Question

If $\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ba& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|=$ square of a determinant $\Delta$ of the third order then $\Delta$ is equal to

• A. $\left|\begin{array}{ccc}0& c& b\\ c& 0& a\\ b& a& 0\end{array}\right|$
• B. $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
• C. $\left|\begin{array}{ccc}0& -c& b\\ c& 0& -a\\ -b& -a& 0\end{array}\right|$
• D. none of these

Answer 1

The correct option is A $\left|\begin{array}{ccc}0& c& b\\ c& 0& a\\ b& a& 0\end{array}\right|$

Checking option wise

Let $\Delta =\left|\begin{array}{ccc}0& c& b\\ c& 0& a\\ b& a& 0\end{array}\right|$

Then

${\Delta }^2=\left|\begin{array}{ccc}0& c& b\\ c& 0& a\\ b& a& 0\end{array}\right|\left|\begin{array}{ccc}0& c& b\\ c& 0& a\\ b& a& 0\end{array}\right|=\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ba& c^2+a^2& bc\\ ca& cb& a^2+b^2\end{array}\right|$

Hence, option 'A' is correct.