<p>The number of positive integral solutions of the equation \(\begin{vmatrix} x^3+1 & x^2y & x^2z \\ xy^2 & y^3+1 & y^2z \\ xz^2 & yz^2 & z^3+1 \end{vmatrix} = 11\) is</p>
Step-by-Step Solution
Key Concept: Factor the determinant by recognizing it equals (x³ + y³ + z³ + 1) when xyz ≠ 0. This reduces the problem to solving x³ + y³ + z³ = 10 for positive integers, which has finitely many solutions.
<p><strong>Step 1:</strong> Recognize the determinant structure. Factor out from rows: Row 1 has factor x, Row 2 has factor y, Row 3 has factor z (in appropriate columns). This gives det = xyz · M where M has a special form.</p><p><strong>Step 2:</strong> Use the identity: When we expand carefully (or recognize the pattern), the determinant equals <strong>(x³ + y³ + z³ + 1)</strong> for the given matrix structure.</p><p><strong>Step 3:</strong> Set up the equation: x³ + y³ + z³ + 1 = 11, so <strong>x³ + y³ + z³ = 10</strong> where x, y, z are positive integers.</p><p><strong>Step 4:</strong> Find all positive integer solutions systematically:</p><ul><li>If x = y = z = 1: 1 + 1 + 1 = 3 ✗</li><li>If x = 2, y = 1, z = 1: 8 + 1 + 1 = 10 ✓</li><li>By symmetry: (2,1,1), (1,2,1), (1,1,2) all work → 3 solutions</li><li>If any variable ≥ 3: that cube alone ≥ 27 > 10 ✗</li><li>If two variables = 2: 8 + 8 > 10 ✗</li></ul><p><strong>Step 5:</strong> Count permutations of (2, 1, 1): There are 3!/2! = 3 distinct ordered triples.</p><p>∴ Answer: <strong>B</strong> (The number of positive integral solutions is <strong>3</strong>)</p>
Correct Answer: B