Q118 / 180 · COMEDK 2014 · Mathematics

$\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2}-b c & b^{2}-c a & c^{2}-a b\end{array}\right|$ is equal to

  • Correct answer
    A
    0
  • B
    1
  • C
    $a b c$
  • D
    $(a-b)(b-c)(c-a)$

Solution

Let $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2}-b c & b^{2}-c a & c^{2}-a b\end{array}\right|$ On applying $C_{3} \rightarrow C_{3}-C_{2}$ and $C_{2} \rightarrow C_{2}-C_{1}$, we get $$ \begin{aligned} \Delta=&\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & c-b \\ a^{2}-b c & +c(b-a) & +a(c-b) \end{array}\right| \\ =& 1[(b-a)(c-b)(c+b+a)\\ &-(c-b)(b-a)(b+a+c)]=0 \end{aligned} $$

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

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