Question

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

• A. $abc$
• B. $a^2{b}^2{c}^2$
• C. $bc+ca+ab$
• D. Zero

Answer 1

The correct option is D Zero

Now,

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

[$R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3-R_1$]

or, $D=\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2(a^2-b^2)& c(a-b)& a-b\\ b^2(a^2-c^2)& b(a-c)& a-c\end{array}\right|$

or, $D=(a-b)(a-c)\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2(a+b)& c& 1\\ b^2(a+c)& b& 1\end{array}\right|$

[$R_3^{\prime }=R_3-R_2$]

or, $D=(a-b)(a-c)\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2(a+b)& c& 1\\ a(b^2-c^2)+bc(b-c)& b-c& 0\end{array}\right|$

or, $D=(a-b)(a-c)(b-c)\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ c^2(a+b)& c& 1\\ ab+ac+bc& 1& 0\end{array}\right|$

or, $D=(a-b)(a-c)(b-c)\left\{\right(ab+bc+ca\left)\right(bc-bc-c^2)-1(b^2{c}^2-c^2(b^2+ab+ac+bc)\left)\right\}$

or, $D=(a-b)(a-c)(b-c)\left\{\right(ab+bc+ca\left)\right(-c^2)+c^2(ab+ac+bc\left)\right\}$

or, $D=0$.