Question

If $a,b,c$ are non-zero real numbers then
$\Delta =\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2{a}^2& ca& c+a\\ a^2{b}^2& ab& a+b\end{array}\right|$ is equal to

• A. $abc$
• B. $a^2{b}^2{c}^2$
• C. $bc+ca+ab$
• D. None of these

Answer 1

The correct option is D None of these

Given, $\Delta =\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2{a}^2& ca& c+a\\ a^2{b}^2& ab& a+b\end{array}\right|$

Applying $R_1\to a{R}_1,R_2\to b{R}_2,R_3\to c{R}_3$ we get

$\Delta =\frac{1}{abc}\left|\begin{array}{ccc}a{b}^2{c}^2& abc& ab+ac\\ a^{2}b{c}^2& abc& bc+ab\\ a^2{b}^{2}c& abc& ac+bc\end{array}\right|$

$\Delta =\frac{a^2{b}^2{c}^2}{abc}\left|\begin{array}{ccc}bc& 1& ab+ac\\ ca& 1& bc+ab\\ ab& 1& ac+bc\end{array}\right|$ Taking common $abc$ from $C_1$ and $C_2$

Applying $C_3\to C_3+C_1$

and taking $bc+ca+ab$ common we get

$\Delta =abc(bc+ca+ab)\left|\begin{array}{ccc}bc& 1& 1\\ ca& 1& 1\\ ab& 1& 1\end{array}\right|=0$ ............... $[\because C_{2}and{C}_{3}areidentical]$