Question

The equation $\left|\begin{array}{ccc}x-a& x-b& x-c\\ x-b& x-c& x-a\\ x-c& x-a& x-b\end{array}\right|=0$ 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 $\left|\begin{array}{ccc}x-a& x-b& x-c\\ x-b& x-c& x-a\\ x-c& x-a& x-b\end{array}\right|=0$ Applying the operation $C_1\to C_1+C_2+C_3,$ we get $\left|\begin{array}{ccc}3x-(a+b+c)& x-b& x-c\\ 3x-(a+b+c)& x-c& x-a\\ 3x-(a+b+c)& x-a& x-b\end{array}\right|=0$ $\Rightarrow$ $[3x-(a+b+c)]\left|\begin{array}{ccc}1& x-b& x-c\\ 1& x-c& x-a\\ 1& x-a& x-b\end{array}\right|=0$ Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ we get $[3x-(a+b+c)]\left|\begin{array}{ccc}1& x-b& x-c\\ 0& b-c& c-a\\ 0& b-a& x-b\end{array}\right|=0$ $\Rightarrow$ $\{3x-(a+b+c)\}\{(b-c)(c-b)$ $-(b-a)(c-a)\}=0$ $\Rightarrow$ $\{3x-(a+b+c)\}\{-b^2-c^2+2bc-bc$ $-bc-ca)\}=0$ $\Rightarrow$ $\{3x-(a+b+c)\}\{-(a^2+b^2+c^2-ab$ $-bc-ca)\}=0$ $\Rightarrow$ $\{3x-(a+b+c)\}(2a^2+2b^2+2c^2-2ab$ $-2bc-2ca)=0$ $\Rightarrow$ $\{3x-(a+b+c)\}\{{(a-b)}^2+{(b-c)}^2$ $+{(c-a)}^2\}=0$ $\Rightarrow$ $3x-(a+b+c)=0$ $[\therefore {(a-b)}^2+{(b-c)}^2+{(c-a)}^2\ne 0$ as a, b, c are different] $\Rightarrow$ $3x=a+b+c$ $x=\frac{1}{3}(a+b+c)$