Discussion
Consider the matrix
\[ A = \begin{bmatrix} X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{23} \\ X_{31} & X_{32} & X_{33} \end{bmatrix}, \]
where \(X_{ij},\ 1 \leq i, j \leq 3\), are independent and identically distributed random variables with the probability mass function
\[ f(x) = \begin{cases} \frac{1}{3}, & x = -1, 0, 1, \\ 0, & \text{otherwise.} \end{cases} \]
The probability that \(A\) is orthogonal equals
Answer: A
The probability is \(\frac{2^{4}}{3^{8}}\).
Each row of an orthogonal matrix has length 1. With entries from \(\{-1, 0, 1\}\), a row of length 1 has exactly one entry \(\pm 1\) and the rest 0. Rows must also be mutually orthogonal, so their non-zero entries sit in different columns. So \(A\) is orthogonal exactly when it is a signed permutation matrix.
There are \(3! = 6\) ways to place the non-zero entries and \(2^{3} = 8\) choices of sign, giving 48 such matrices. All \(3^{9}\) matrices are equally likely, so
\[ \begin{aligned} P &= \frac{48}{3^{9}} = \frac{2^{4} \cdot 3}{3^{9}} \\ &= \frac{2^{4}}{3^{8}} \end{aligned} \]
\(\frac{2^{5}}{3^{9}}\) counts 32 matrices and \(\frac{2^{3}}{3^{7}}\) counts 72, so both miscount the signed permutations; \(\frac{1}{3^{5}}\) corresponds to 81.
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