If `A = [a_(ij)]` is a skew-symmetric matrix of order n, then `a_(ij)=`

Options

A. 0 for some i
B. 0 for all I = 1,2,…,n
C. 1 for some i
D. 1 for all I = 1,2,…,n

Answer

To solve the problem, we need to understand the properties of a skew-symmetric matrix. A skew-symmetric matrix \( A \) has the property that for any elements \( a_{ij} \) of the matrix, the following holds:

\[ a_{ij} = -a_{ji} \]

This means that the element in the \( i \)-th row and \( j \)-th column is the negative of the element in the \( j \)-th row and \( i \)-th column.

### Step-by-step Solution:

1. **Definition of Skew-Symmetric Matrix**:
A matrix \( A = [a_{ij}] \) is skew-symmetric if \( a_{ij} = -a_{ji} \) for all \( i, j \).

2. **Diagonal Elements**:
For the diagonal elements where \( i = j \), we have:
\[ a_{ii} = -a_{ii} \]
This implies that:
\[ 2a_{ii} = 0 \implies a_{ii} = 0 \]
Therefore, all diagonal elements of a skew-symmetric matrix are zero.

3. **Off-Diagonal Elements**:
For off-diagonal elements where \( i \neq j \), we can express \( a_{ij} \) in terms of \( a_{ji} \):
\[ a_{ij} = -a_{ji} \]
This shows that the elements are related by a negative sign.

4. **Conclusion**:
Thus, for any skew-symmetric matrix \( A \):
- The diagonal elements \( a_{ii} = 0 \) for all \( i \).
- The off-diagonal elements satisfy \( a_{ij} = -a_{ji} \).

### Final Answer:
The elements of the skew-symmetric matrix \( A \) can be summarized as:
\[ a_{ij} = \begin{cases} 0 & \text{if } i = j \\ -a_{ji} & \text{if } i \neq j \end{cases} \]