Mathematics - Determinant Question with Solution | TestHub

$\left|\begin{array}{ccc}{\left(1+\mathrm{\alpha }\right)}^2& {\left(1+2\mathrm{\alpha }\right)}^2& {\left(1+3\mathrm{\alpha }\right)}^2\\ {\left(2+\mathrm{\alpha }\right)}^2& {\left(2+2\mathrm{\alpha }\right)}^2& {\left(2+3\mathrm{\alpha }\right)}^2\\ {\left(3+\mathrm{\alpha }\right)}^2& {\left(3+2\mathrm{\alpha }\right)}^2& {\left(3+3\mathrm{\alpha }\right)}^2\end{array}\right|$

${\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1$(Take common)
${\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1$

$2\left|\begin{array}{ccc}{\left(1+\mathrm{\alpha }\right)}^2& {\left(1+2\mathrm{\alpha }\right)}^2& {\left(1+3\mathrm{\alpha }\right)}^2\\ 3+2\mathrm{\alpha }& \left(3+4\mathrm{\alpha }\right)& \left(3+6\mathrm{\alpha }\right)\\ \left(4+2\mathrm{\alpha }\right)& \left(4+4\mathrm{\alpha }\right)& \left(4+6\mathrm{\alpha }\right)\end{array}\right|$

${\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_2$

$2\left|\begin{array}{ccc}{\left(1+\mathrm{\alpha }\right)}^2& {\left(1+2\mathrm{\alpha }\right)}^2& {\left(1+3\mathrm{\alpha }\right)}^2\\ 3+2\mathrm{\alpha }& 3+4\mathrm{\alpha }& 3+6\mathrm{\alpha }\\ 1& 1& 1\end{array}\right|$

${\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1$(Take common)
${\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1$

$\therefore 2\mathrm{\alpha }2.\mathrm{\alpha }\left|\begin{array}{ccc}{\left(1+\mathrm{\alpha }\right)}^2& \left(2+3\mathrm{\alpha }\right)& \left(2+4\mathrm{\alpha }\right)\\ 3+2\mathrm{\alpha }& 2& 2\\ 1& 0& 0\end{array}\right|$

$4{\mathrm{\alpha }}^2\left|-2\mathrm{\alpha }\right|=-8{\mathrm{\alpha }}^3=-648\mathrm{\alpha }$
$\mathrm{\alpha }=0,\mathrm{ }{\mathrm{\alpha }}^2=81,\mathrm{ }\mathrm{\alpha }=\pm 9$