Mathematics - Determinant Question with Solution | TestHub

Let $\Delta =\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+\lambda {)}^2& (b+\lambda {)}^2& (c+\lambda {)}^2\\ (a-\lambda {)}^2& (b-\lambda {)}^2& (c-\lambda {)}^2\end{array}\right|$ $$\begin{array}{rl}& \text{ Apply }{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3\\ & \Delta =\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ (a+\lambda {)}^2-(a-\lambda {)}^2& (b+\lambda {)}^2-(b-\lambda {)}^2& (c+\lambda {)}^2-(c-\lambda {)}^2\\ (a-\lambda {)}^2& (b-\lambda {)}^2& (c-\lambda {)}^2\end{array}\right|\\ & =\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 4a\lambda & 4b\lambda & 4c\lambda \\ (a-\lambda {)}^2& (b-\lambda {)}^2& (c-\lambda {)}^2\end{array}\right|\\ & \left(\because (x+y{)}^2-(x-y{)}^2=4xy\right)\\ & \end{array}$$ Taking out 4 common from ${\mathrm{R}}_2$ $=4\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a\lambda & b\lambda & c\lambda \\ a^2+{\lambda }^2-2a\lambda & b^2+{\lambda }^2-2b\lambda & c^2+{\lambda }^2-2c\lambda \end{array}\right|$ $$\begin{array}{rl}& \text{ Apply }{\mathrm{R}}_3\to \left[{\mathrm{R}}_3-\left({\mathrm{R}}_1-2{\mathrm{R}}_2\right)\right]\\ & =4\left|\begin{array}{lll}{a}^2& b^2& c^2\\ a\lambda & b\lambda & c\lambda \\ {\lambda }^2& {\lambda }^2& {\lambda }^2\end{array}\right|\end{array}$$ Taking out $\lambda$ common from ${\mathrm{R}}_2$ and ${\lambda }^2$ from ${\mathrm{R}}_3$. $$=4\lambda \left({\lambda }^2\right)\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|$$ $$\begin{array}{rl}& =k\lambda \left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a& b& c\\ 1& 1& 1\end{array}\right|\\ & \Rightarrow k=4{\lambda }^2\end{array}$$