Elements of a matrix A of order 10 x 10 are defined as `a_(ij)=omega^(i+j)` (where omega is cube root unity), then tr(A) of matrix is

Options

A. 0
B. 1
C. 3
D. none of these

Explanation

To solve the problem, we need to find the trace of the matrix \( A \) defined by its elements \( a_{ij} = \omega^{i+j} \), where \( \omega \) is a cube root of unity. The trace of a matrix is the sum of its diagonal elements.

### Step-by-Step Solution:

1. **Understanding the Matrix Elements**:
The elements of the matrix \( A \) are given by \( a_{ij} = \omega^{i+j} \). Therefore, the diagonal elements (where \( i = j \)) will be \( a_{ii} = \omega^{i+i} = \omega^{2i} \).

2. **Identifying the Diagonal Elements**:
For a \( 10 \times 10 \) matrix, the diagonal elements will be:
\[ a_{11} = \omega^2, \quad a_{22} = \omega^4, \quad a_{33} = \omega^6, \quad a_{44} = \omega^8, \quad a_{55} = \omega^{10}, \quad a_{66} = \omega^{12}, \quad a_{77} = \omega^{14}, \quad a_{88} = \omega^{16}, \quad a_{99} = \omega^{18}, \quad a_{1010} = \omega^{20} \]

3. **Calculating the Trace**:
The trace \( \text{tr}(A) \) is the sum of these diagonal elements:
\[ \text{tr}(A) = \omega^2 + \omega^4 + \omega^6 + \omega^8 + \omega^{10} + \omega^{12} + \omega^{14} + \omega^{16} + \omega^{18} + \omega^{20} \]

4. **Using Properties of Cube Roots of Unity**:
Since \( \omega \) is a cube root of unity, we know that \( \omega^3 = 1 \). Thus, we can simplify the powers of \( \omega \):
- \( \omega^{10} = \omega^{3 \cdot 3 + 1} = \omega^1 = \omega \)
- \( \omega^{12} = \omega^{3 \cdot 4} = 1 \)
- \( \omega^{14} = \omega^{3 \cdot 4 + 2} = \omega^2 \)
- \( \omega^{16} = \omega^{3 \cdot 5 + 1} = \omega \)
- \( \omega^{18} = \omega^{3 \cdot 6} = 1 \)
- \( \omega^{20} = \omega^{3 \cdot 6 + 2} = \omega^2 \)

5. **Grouping the Terms**:
Now, we can rewrite the trace:
\[ \text{tr}(A) = \omega^2 + \omega^4 + \omega^6 + \omega^8 + \omega + 1 + \omega^2 + \omega + 1 + \omega^2 \]
Simplifying this gives:
\[ \text{tr}(A) = 4\omega^2 + 3\omega + 3 \]

6. **Final Calculation**:
Using the fact that \( 1 + \omega + \omega^2 = 0 \), we can express \( \omega + \omega^2 = -1 \):
\[ \text{tr}(A) = 4\omega^2 + 3(-1) = 4\omega^2 - 3 \]

7. **Conclusion**:
The final result for the trace of the matrix \( A \) is:
\[ \text{tr}(A) = \omega^2 \]