Question

The determinant $\left|\begin{matrix}b^{2}-ab&b-c&bc-ac\\ ab-a^{2}&a-b&b^{2}-ab\\ bc-ac&c-a&ab-a^{2}\end{matrix}\right|$ 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 $\Delta=\left|\begin{matrix}b^{2}-ab&b-c&bc-ac\\ ab-a^{2}&a-c&b^{2}-ab\\ bc-ac&c-a&ab-a^{2}\end{matrix}\right|$

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

Taking $\left(b - a\right)$ common from $C_{1}$ and $C_{3}$, we get
$\Delta=\left(b-a\right)^{2}\left|\begin{matrix}b&b-c&c\\ a&a-b&b\\ c&c-a&a\end{matrix}\right|$

Applying $C_{1} \rightarrow C_{1}- C_{2}$, we get
$\Delta=\left(b-a\right)^{2} \left|\begin{matrix}c&b-c&c\\ b&a-b&b\\ a&c-a&a\end{matrix}\right|$

Since, $C_{1}$ and $C_{3}$ are identical
$\Rightarrow \quad\Delta=0$