Question

The determinant $\left|\begin{array}{ccc}{b}^2-ab& b-c& bc-ac\\ ab-a^2& a-b& b^2-ab\\ bc-ca& c-a& ab-a^2\end{array}\right|$ equals

(a) $abc\left(b-c\right)\left(c-a\right)\left(a-b\right)$
(b) $\left(b-c\right)\left(c-a\right)\left(a-b\right)$
(c) $\left(a+b+c\right)\left(b-c\right)\left(c-a\right)\left(a-b\right)$
(d) none of these

Answer 1

$$\begin{gathered} \left|\begin{array}{ccc}{b}^2-ab& b-c& bc-ac\\ ab-a^2& a-b& b^2-ab\\ bc-ca& c-a& ab-a^2\end{array}\right| \\ =\left|\begin{array}{ccc}b\left(b-a\right)& b-c& c\left(b-a\right)\\ a\left(b-a\right)& a-b& b\left(b-a\right)\\ c\left(b-a\right)& c-a& a\left(b-a\right)\end{array}\right| \\ ={\left(b-a\right)}^2\left|\begin{array}{ccc}b& b-c& c\\ a& a-b& b\\ c& c-a& a\end{array}\right|\left[\mathrm{Taking}\left(b-a\right)\mathrm{common}\mathrm{from}{C}_1\mathrm{and}{C}_3\right] \\ ={\left(b-a\right)}^2\left|\begin{array}{ccc}0& b-c& c\\ 0& a-b& b\\ 0& c-a& a\end{array}\right|\left[\mathrm{Applying}{C}_1\to C_1-C_2-C_3\right] \\ =0 \end{gathered}$$

Hence, the correct option is (d).