Let $A = [a_{ij}]$ be a square matrix of order 2 with entries either 0 or 1. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:
Step-by-Step Solution
Key Concept: Total $2 \times 2$ matrices with 0/1 entries: $2^4 = 16$. Count non-invertible (singular) matrices (where $ad - bc = 0$), then subtract.
Total $= 2^4 = 16$. Non-invertible cases ($ad-bc=0$): Case I $ad=bc=1$: only $a=b=c=d=1$ (1 case). Case II $ad=bc=0$: 9 cases. Total non-invertible: 10. Invertible: 6. $P(E) = \frac{6}{16} = \frac{3}{8}$.
<div class="key-concept"><strong>Key Concept:</strong> Total $2 \times 2$ matrices with 0/1 entries: $2^4 = 16$. Count non-invertible (singular) matrices (where $ad - bc = 0$), then subtract.</div>
<div class="trap-box"><strong>Trap:</strong> Systematically list all cases where $ad = bc$ with $a,b,c,d \in \{0,1\}$; it's easy to miss or double-count cases like $a=d=0$ or all-zero rows.</div>
Correct Answer: $\frac{3}{8}$