Mathematics - Determinant Question with Solution | TestHub

$A_r=\left|\begin{array}{ccc}r& 1& \frac{n^2}{2}+\alpha \\ 2r& 2& n^2-\beta \\ 3r-2& 3& \frac{n(3n-1)}{2}\end{array}\right|$
$2{\mathrm{A}}_{10}-{\mathrm{A}}_8=\left|\begin{array}{ccc}20& 1& \frac{{\mathrm{n}}^2}{2}+\alpha \\ 40& 2& {\mathrm{n}}^2-\beta \\ 56& 3& \frac{\mathrm{n}(3\mathrm{n}-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8& 1& \frac{{\mathrm{n}}^2}{2}+\alpha \\ 16& 2& {\mathrm{n}}^2-\beta \\ 22& 3& \frac{\mathrm{n}(3\mathrm{n}-1)}{2}\end{array}\right|$ $\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}12& 1& \frac{n^2}{2}+\alpha \\ 24& 2& n^2-\beta \\ 34& 3& \frac{n(3n-1)}{2}\end{array}\right|\\ & \Rightarrow \left|\begin{array}{ccc}0& 1& \frac{n^2}{2}+\alpha \\ 0& 2& n^2-\beta \\ -2& 3& \frac{n(3n-1)}{2}\end{array}\right|\\ & \Rightarrow -2\left(\left(n^2-\beta \right)-\left(n^2+2\alpha \right)\right)\\ & \Rightarrow -2(-\beta -2\alpha )\Rightarrow 4\alpha +2\beta \end{array}$