Question

The value of determinant $\left|\begin{array}{ccc}1/a& bc& a^3\\ 1/b& ca& b^3\\ 1/c& ab& c^3\end{array}\right|$

• A. $a^2{b}^2{c}^2(a-b)(b-c)(c-a)$
• B. $0$
• C. $(a-b)(b-c)(c-a)$
• D. None of the above

Answer 1

The correct option is B $0$

Let $\Delta =\left|\begin{array}{ccc}1/a& bc& a^3\\ 1/b& ca& b^3\\ 1/c& ab& c^3\end{array}\right|$
Applying $R_1\to a{R}_1$, $R_2\to b{R}_2$, $R_3\to c{R}_3$, we get
$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}1& abc& a^4\\ 1& abc& b^4\\ 1& abc& c^4\end{array}\right|$
Taking $abc$ common from $C_2$, we get
$\Delta =\frac{abc}{abc}\left|\begin{array}{ccc}1& 1& a^4\\ 1& 1& b^4\\ 1& 1& c^4\end{array}\right|$
$=0$ $[\because C_1{C}_2]$.