Question

An \( n \times n \) matrix is formed using \( 0, 1 \) and \( -1 \) as its elements. The number of such matrices which are skew symmetric is:

• A. \( \frac{n(n-1)}{2} \)
• B. \( (n-1)^2 \)
• C. \( 2^{\frac{n(n-1)}{2}} \)
• D. \( 3^{\frac{n(n-1)}{2}} \)

Hint:

For skew symmetric matrices: \begin{itemize} \item Diagonal elements are always zero. \item Only upper triangular entries are independent. \item Number of independent entries = \( \frac{n(n-1)}{2} \). \end{itemize}

Answer 1

The Correct Option is D

Concept: A matrix \( A \) is skew symmetric if: \[ A^T = -A \quad \Rightarrow \quad a_{ij} = -a_{ji} \] Important properties: \begin{itemize} \item All diagonal elements must be zero: \( a_{ii} = 0 \) \item Elements below diagonal are determined by elements above diagonal \end{itemize} So only upper triangular entries (excluding diagonal) are independent. Step 1: {\color{red}Count independent positions.} Total entries in an \( n \times n \) matrix: \[ n^2 \] Diagonal elements: \[ n \quad (\text{fixed as } 0) \] Remaining positions: \[ n^2 - n = n(n-1) \] Since entries are paired: \[ a_{ij} = -a_{ji} \] Independent positions: \[ \frac{n(n-1)}{2} \] Step 2: {\color{red}Choices for each independent entry.} Each independent entry can be chosen from: \[ \{ -1, 0, 1 \} \] So each has 3 choices. Lower triangle entries are automatically fixed by skew symmetry. Step 3: {\color{red}Total number of skew symmetric matrices.} \[ \text{Total} = 3^{\frac{n(n-1)}{2}} \]