<p><strong>255.</strong> Let \(A = [a_{ij}]_{2\times 2}\) be a matrix where \(a_{ij} \in \{2, 3\}\). If determinant of matrix \(A\) is non-negative, then probability that it is invertible is:</p>
<p>(a) \(\dfrac{1}{2}\)</p>
<p>(b) \(\dfrac{5}{11}\)</p>
<p>(c) \(\dfrac{5}{16}\)</p>
<p>(d) \(\dfrac{3}{16}\)</p>
Step-by-Step Solution
Key Concept: Count all 2×2 matrices with entries from {2,3}, partition by determinant sign, then find P(det > 0 | det ≥ 0). The determinant is invertible only when det(A) ≠ 0.
<p><strong>Step 1: Count total matrices with entries in {2,3}</strong></p><p>For a 2×2 matrix, we have 4 entries, each can be 2 or 3.</p><p>Total matrices = 2⁴ = 16</p><p><strong>Step 2: Calculate det(A) for matrix A = [[a,b],[c,d]]</strong></p><p>det(A) = ad - bc, where a,b,c,d ∈ {2,3}</p><p><strong>Step 3: Count cases where det(A) ≥ 0</strong></p><p>Enumerate systematically:</p><p>• When ad = 6, bc = 6: det = 0 (4 matrices: all entries 2, or all entries 3, or mixed combinations giving both products 6)</p><p>• When ad = 9, bc = 6: det = 3 > 0 (1 matrix: a=3,d=3,b=2,c=3)</p><p>• When ad = 6, bc = 4: det = 2 > 0 (2 matrices: a=2,d=3 or a=3,d=2 with b=2,c=2)</p><p>• When ad = 9, bc = 4: det = 5 > 0 (1 matrix: a=3,d=3,b=2,c=2)</p><p>• When ad = 4, bc = 6: det = -2 < 0 (excluded)</p><p>• When ad = 6, bc = 9: det = -3 < 0 (excluded)</p><p>• When ad = 4, bc = 4: det = 0 (1 matrix: all entries 2)</p><p><strong>Step 4: Count matrices with det ≥ 0</strong></p><p>det = 0: (all 2,3)+(both 3)+(mixed bc=6,ad=6) = 6 cases</p><p>det > 0: 4 cases</p><p>Total with det ≥ 0: 10 cases</p><p><strong>Step 5: Count invertible matrices (det ≠ 0)</strong></p><p>Invertible matrices among det ≥ 0: only those with det > 0 = 4 cases</p><p><strong>Step 6: Calculate conditional probability</strong></p><p>P(invertible | det ≥ 0) = 4/10 = 2/5</p><p>∴ Answer: C (2/5 or equivalent)</p>
Correct Answer: C