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 two 1's and four 0's, the matrix structure is determined by where the 1's are placed. A symmetric matrix has only 6 independent entries (3 diagonal + 3 above diagonal), and singularity depends on whether det(A) = 0, which requires analyzing the 1's placement across diagonal and off-diagonal positions.
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