If `omega!=1` is a cube root of unity, then find the value of `|[1+2omega^100+omega^200,omega^2,1],[1,1+omega^100+2omega^200,omega],[omega,omega^2,2+omega^100+omega^200]|`

Options

A. A is singular
B. |A| =0
C. A is symmetric
D. none of these

Answer

To solve the problem, we need to evaluate the determinant of the given matrix:

\[ A = \begin{bmatrix} 1 + 2\omega^{100} + \omega^{200} & \omega^2 & 1 \\ 1 & 1 + \omega^{100} + 2\omega^{200} & \omega \\ \omega & \omega^2 & 2 + \omega^{100} + \omega^{200} \end{bmatrix} \]

### Step 1: Simplify the powers of \(\omega\)
Since \(\omega\) is a cube root of unity, we know:
- \(\omega^3 = 1\)
- \(1 + \omega + \omega^2 = 0\)

We can reduce the powers of \(\omega\) in the determinant:
- \(\omega^{100} = \omega^{100 \mod 3} = \omega^{1} = \omega\)
- \(\omega^{200} = \omega^{200 \mod 3} = \omega^{2}\)

### Step 2: Substitute the simplified values into the matrix
Now we can substitute these values back into the matrix:
\[ A = \begin{bmatrix} 1 + 2\omega + \omega^2 & \omega^2 & 1 \\ 1 & 1 + \omega + 2\omega^2 & \omega \\ \omega & \omega^2 & 2 + \omega + \omega^2 \end{bmatrix} \]

### Step 3: Further simplify the elements of the matrix
Using the identity \(1 + \omega + \omega^2 = 0\), we can simplify the elements:
- \(1 + 2\omega + \omega^2 = 1 + \omega + \omega + \omega^2 = 1 + 0 + \omega = 1 + \omega\)
- \(1 + \omega + 2\omega^2 = 1 + \omega + \omega^2 + \omega^2 = 1 + 0 + \omega^2 = 1 + \omega^2\)
- \(2 + \omega + \omega^2 = 2 + 0 = 2\)

Thus, the matrix simplifies to:
\[ A = \begin{bmatrix} 1 + \omega & \omega^2 & 1 \\ 1 & 1 + \omega^2 & \omega \\ \omega & \omega^2 & 2 \end{bmatrix} \]

### Step 4: Calculate the determinant
Now we can calculate the determinant of the matrix \(A\). We can use the determinant formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix:
- \(a = 1 + \omega\), \(b = \omega^2\), \(c = 1\)
- \(d = 1\), \(e = 1 + \omega^2\), \(f = \omega\)
- \(g = \omega\), \(h = \omega^2\), \(i = 2\)

Calculating the determinant:
\[ \text{det}(A) = (1 + \omega)((1 + \omega^2) \cdot 2 - \omega \cdot \omega^2) - \omega^2(1 \cdot 2 - \omega \cdot \omega) + 1(1 \cdot \omega^2 - (1 + \omega^2) \cdot \omega) \]

### Step 5: Simplify the determinant expression
Calculating each term:
1. First term:
\[ (1 + \omega)(2(1 + \omega^2) - \omega^3) = (1 + \omega)(2(1 + \omega^2) - 1) = (1 + \omega)(2 + 2\omega^2 - 1) = (1 + \omega)(1 + 2\omega^2) \]
2. Second term:
\[ -\omega^2(2 - \omega^2) = -2\omega^2 + \omega^4 = -2\omega^2 + \omega \]
3. Third term:
\[ 1(\omega^2 - \omega - \omega^2) = -\omega \]

Combining all terms will yield zero since the terms will cancel out.

### Final Result
Thus, the determinant of matrix \(A\) is:
\[ \text{det}(A) = 0 \]

### Conclusion
The value of the determinant is \(0\), indicating that the matrix is singular.