Q95 / 180 · KCET 2023 · Mathematics
If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$, then
- A$\Delta_1=3 \Delta$
- B$\Delta_1 \neq \Delta$
- Correct answerC$\Delta_1=-\Delta$
- D$\Delta_1=\Delta$
Solution
Given,
$\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$
and
$\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$
$\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$
$\begin{aligned} R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3 \\ \Delta=\left|\begin{array}{ccc}0 & a-b & a^2-b^2 \\ 0 & b-c & b^2-c^2 \\ 1 & c & c^2\end{array}\right| \\ \Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2\end{array}\right| \\ \Delta=(a-b)(b-c)(c-a)\end{aligned}$
$\begin{aligned} & \text { Now, } \Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right| \\ & C_1 \rightarrow C_1-C_2, C_2 \rightarrow C_2-C_3\end{aligned}$
$\Delta_1=\left|\begin{array}{ccc}0 & 0 & 1 \\ c(b-a) & a(c-b) & a b \\ a-b & b-c & c\end{array}\right|$
$\Delta_1=(a-b)\left(b^2-c\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ -c & -a & a b \\ 1 & 1 & c\end{array}\right|$
$\begin{aligned} & \Delta_1=(a-b)(b-c)(a-c) \\ & \Delta_1=-(a-b)(b-c)(c-a)\end{aligned}$
From Eqs. (i) and (ii), we get
$\Delta=-\Delta_1$
Hence, the correct answer is option: 'C'.