For and a natural number n , let A r = | r 1 n 2 2 + 2 r 2 n 2 - 3 r - 2 3 n ( 3 n - 1 ) 2 | . Then 2 A 10 - A 8 is [2024]

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Solution

Q.
For $α, β \in R$ and a natural number $n$ , let $A_{r} =$ $| \begin{matrix} r & 1 & \frac{n^{2}}{2} + α \\ 2 r & 2 & n^{2} - β \\ 3 r - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ . Then $2 A_{10} - A_{8}$ is [2024]

1 $0$

2 $4 α + 2 β$

3 $2 n$

4 $2 α + 4 β$

Ans.
(2)

We have, $2 A_{10} - A_{8} =$ $| \begin{matrix} 20 & 1 & \frac{n^{2}}{2} + α \\ 40 & 2 & n^{2} - β \\ 56 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $-$ $| \begin{matrix} 8 & 1 & \frac{n^{2}}{2} + α \\ 16 & 2 & n^{2} - β \\ 22 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$

$=$ $| \begin{matrix} 20 - 8 & 1 & \frac{n^{2}}{2} + α \\ 40 - 16 & 2 & n^{2} - β \\ 56 - 22 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $=$ $| \begin{matrix} 12 & 1 & \frac{n^{2}}{2} + α \\ 24 & 2 & n^{2} - β \\ 34 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$

$=$ $| \begin{matrix} 0 & 1 & \frac{n^{2}}{2} + α \\ 0 & 2 & n^{2} - β \\ - 2 & 3 & \frac{n (3 n - 1)}{2} \end{matrix} |$ $(Applying C_{1} \to C_{1} - 12 C_{2})$

$= - 2 (n^{2} - β - n^{2} - 2 α)$

$= - 2 (- β - 2 α) = 4 α + 2 β$

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