If `A_1, A_2, , A_(2n-1)a r en` skew-symmetric matrices of same order, then `B=sum_(r=1)^n(2r-1)(A^(2r-1))^(2r-1)` will be i) symmetric ii) skew-symmetric iii) neither symmetric nor skew-symmetric iv) data not adequate

Options

A. symmetric
B. skew-symmetric
C. neither symmetric nor skew-symmetric
D. data not adequate

Explanation

To solve the given problem, we need to analyze the properties of the skew-symmetric matrices and the expression for \( B \).

### Step-by-step Solution:

1. **Understanding Skew-Symmetric Matrices**:
A matrix \( A \) is skew-symmetric if \( A^T = -A \). Given that \( A_1, A_3, \ldots, A_{2n-1} \) are skew-symmetric matrices, we have:
\[ A_{2r-1}^T = -A_{2r-1} \]
for \( r = 1, 2, \ldots, n \).

**Hint**: Recall the definition of skew-symmetric matrices and their properties.

2. **Expression for \( B \)**:
The expression for \( B \) is given by:
\[ B = \sum_{r=1}^{n} (2r-1)(A_{2r-1})^{2r-1} \]
We need to analyze the transpose of \( B \).

**Hint**: Write down the expression for \( B \) clearly to see how it is formed.

3. **Taking the Transpose of \( B \)**:
We take the transpose of \( B \):
\[ B^T = \left( \sum_{r=1}^{n} (2r-1)(A_{2r-1})^{2r-1} \right)^T \]
Using the property of transpose, we can distribute it:
\[ B^T = \sum_{r=1}^{n} (2r-1) \left( (A_{2r-1})^{2r-1} \right)^T \]

**Hint**: Remember that the transpose of a product of matrices follows the rule \( (XY)^T = Y^T X^T \).

4. **Applying the Transpose Property**:
Since \( A_{2r-1} \) is skew-symmetric:
\[ (A_{2r-1})^{2r-1} \text{ is skew-symmetric if } 2r-1 \text{ is odd.} \]
Therefore, we have:
\[ \left( (A_{2r-1})^{2r-1} \right)^T = - (A_{2r-1})^{2r-1} \]
Thus:
\[ B^T = \sum_{r=1}^{n} (2r-1)(- (A_{2r-1})^{2r-1}) = -\sum_{r=1}^{n} (2r-1)(A_{2r-1})^{2r-1} = -B \]

**Hint**: Use the property of skew-symmetric matrices to simplify the expression.

5. **Conclusion**:
Since we have shown that \( B^T = -B \), it follows that \( B \) is skew-symmetric.

**Final Answer**: The correct option is (ii) skew-symmetric.