Question

For \( \alpha, \beta \in \mathbb{R} \) and a natural number \( n \), let \[A_r = \begin{vmatrix} r & 1 & \frac{n^2}{2} + \alpha \\ 2r & 2 & n^2 - \beta \\3r - 2 & 3 & \frac{n(3n - 1)}{2} \end{vmatrix}.\]Then \( 2A_{10} - A_8 \) is:

• A. \( 4\alpha + 2\beta \)
• B. \( 2\alpha + 4\beta \)
• C. \( 2n \)
• D. \( 0 \)

Answer 1

The Correct Option is A

Step 1: Write the determinant for \( A_r \):

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

Step 2: Expand \( 2A_{10} \): Substitute \( r = 10 \):

\[ 2A_{10} = 2 \cdot \begin{vmatrix} 10 & 1 & \frac{n^2}{2} + \alpha \\ 20 & 2 & \frac{n^2}{2} - \beta \\ 28 & 3 & n\frac{3n-1}{2} \end{vmatrix}. \]

Step 3: Expand \( A_5 \): Substitute \( r = 5 \):

\[ A_5 = \begin{vmatrix} 5 & 1 & \frac{n^2}{2} + \alpha \\ 10 & 2 & \frac{n^2}{2} - \beta \\ 13 & 3 & n\frac{3n-1}{2} \end{vmatrix}. \]

Step 4: Compute \( 2A_{10} - A_5 \):

\[ 2A_{10} - A_5 = \begin{vmatrix} 20 & 1 & \frac{n^2}{2} + \alpha \\ 40 & 2 & \frac{n^2}{2} - \beta \\ 56 & 3 & n\frac{3n-1}{2} \end{vmatrix} - \begin{vmatrix} 8 & 1 & \frac{n^2}{2} + \alpha \\ 16 & 2 & \frac{n^2}{2} - \beta \\ 22 & 3 & n\frac{3n-1}{2} \end{vmatrix}. \]

Step 5: Simplify: Subtract the rows:

\[ 2A_{10} - A_5 = \begin{vmatrix} 12 & 1 & \frac{n^2}{2} + \alpha \\ 24 & 2 & \frac{n^2}{2} - \beta \\ 34 & 3 & n\frac{3n-1}{2} \end{vmatrix}. \]

Factor and simplify further:

\[ = -2 \left[ (n^2 - \beta) - (n^2 + 2\alpha) \right] = -2(-\beta - 2\alpha). \]

Therefore:

\[ 2A_{10} - A_5 = 4\alpha + 2\beta. \]