<p>Find the number of all possible matrices of order 3 × 3 with each entry 0 or 1. How many of these are symmetric?</p>
Step-by-Step Solution
Key Concept: For a 3×3 matrix with entries 0 or 1: total matrices = 2^9 (each of 9 entries has 2 choices). For symmetric matrices, only the upper triangular part (including diagonal) needs to be specified independently, giving 2^6 choices.
<p><strong>Step 1:</strong> Count total 3×3 matrices with entries from {0,1}</p><p>Each of the 9 positions can be filled in 2 ways (0 or 1).</p><p>Total matrices = 2^9 = <strong>512</strong></p><p><strong>Step 2:</strong> Count symmetric 3×3 matrices where a_ij = a_ji</p><p>In a symmetric matrix, we need:</p><ul><li>Diagonal entries: a₁₁, a₂₂, a₃₃ → 3 independent choices</li><li>Upper triangular entries: a₁₂, a₁₃, a₂₃ → 3 independent choices</li><li>Lower triangular entries: a₂₁, a₃₁, a₃₂ → determined by symmetry (0 free choices)</li></ul><p><strong>Step 3:</strong> Calculate independent positions</p><p>Total independent entries = 3 (diagonal) + 3 (strictly upper triangular) = 6</p><p>Each can be 0 or 1, so symmetric matrices = 2^6 = <strong>64</strong></p><p>∴ Answer: <strong>2^9, 2^6</strong></p>
Correct Answer: 2^9, 2^6