Let $S$ be the set of all $3 \times 3$ symmetric matrices whose entries are either $0$ or $1$. Two of these entries are $1$ and four of them are $0$. A matrix is selected from set $S$, what is the probability that the selected matrix is non singular
Step-by-Step Solution
Key Concept: For a 3×3 symmetric matrix with exactly 2 ones and 4 zeros, the total count is C(6,2)=15 matrices (choosing 2 positions from 6 independent entries in upper triangle and diagonal). Non-singularity requires det(A)≠0, which depends on the specific placement of ones relative to the diagonal structure.
A $3 \times 3$ symmetric matrix with all diagonal elements equal to 1 has the form $\begin{pmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{pmatrix}$. For the matrix to be nonsingular, there are 12 such symmetric matrices total. Of these, 6 are nonsingular (those with determinant non-zero). Therefore, the probability that a randomly selected such matrix is nonsingular is $\frac{6}{12} = \frac{1}{2}$.
Correct Answer: 3