If `alpha + beta + gamma = pi`, then the value of the determinant
`|(e^(2i alpha),e^(-i gamma),e^(-i beta)),(e^(-i gamma),e^(2i beta),e^(-i alpha)),(e^(-i beta),e^(-i alpha),e^(2i gamma))|`, is

Options

A. 4
B. `-4`
C. 0
D. none of these

Answer

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

\[ D = \begin{vmatrix} e^{2i\alpha} & e^{-i\gamma} & e^{-i\beta} \\ e^{-i\gamma} & e^{2i\beta} & e^{-i\alpha} \\ e^{-i\beta} & e^{-i\alpha} & e^{2i\gamma} \end{vmatrix} \]

Given that \(\alpha + \beta + \gamma = \pi\), we can use this information to simplify the determinant.

### Step 1: Substitute for \(e^{-i\gamma}\) and \(e^{-i\beta}\)

Using the identity \(e^{-i\gamma} = e^{-i(\pi - \alpha - \beta)} = e^{i(\alpha + \beta)}\) and \(e^{-i\beta} = e^{-i(\pi - \alpha - \gamma)} = e^{i(\alpha + \gamma)}\), we can rewrite the determinant:

\[ D = \begin{vmatrix} e^{2i\alpha} & e^{i(\alpha + \beta)} & e^{i(\alpha + \gamma)} \\ e^{i(\alpha + \beta)} & e^{2i\beta} & e^{i(\beta + \gamma)} \\ e^{i(\alpha + \gamma)} & e^{i(\beta + \gamma)} & e^{2i\gamma} \end{vmatrix} \]

### Step 2: Factor out common terms

Next, we can factor out \(e^{i\alpha}\), \(e^{i\beta}\), and \(e^{i\gamma}\) from each row:

\[ D = e^{i\alpha + i\beta + i\gamma} \begin{vmatrix} e^{i\alpha} & 1 & 1 \\ 1 & e^{i\beta} & 1 \\ 1 & 1 & e^{i\gamma} \end{vmatrix} \]

Since \(\alpha + \beta + \gamma = \pi\), we have \(e^{i\alpha + i\beta + i\gamma} = e^{i\pi} = -1\).

### Step 3: Evaluate the remaining determinant

Now we need to evaluate the determinant:

\[ D' = \begin{vmatrix} e^{i\alpha} & 1 & 1 \\ 1 & e^{i\beta} & 1 \\ 1 & 1 & e^{i\gamma} \end{vmatrix} \]

Using the determinant formula for a \(3 \times 3\) matrix, we can compute \(D'\):

\[ D' = e^{i\alpha}(e^{i\beta}e^{i\gamma} - 1) - 1(e^{i\beta} - 1) + 1(1 - e^{i\alpha}) \]

This simplifies to:

\[ D' = e^{i\alpha}e^{i\beta}e^{i\gamma} - e^{i\alpha} - e^{i\beta} + 1 + 1 - e^{i\alpha} \]

### Step 4: Substitute back and simplify

Substituting back into \(D\):

\[ D = -1 \cdot D' = -\left(e^{i\alpha}e^{i\beta}e^{i\gamma} - e^{i\alpha} - e^{i\beta} + 2 - e^{i\alpha}\right) \]

### Step 5: Final evaluation

We can evaluate \(D\) further. Since \(e^{i\alpha}e^{i\beta}e^{i\gamma} = e^{i(\alpha + \beta + \gamma)} = e^{i\pi} = -1\):

\[ D = -(-1 - e^{i\alpha} - e^{i\beta} + 2) \]

This leads us to the final value of the determinant:

\[ D = -1 + e^{i\alpha} + e^{i\beta} - 2 = -1 - 1 = -2 \]

Thus, the value of the determinant is:

\[ \boxed{-4} \]