<p>How many \(3 \times 3\) matrices \(M\) with entries from \(\{0, 1, 2\}\) are there, for which the sum of the diagonal entries of \(M^T M\) is 5?</p>
Step-by-Step Solution
Key Concept: The sum of diagonal entries of M^T M equals the sum of squares of all entries in M. We need entries from {0,1,2} such that their squares sum to 5.
<p><strong>Step 1:</strong> Find the trace formula. The diagonal entries of M^T M are: (M^T M)_{ii} = (column i of M)·(column i of M) = sum of squares of entries in column i.</p><p>Therefore, trace(M^T M) = sum of all entries of M squared = Σ m_{ij}² = 5</p><p><strong>Step 2:</strong> With entries from {0,1,2}, find all non-negative integer solutions where each entry squared contributes 0, 1, or 4.</p><p>Case 1: One entry is 2, one entry is 1, rest are 0 → 2² + 1² = 4 + 1 = 5 ✓</p><p>Case 2: Five entries are 1, rest are 0 → 1² + 1² + 1² + 1² + 1² = 5 ✓</p><p><strong>Step 3:</strong> Count arrangements.</p><p>Case 1 (one 2, one 1, seven 0s): Choose position for 2 (9 ways) × choose position for 1 from remaining 8 (8 ways) = 9 × 8 = 72</p><p>Case 2 (five 1s, four 0s): C(9,5) = 126</p><p><strong>Step 4:</strong> Total = 72 + 126 = 198</p><p>∴ Answer: A (198)</p>
Correct Answer: A