Q60 / 160 · TS EAMCET 2008 · Mathematics

$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ is equal to

  • A
    $0$
  • B
    $a+b+c$
  • C
    $(a+b+c)^2$
  • Correct answer
    D
    $(a+b+c)^3$

Solution

Let $\Delta=\left[\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right]$ Applying $\quad R_1 \rightarrow R_1+R_2+R_3 \quad$ and taking common $(a+b+c)$ from $R_1$ $$ =(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| $$ Applying $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$, $$ \begin{aligned} & =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -b-c-a & 0 \\ 2 c & 0 & -a-b-c \end{array}\right| \\ & =(a+b+c)[(-b-c-a)(-a-b-c)] \\ & =(a+b+c)^3 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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