If `A_(1), A_(2), A_(3)...........A_(20)` are 20 skew - symmetric matrices of same order and `B=Sigma_(r=1)^(20)2r(A_(r))^((2r+1))`, then the sum of the principal diagonal elements of matrix B is equal to

Answer

To solve the problem, we need to find the sum of the principal diagonal elements (trace) of the matrix \( B \) defined as:

\[ B = \sum_{r=1}^{20} 2r (A_r)^{(2r+1)} \]

where \( A_1, A_2, \ldots, A_{20} \) are skew-symmetric matrices.

### Step 1: Understand the properties of skew-symmetric matrices
A skew-symmetric matrix \( A \) has the property that:

\[ A^T = -A \]

and all diagonal entries of a skew-symmetric matrix are zero.

**Hint:** Remember that for any skew-symmetric matrix, the diagonal elements are always zero.

### Step 2: Analyze the matrix \( B \)
The matrix \( B \) is a summation of terms involving the skew-symmetric matrices \( A_r \). Each term in the summation is of the form \( 2r (A_r)^{(2r+1)} \).

**Hint:** Focus on the exponent \( (2r + 1) \) which is always odd.

### Step 3: Determine the transpose of \( B \)
Taking the transpose of \( B \):

\[ B^T = \left( \sum_{r=1}^{20} 2r (A_r)^{(2r+1)} \right)^T = \sum_{r=1}^{20} 2r \left( (A_r)^{(2r+1)} \right)^T \]

Using the property of skew-symmetric matrices, we have:

\[ (A_r)^{(2r+1)} = (A_r^T)^{(2r+1)} = (-A_r)^{(2r+1)} = - (A_r)^{(2r+1)} \]

Thus:

\[ B^T = \sum_{r=1}^{20} 2r \left( - (A_r)^{(2r+1)} \right) = -B \]

**Hint:** The property \( (A^T)^n = (A^n)^T \) can be useful here.

### Step 4: Conclude that \( B \) is skew-symmetric
Since \( B^T = -B \), we conclude that \( B \) is a skew-symmetric matrix.

**Hint:** Recall that if a matrix is skew-symmetric, its diagonal entries must be zero.

### Step 5: Find the trace of \( B \)
The trace of a matrix is the sum of its diagonal elements. Since \( B \) is skew-symmetric, all of its diagonal elements are zero. Therefore, the trace of \( B \) is:

\[ \text{Trace}(B) = 0 \]

### Final Answer
The sum of the principal diagonal elements of matrix \( B \) is equal to:

\[ \boxed{0} \]