A) \[4\,\left| \,\begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix}\, \right|\]
B) \[3\,\,\left| \,\begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix}\, \right|\]
C) \[2\,\,\left| \,\begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix}\, \right|\]
D) None of these
Correct Answer: A
Solution :
Apply \[{{R}_{2}}-{{R}_{3}}\] and note that \[{{(x+y)}^{2}}-{{(x-y)}^{2}}=4xy\] \[\therefore \] \[\Delta =4\,\left| \,\begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ {{(a-1)}^{2}} & {{(b-1)}^{2}} & {{(c-1)}^{2}} \\ \end{matrix}\, \right|\] = \[4\,\left| \,\begin{matrix} {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\ a & b & c \\ 1 & 1 & 1 \\ \end{matrix}\, \right|\] {Applying\[{{R}_{3}}-({{R}_{1}}-2{{R}_{2}})\}\].
You need to login to perform this action.
You will be redirected in
3 sec
