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

• 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

Answer: (D)

Let $\Delta =\left|\begin{array}{ccc}{b}^2-ab& b-c& bc-ac\\ ab-a^2& a-c& b^2-ab\\ bc-ac& c-a& ab-a^2\end{array}\right|$

$=\left|\begin{array}{ccc}b\left(b-a\right)& b-c& c\left(b-a\right)\\ a\left(b-a\right)& a-b& b\left(b-a\right)\\ c\left(b-a\right)& c-a& a\left(b-a\right)\end{array}\right|$

Taking $\left(b-a\right)$ common from $C_1$ and $C_3$, we get
$\Delta ={\left(b-a\right)}^2\left|\begin{array}{ccc}b& b-c& c\\ a& a-b& b\\ c& c-a& a\end{array}\right|$

Applying $C_1\to C_1-C_2$, we get
$\Delta ={\left(b-a\right)}^2\left|\begin{array}{ccc}c& b-c& c\\ b& a-b& b\\ a& c-a& a\end{array}\right|$

Since, $C_1$ and $C_3$ are identical
$\Rightarrow \Delta =0$