Question

If $a,b$ and $c$ 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

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

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

$=bca\left|\begin{array}{ccc}{b}^2{c}^2& c& 1\\ a^2{c}^2& a& 1\\ a^2{b}^2& b& 1\end{array}\right|+bac\left|\begin{array}{ccc}{b}^2{c}^2& b& 1\\ a^2{c}^2& c& 1\\ a^2{b}^2& a& 1\end{array}\right|$

$=abc\left[b{c}^2\right(a-b)-c(a^{2}c-a{b}^2)+1(a^{2}cb-a^2{b}^2\left)\right]+abc\left[b^{2}c\right(c-a)-b(a{c}^2-a^{2}b)+1(a^2{c}^2-a^{2}bc\left)\right]$

$=abc[ba{c}^2-b^2{c}^2-a^2{c}^2+a{b}^{2}c+a^{2}cb-a^2{b}^2+b^2{c}^2-a{b}^{2}c-ba{c}^2+a^2{b}^2+a^2{c}^2-a^{2}bc]$

$=0$

$\therefore D=0$