<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 matrix A equals a + b + c = 10. Since a, b, c are distinct elements from {1,2,3,4,5,6}, we must find all unordered triples summing to 10, then count ordered arrangements and choices for x, y, z.
<p><strong>Step 1:</strong> Find all unordered triples {a, b, c} of distinct elements from {1,2,3,4,5,6} with sum = 10.</p><p>Possible triples: {1,3,6}, {1,4,5}, {2,3,5}</p><p>That's 3 unordered triples.</p><p><strong>Step 2:</strong> For each unordered triple, count ordered arrangements on the diagonal. Each unordered triple {a,b,c} can be placed on the diagonal in 3! = 6 ways.</p><p>Total diagonal arrangements: 3 × 6 = 18</p><p><strong>Step 3:</strong> For each diagonal arrangement, the remaining 3 values from {1,2,3,4,5,6} must fill positions x, y, z. These 3 remaining distinct values can be arranged in the symmetric positions in 3! = 6 ways.</p><p>Total matrices: 18 × 6 = 108</p><p>∴ Answer: B</p>
Correct Answer: B