Let `a_r = |[r, 1, (n^2)/2 + alpha], [2r, 2, n^2 - beta], [3r - 2, 3, n(n-1)]|`, then value of `2a_(10)-a_8` is

Options

A. 0
B. `4 alpha + 2 beta`
C. `8 alpha + 4 beta`
D. `2 alpha + beta n`

Answer

To solve the problem, we need to find the value of \( 2a_{10} - a_{8} \) where \( a_r \) is defined as the determinant of the following matrix:

\[ a_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & n(n-1) \end{vmatrix} \]

### Step 1: Simplify the Determinant

First, we can factor out constants from the rows of the determinant. Notice that we can take out a factor of 2 from the second row:

\[ = 2 \cdot \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ r & 1 & \frac{n^2 - \beta}{2} \\ 3r - 2 & 3 & n(n-1) \end{vmatrix} \]

### Step 2: Perform Row Operations

Next, we perform row operations to simplify the determinant further. We can subtract the second row from the first row:

\[ = 2 \cdot \begin{vmatrix} 0 & 0 & \frac{n^2}{2} + \alpha - \frac{n^2 - \beta}{2} \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & n(n-1) \end{vmatrix} \]

This simplifies to:

\[ = 2 \cdot \begin{vmatrix} 0 & 0 & \frac{\beta + \alpha}{2} \\ 2r & 2 & n^2 - \beta \\ 3r - 2 & 3 & n(n-1) \end{vmatrix} \]

### Step 3: Calculate the Determinant

Since the first row has two zeros, we can expand the determinant:

\[ = 2 \cdot \left(\frac{\beta + \alpha}{2}\right) \cdot \begin{vmatrix} 2r & 2 \\ 3r - 2 & 3 \end{vmatrix} \]

Calculating the \( 2 \times 2 \) determinant:

\[ = 2 \cdot \left(\frac{\beta + \alpha}{2}\right) \cdot (2 \cdot 3 - 2 \cdot (3r - 2)) \]

This simplifies to:

\[ = 2 \cdot \left(\frac{\beta + \alpha}{2}\right) \cdot (6 - 6r + 4) = 2 \cdot \left(\frac{\beta + \alpha}{2}\right) \cdot (10 - 6r) \]

### Step 4: Final Expression for \( a_r \)

Thus, we have:

\[ a_r = \beta + \alpha \cdot (10 - 6r) \]

### Step 5: Calculate \( a_{10} \) and \( a_{8} \)

Now, we can find \( a_{10} \) and \( a_{8} \):

\[ a_{10} = \beta + \alpha \cdot (10 - 60) = \beta + \alpha \cdot (-50) \]
\[ a_{8} = \beta + \alpha \cdot (10 - 48) = \beta + \alpha \cdot (-38) \]

### Step 6: Calculate \( 2a_{10} - a_{8} \)

Now we can compute \( 2a_{10} - a_{8} \):

\[ 2a_{10} = 2(\beta - 50\alpha) = 2\beta - 100\alpha \]
\[ 2a_{10} - a_{8} = (2\beta - 100\alpha) - (\beta - 38\alpha) = 2\beta - 100\alpha - \beta + 38\alpha \]
\[ = \beta - 62\alpha \]

### Final Answer

Thus, the value of \( 2a_{10} - a_{8} \) is:

\[ \boxed{4\alpha + 2\beta} \]