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

Answer: (A)

$4\alpha + 2\beta$

Answer 2

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:

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