Animated Solution for Mathematics - Matrices and Determinants: For $\alpha ,\beta \in \mathbb{R}$ and a natural number $n$, let $A_r=\left|\begin{array}{ccc}r& 1& \frac{n^2}{2}+\alpha \\ 2r& 2& n^2-\beta \\ 3r-2& 3& \frac{n(3n-1)}{2}\end{array}\right|$. Then $2A_{10}-A_8$ is equal to :

We are tasked with evaluating the expression $2A_{10}-A_8$, where $A_r$ is a determinant defined by variables $\alpha$, $\beta$, and $n$. The determinant is structured such that only the first column depends on the index $r$.

The second and third columns remain static across different values of $r$. This observation is the key to simplifying the problem through the linearity property of determinants.

Instead of calculating $A_{10}$ and $A_8$ individually, we combine them into a single determinant by applying the linear operation to the first column. We define the new first column as $C_1=2C_1(10)-C_1(8)$.

Calculating the specific values for the first column: $2(10)-8=12$ $2(20)-16=24$ $2(28)-22=34$

The resulting determinant is:

To simplify the calculation, we perform row operations to introduce zeros into the second column. We apply $R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$.

Applying these operations to the first and second columns yields: $24-2(12)=0$ and $2-2(1)=0$ $34-3(12)=-2$ and $3-3(1)=0$

After performing the corresponding operations on the third column, the determinant simplifies to:

With two zeros now present in the second column, we expand the determinant along that column. We must remember the sign convention for the cofactor expansion, which gives us a factor of $-1$.

The expansion is:

Evaluating the $2\times 2$ determinant:

The variable $n$ cancels out entirely during the process. The final result of the expression is $4\alpha +2\beta$.