1. Key Concepts and Formulas

• Properties of Determinants: Applying elementary row or column operations of the form $R_i \to R_i + kR_j$ (or $C_i \to C_i + kC_j$) to a determinant does not change its value. This property is crucial for simplifying complex determinants.
• Expansion of a $3 \times 3$ Determinant: A $3 \times 3$ determinant can be expanded along any row or column. If expanding along the $j$-th column, the formula is $\det(A) = \sum_{i=1}^{3} (-1)^{i+j} a_{ij} M_{ij}$, where $a_{ij}$ is the element in the $i$-th row and $j$-th column, and $M_{ij}$ is its corresponding $2 \times 2$ minor. Expanding along a row or column with multiple zeros significantly reduces computation.
• Calculation of a $2 \times 2$ Determinant: For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant is $ad - bc$.

2. Step-by-Step Solution

Step 1: Analyze the Given Determinant and Identify a Simplification Strategy. We are given the determinant $A_r$: $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$ Our goal is to find $2A_{10} - A_8$. Directly expanding this determinant would be cumbersome due to the complex expressions in the third column. A more efficient approach is to use row operations to create zeros in a column (or row), thereby simplifying the expansion. Observe the second column: it contains the simple sequence $1, 2, 3$. This makes it an ideal column to target for creating zeros. We can use the first element ($a_{12}=1$) to eliminate the other elements in this column.

Step 2: Apply Row Operations to Create Zeros in the Second Column. We will perform two row operations to make $a_{22}$ and $a_{32}$ zero.

• Operation 1: $R_2 \to R_2 - 2R_1$. This operation subtracts two times the elements of the first row from the corresponding elements of the second row.

  • For the first element: $(2r) - 2(r) = 2r - 2r = 0$.
  • For the second element: $(2) - 2(1) = 2 - 2 = 0$.
  • For the third element: $(n^2 - \beta) - 2\left(\frac{n^2}{2}+\alpha\right) = n^2 - \beta - n^2 - 2\alpha = -2\alpha - \beta$. The new second row is $\begin{pmatrix} 0 & 0 & -2\alpha - \beta \end{pmatrix}$.

• Operation 2: $R_3 \to R_3 - 3R_1$. This operation subtracts three times the elements of the first row from the corresponding elements of the third row.

  • For the first element: $(3r - 2) - 3(r) = 3r - 2 - 3r = -2$.
  • For the second element: $(3) - 3(1) = 3 - 3 = 0$.
  • For the third element: $\frac{n(3 n-1)}{2} - 3\left(\frac{n^2}{2}+\alpha\right) = \frac{3n^2 - n}{2} - \frac{3n^2}{2} - 3\alpha = \frac{3n^2 - n - 3n^2}{2} - 3\alpha = -\frac{n}{2} - 3\alpha$. The new third row is $\begin{pmatrix} -2 & 0 & -\frac{n}{2} - 3\alpha \end{pmatrix}$.

After these operations, the determinant $A_r$ becomes: $A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 0 & 0 & -2\alpha - \beta \\ -2 & 0 & -\frac{n}{2} - 3\alpha\end{array}\right|$

Step 3: Expand the Simplified Determinant. Now, the second column has two zeros, making it ideal for expansion. We will expand $A_r$ along the second column ($C_2$). $A_r = (-1)^{1+2} (1) \left|\begin{array}{cc} 0 & -2\alpha - \beta \\ -2 & -\frac{n}{2} - 3\alpha \end{array}\right| + (-1)^{2+2} (0) M_{22} + (-1)^{3+2} (0) M_{32}$ The terms with $0$ will vanish, so we only need to calculate the first term: $A_r = -1 \cdot \left( (0) \left(-\frac{n}{2} - 3\alpha\right) - (-2) (-2\alpha - \beta) \right)$ $A_r = -1 \cdot (0 - (4\alpha + 2\beta))$ $A_r = -1 \cdot (-4\alpha - 2\beta)$ $A_r = 4\alpha + 2\beta$ Notice that the value of $A_r$ is $4\alpha + 2\beta$, which is a constant and does not depend on $r$ or $n$. This is a crucial simplification.

Step 4: Evaluate the Expression $2A_{10} - A_8$. Since $A_r = 4\alpha + 2\beta$ for any value of $r$:

• $A_{10} = 4\alpha + 2\beta$
• $A_8 = 4\alpha + 2\beta$

Substitute these values into the required expression: $2A_{10} - A_8 = 2(4\alpha + 2\beta) - (4\alpha + 2\beta)$ $2A_{10} - A_8 = 8\alpha + 4\beta - 4\alpha - 2\beta$ $2A_{10} - A_8 = (8\alpha - 4\alpha) + (4\beta - 2\beta)$ $2A_{10} - A_8 = 4\alpha + 2\beta$

3. Common Mistakes & Tips

• Algebraic Errors: The third column contains more complex expressions. Be extremely careful with signs and distribution when performing row operations and combining terms. A single sign error can lead to an incorrect answer.
• Premature Expansion: Avoid expanding the determinant directly without first attempting to simplify it using row/column operations. This can save significant time and reduce the chances of calculation errors.
• Missing Independence: Recognize when a variable (like $r$ in this case) cancels out or disappears during simplification. This often means the final calculation becomes much simpler than it initially appears.

4. Summary

The problem required evaluating an expression involving a determinant $A_r$. By strategically applying row operations ($R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$), we transformed the determinant into a simpler form with two zeros in the second column. Expanding along this column revealed that $A_r$ simplifies to a constant value, $4\alpha + 2\beta$, independent of $r$ and $n$. Substituting this constant value into the expression $2A_{10} - A_8$ yielded the final result of $4\alpha + 2\beta$.

The final answer is $\boxed{4\alpha+2\beta}$, which corresponds to option (A).