Entries of a $2 \times 2$ determinant are chosen from the set $\{-1, 1\}$. The probability that determinant has zero value is:
Step-by-Step Solution
Key Concept: A determinant equals zero when $ad = bc$, which with binary entries $\{-1, 1\}$ has a countable number of solutions.
For the determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc = 0$, we need $ad = bc$. With entries from $\{-1, 1\}$, this occurs when $ad = 1, bc = 1$ or $ad = -1, bc = -1$, giving 8 favorable outcomes. Since total $2 \times 2$ determinants from $\{-1, 1\}$ is 16, the probability is $\frac{8}{16} = \frac{1}{2}$.
Correct Answer: 3