<p><strong>258.</strong> If \(A = \begin{bmatrix} a & x & y \\ x & b & z \\ y & z & c \end{bmatrix}\) where \(a, b, c, x, y, z \in \{1, 2, 3, 4, 5, 6\}\) and also \(a, b, c, x, y, z\) are distinct, then number of matrices in \(A\) with trace equal to 10 are:</p>
Step-by-Step Solution
Key Concept: The trace of a matrix equals the sum of diagonal elements (a + b + c), so we need distinct values from {1,2,3,4,5,6} that sum to 10. Once (a,b,c) is fixed, the remaining 3 values can be assigned to x, y, z in any order.
<p><strong>Step 1:</strong> Find all distinct triples (a,b,c) from {1,2,3,4,5,6} with a + b + c = 10.</p><p>Systematic check:</p><ul><li>(1,3,6): 1+3+6 = 10 ✓</li><li>(1,4,5): 1+4+5 = 10 ✓</li><li>(2,3,5): 2+3+5 = 10 ✓</li></ul><p>These are the <strong>only 3 triples</strong>.</p><p><strong>Step 2:</strong> For each valid triple (a,b,c), the remaining 3 elements from {1,2,3,4,5,6} must fill positions x, y, z.</p><p>Example: If (a,b,c) = (1,3,6), then {x,y,z} = {2,4,5}.</p><p><strong>Step 3:</strong> The 3 remaining distinct values can be assigned to x, y, z positions in <strong>3! = 6 ways</strong>.</p><p><strong>Step 4:</strong> Total matrices = (Number of valid (a,b,c) triples) × (Arrangements of remaining elements)</p><p>= 3 × 6 = <strong>18</strong></p><p>∴ Answer: A</p>
Correct Answer: A