Q45 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

If $\Delta_1=\left|\begin{array}{lll}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{array}\right|$ and $\Delta_2=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|$, then $\frac{\Delta_1}{\Delta_2}=$

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

Solution

$$ \begin{aligned} & \text { (a) } \Delta_1=\left|\begin{array}{lll} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{array}\right| \\ & R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1 \\ & =\left|\begin{array}{ccc} 1 & a^2 & a^3 \\ 0 & b^2-a^2 & b^3-a^3 \\ 0 & c^2-a^2 & c^3-a^3 \end{array}\right| \\ & =(b-a)(c-a)\left|\begin{array}{ccc} 1 & a^2 & a^3 \\ 0 & b+a & b^2+a^2+a b \\ 0 & c+a & c^2+a^2+a c \end{array}\right| \\ & =(b-a)(c-a)\left[(b+a)\left(c^2+a^2+a c\right)\right. \\ & \left.-(c+a)\left(b^2+a^2+a b\right)\right] \\ & \end{aligned} $$ $$ \begin{aligned} = & (b-a)(c-a)\left[b c^2+a^2 b+a b c+a c^2\right. \\ & \left.+a^3+a^2 c-b^2 c-a^2 c-a b c-b^2 a-a^3-a^2 b\right] \\ = & (b-a)(c-a)[(b-c)(a b+b c+c a)] \\ = & -(a-b)(b-c)(c-a)(a b+b c+c a) \end{aligned} $$ Now, $$ \Delta_2=\left|\begin{array}{lll} b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1 \end{array}\right| $$ $$ R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3-R_1 $$ $$ \begin{aligned} & =\left|\begin{array}{ccc} b c & b+c & 1 \\ c(a-b) & a-b & 0 \\ b(c-a) & a-c & 0 \end{array}\right|=(a-b)(c-a)\left|\begin{array}{ccc} b c & b+c & 1 \\ c & 1 & 0 \\ a & 1 & 0 \end{array}\right| \\ & =(a-b)(c-a)[1(c-b)] \\ & =(a-b)(c-b)(c-a) \\ & \end{aligned} $$ Now, $\frac{\Delta_1}{\Delta_2}=\frac{-(a-b)(b-c)(c-a)(a b+b c+c a)}{-(a-b)(b-c)(c-a)}$ $$ \frac{\Delta_1}{\Delta_2}=(a b+b c+c a) $$

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

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