If `alpha,beta,gamma` are the roots of `x^(3)+2x^(2)-x-3=0` The value of `|{:(alpha, beta ,gamma),(gamma,alpha ,beta),(beta,gamma ,alpha):}|` is equal to

Options

A. 14
B. -2
C. 10
D. -14

Answer

To solve the problem, we need to evaluate the determinant given the roots of the polynomial \( x^3 + 2x^2 - x - 3 = 0 \). The roots are denoted as \( \alpha, \beta, \gamma \). We will denote the determinant as:

\[ D = \begin{vmatrix} \alpha & \beta & \gamma \\ \gamma & \alpha & \beta \\ \beta & \gamma & \alpha \end{vmatrix} \]

### Step 1: Find the roots' relationships
From Vieta's formulas, for the polynomial \( x^3 + 2x^2 - x - 3 = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{2}{1} = -2 \) (Equation 1)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -\frac{-1}{1} = -1 \) (Equation 2)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{1} = 3 \) (Equation 3)

### Step 2: Set up the determinant
We can express the determinant \( D \) as follows:

\[ D = \begin{vmatrix} \alpha & \beta & \gamma \\ \gamma & \alpha & \beta \\ \beta & \gamma & \alpha \end{vmatrix} \]

### Step 3: Simplify the determinant
To simplify the determinant, we can perform row operations. We can add all three rows together:

\[ R_1 + R_2 + R_3 \rightarrow R_1 \]

This gives us:

\[ D = \begin{vmatrix} \alpha + \beta + \gamma & \alpha + \beta + \gamma & \alpha + \beta + \gamma \\ \gamma & \alpha & \beta \\ \beta & \gamma & \alpha \end{vmatrix} \]

### Step 4: Factor out the common term
We can factor out \( \alpha + \beta + \gamma \) from the first row:

\[ D = (\alpha + \beta + \gamma) \begin{vmatrix} 1 & 1 & 1 \\ \gamma & \alpha & \beta \\ \beta & \gamma & \alpha \end{vmatrix} \]

### Step 5: Compute the remaining determinant
Now we need to compute the determinant:

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

Expanding this determinant along the first row:

\[ D' = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \gamma & \alpha \end{vmatrix} - 1 \cdot \begin{vmatrix} \gamma & \beta \\ \beta & \alpha \end{vmatrix} + 1 \cdot \begin{vmatrix} \gamma & \alpha \\ \beta & \gamma \end{vmatrix} \]

Calculating these 2x2 determinants:

1. \( \begin{vmatrix} \alpha & \beta \\ \gamma & \alpha \end{vmatrix} = \alpha^2 - \beta\gamma \)

2. \( \begin{vmatrix} \gamma & \beta \\ \beta & \alpha \end{vmatrix} = \gamma\alpha - \beta^2 \)

3. \( \begin{vmatrix} \gamma & \alpha \\ \beta & \gamma \end{vmatrix} = \gamma^2 - \beta\alpha \)

Combining these results:

\[ D' = (\alpha^2 - \beta\gamma) - (\gamma\alpha - \beta^2) + (\gamma^2 - \beta\alpha) \]

### Step 6: Substitute the known values
Now substituting \( \alpha + \beta + \gamma = -2 \) and \( \alpha\beta + \beta\gamma + \gamma\alpha = -1 \):

\[ D' = \alpha^2 + \beta^2 + \gamma^2 - (\beta\gamma + \gamma\alpha + \beta\alpha) \]

Using the identity \( \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \):

\[ \alpha^2 + \beta^2 + \gamma^2 = (-2)^2 - 2(-1) = 4 + 2 = 6 \]

Thus:

\[ D' = 6 - (-1) = 6 + 1 = 7 \]

### Step 7: Final calculation
Now substituting back into our expression for \( D \):

\[ D = (-2) \cdot 7 = -14 \]

### Conclusion
The value of the determinant is:

\[ \boxed{-14} \]