²+AB+B² = 0, multiply by (A-B) on the right: (A²+AB+B²)(A-B) = 0 gives A³ - A²B + A²B - AB² + AB² - B³ = 0, confirming consistency.</p><p><strong>Step 4:</strong> From A²+AB+B² = 0, we can write A²+B² = -AB. The condition A²B = B²A (commutativity of specific combinations) ensures this system is consistent and forces det(A²+AB+B²) = 0.</p><p><strong>Step 5:</strong> Since A²+AB+B² = 0 is singular, and using A²+B² = -AB, we find that A²+B² is also singular when the conditions are satisfied.</p><p>∴ det(A² + B²) = <strong>0</strong></p>
Correct Answer: 0