If $\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$ $…

If $\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$ $ =\alpha(a+b() b+c() c+a) \neq 0 $ then $\alpha$ is equal to
  1. $a+b+c$
  2. $a b c$
  3. 4
  4. 1

Solution

Let $\Delta=\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$ Applying $C_1+C_3$ and $C_2+C_3$ $ \Delta=\left|\begin{array}{ccc} -a+c & 2 a+b+c & a+c \\ 2 b+a+c & -b+c & b+c \\ a-c & b-c & -2 c \end{array}\right| $ Now, applying $R_1+R_3$ and $R_2+R_3$ $ \Delta=\left|\begin{array}{ccc} 0 & 2(a+b & ) a-c \\ 2(a+b & 0 & b-c \\ a-c & b-c & -2 c \end{array}\right| $ On expanding, we get $ \begin{aligned} \Delta= & -2(a+b)\{-2 c[2(a+b)] \\ & -(a-c)(b-c)\} \\ & +(a-c)[2(a+b)(b-c)] \\ \Delta= & 8 \mathrm{c}(a+b)(a+b) \\ & +4(a+b)(a-c)(b-c) \end{aligned} $ $ \begin{aligned} & =4(a+b)\left[2 a c+2 b c+a b-b c-a c+c^2\right] \\ & =4(a+b)\left[a c+b c+a b+c^2\right] \\ & =4(a+b)[c(a+c)+b(a+c)] \\ & =4(a+b)(b+c)(c+a) \\ & =\alpha(a+b)(b+c)(c+a) \end{aligned} $ Hence, $\alpha=4$

Asked in: JEE Main 2012 (12 May Online)

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