Question

Consider the matrix

• A. \(\dfrac{2^4}{3^8}\)
• B. \(\dfrac{1}{3^5}\)
• C. \(\dfrac{2^5}{3^9}\)
• D. \(\dfrac{2^3}{3^7}\)

Hint:

A \(3\times 3\) matrix with entries from \(\{-1,0,1\}\) is orthogonal only when it is a signed permutation matrix.

Answer 1

The Correct Option is A

Step 1: See which columns work.
Entries come from $\{-1,0,1\}$. For an orthogonal matrix each column has length $1$, so a column must be all zeros except one $\pm1$.

Step 2: Build a valid matrix.
The non-zero spots in the three columns must sit in different rows, that is a permutation.

Step 3: Count them.
Choose the permutation in $3!=6$ ways and the signs in $2^3=8$ ways, giving $48$ orthogonal matrices.

Step 4: Count all matrices.
Each of $9$ entries has $3$ choices, so $3^9$ in total, all equally likely.

Step 5: Form the probability.
\[ P=\frac{48}{3^9}=\frac{2^4\cdot3}{3^9}=\frac{2^4}{3^8}. \]
\[ \boxed{\frac{2^4}{3^8}} \]