Question

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

(a) abc(b-c)(c-a)(a-b)
(b) (b-c)(c-a)(a-b)
(c) (a+b+c)(b-c)(c-a)(a-b)
(d) None of these

Answer 1

(d) We have,
$\left|\begin{array}{ccc}{b}^2-ab& b-c& bc-ac\\ ab-a^2& a-b& b^2-ab\\ bc-ac& c-a& ab-a^2\end{array}\right|=\left|\begin{array}{ccc}b(b-a)& b-c& c(b-a)\\ a(b-a)& a-b& b(b-a)\\ c(b-a)& c-a& a(b-a)\end{array}\right|=(b-a{)}^2\left|\begin{array}{ccc}b& b-c& c\\ a& a-b& b\\ c& c-a& a\end{array}\right|$
[on taking (b-a) common from $C_1$ and $C_3$ each]
$=(b-a{)}^2\left|\begin{array}{ccc}b-c& b-c& c\\ a-b& a-b& b\\ c-a& c-a& a\end{array}\right|$ $[\because C_1\to C_1-C_3]$
$=0$
[since two columns $C_1$ and $C_2$ are identical, the value of determinant is zero]