Question

The equation $ \left| \begin{matrix} x-a & x-b & x-c \\ x-b & x-c & x-a \\ x-c & x-a & x-b \\ \end{matrix} \right|=0 $ where a, b, c are different is satisfied by

• A. $ x=0 $
• B. $ x=a $
• C. $ x=\frac{1}{3}(a+b+c) $
• D. $ x=a+b+c $

Answer 1

Answer: (C)

We have the equation $ \left| \begin{matrix} x-a & x-b & x-c \\ x-b & x-c & x-a \\ x-c & x-a & x-b \\ \end{matrix} \right|=0 $ Applying the operation $ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}, $ we get $ \left| \begin{matrix} 3x-(a+b+c) & x-b & x-c \\ 3x-(a+b+c) & x-c & x-a \\ 3x-(a+b+c) & x-a & x-b \\ \end{matrix} \right|=0 $ $ \Rightarrow $ $ [3x-(a+b+c)]\left| \begin{matrix} 1 & x-b & x-c \\ 1 & x-c & x-a \\ 1 & x-a & x-b \\ \end{matrix} \right|=0 $ Applying $ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} $ and $ {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} $ we get $ [3x-(a+b+c)]\left| \begin{matrix} 1 & x-b & x-c \\ 0 & b-c & c-a \\ 0 & b-a & x-b \\ \end{matrix} \right|=0 $ $ \Rightarrow $ $ \{3x-(a+b+c)\}\{(b-c)(c-b) $ $ -(b-a)(c-a)\}=0 $ $ \Rightarrow $ $ \{3x-(a+b+c)\}\{-{{b}^{2}}-{{c}^{2}}+2bc-bc $ $ -bc-ca)\}=0 $ $ \Rightarrow $ $ \{3x-(a+b+c)\}\{-({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab $ $ -bc-ca)\}=0 $ $ \Rightarrow $ $ \{3x-(a+b+c)\}(2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab $ $ -2bc-2ca)=0 $ $ \Rightarrow $ $ \{3x-(a+b+c)\}\{{{(a-b)}^{2}}+{{(b-c)}^{2}} $ $ +{{(c-a)}^{2}}\}=0 $ $ \Rightarrow $ $ 3x-(a+b+c)=0 $ $ [\therefore {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\ne 0 $ as a, b, c are different] $ \Rightarrow $ $ 3x=a+b+c $ $ x=\frac{1}{3}(a+b+c) $