Question

If $\Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=k(a-b)(b-c)(c-a)$, then $k$ is equal to:

• A. $-2$
• B. $1$
• C. $2$
• D. $abc$

Answer 1

Answer: (B)

$1$

$\Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=\left|\begin{array}{ccc}0& a-b& a^2-b^2\\ 0& b-c& b^2-c^2\\ 1& c& c^2\end{array}\right|,R_1\to R_1-R_2,R_2\to R_2-R_3$

$=(a-b)(b-c)\left|\begin{array}{ccc}0& 1& a+b\\ 0& 1& b+c\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)\left[1\right(b+c)-(a+b\left)\right]=(a-b)(b-c)(c-a)$

$\therefore k=1$