A) \[4\lambda abc\]
B) \[-4\lambda abc\]
C) \[4{{\lambda }^{2}}\]
D) \[-4{{\lambda }^{2}}\]
Correct Answer: C
Solution :
Let \[\Delta =\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{(a+\lambda )}^{2}} & {{(b+\lambda )}^{2}} & {{(c+\lambda )}^{2}} \\ {{(a-\lambda )}^{2}} & {{(b-\lambda )}^{2}} & {{(c-\lambda )}^{2}} \\ \end{matrix} \right|\] Apply \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] \[\Delta =\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{(a+\lambda )}^{2}}-{{(a-\lambda )}^{2}} & {{(b+\lambda )}^{2}}-{{(b-\lambda )}^{2}} & {{(c+\lambda )}^{2}}-{{(c-\lambda )}^{2}} \\ {{(a-\lambda )}^{2}} & {{(b-\lambda )}^{2}} & {{(a-\lambda )}^{2}} \\ \end{matrix} \right|\]\[=\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ 4a\lambda & 4b\lambda & 4c\lambda \\ {{(a-\lambda )}^{2}} & {{(b-\lambda )}^{2}} & {{(a-\lambda )}^{2}} \\ \end{matrix} \right|\] \[(\because {{(x+y)}^{2}}-{{(x-y)}^{2}}=4xy)\] Taking out 4 common from \[{{R}_{2}}\] \[=4\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a\lambda & b\lambda & c\lambda \\ {{a}^{2}}+{{\lambda }^{2}}-2a\lambda & {{b}^{2}}+{{\lambda }^{2}}-2b\lambda & {{c}^{2}}+{{\lambda }^{2}}-2c\lambda \\ \end{matrix} \right|\]Apply \[{{R}_{3}}\to [{{R}_{3}}-({{R}_{1}}-2{{R}_{2}})]\]\[=4\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a\lambda & b\lambda & c\lambda \\ {{\lambda }^{2}} & {{\lambda }^{2}} & {{\lambda }^{2}} \\ \end{matrix} \right|\] Taking out \[\lambda \] common from \[{{R}_{2}}\]and \[{{\lambda }^{2}}\]from \[{{R}_{3}}.\] \[=4\lambda ({{\lambda }^{2}})\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix} \right|\]\[=k\lambda \left| \begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix} \right|\] \[\Rightarrow \]\[k=4{{\lambda }^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
