Question

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

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

Answer 1

The correct option is D none of these

$\Delta =\left|\begin{array}{ccc}{b}^2{c}^2& bc& b+c\\ a^2{c}^2& ac& a+c\\ b^2{a}^2& ab& a+b\end{array}\right|\Delta =\frac{1}{b^2{a}^2{c}^2}\left|\begin{array}{ccc}{b}^2{a}^2{c}^2& a^{2}bc& (b+c)a^2\\ b^2{a}^2{c}^2& a{b}^{2}c& (a+c)b^2\\ b^2{a}^2{c}^2& ab{c}^2& (b+a)c^2\end{array}\right|=\frac{b^2{a}^2{c}^2\left(abc\right)}{b^2{a}^2{c}^2}\left|\begin{array}{ccc}1& a& (b+c)a^2\\ 1& b& (a+c)b^2\\ 1& c& (b+a)c^2\end{array}\right|$
Applying $R_2\to R_2-R_1;R_3\to R_3-R_1$
$=abc\left|\begin{array}{ccc}1& a& (b+c)a^2\\ 0& b-a& (a+c)b^2-(b+c)a^2\\ 0& c-a& (b+a)c^2-(b+c)a^2\end{array}\right|=abc\{(b-a)(a{c}^2+b{c}^2-b{a}^2-c{a}^2)-(c-a)(a{b}^2+c{b}^2-b{a}^2-c{a}^2)\}$