Question

If $a^2+b^2+c^2=0$ and $\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ac& bc& a^2+b^2\end{array}\right|=k{a}^2{b}^2{c}^2$ then k is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

Given determinant = $\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|$
$\left(\because a^2+b^2+c^2=0\right)$
$=abc\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$
(taking out a from $R_1$, b from $R_2$ and c from $R_3$)
$=\left(abc\right)\left(abc\right)\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$ ,
operate $C_2\to C_2+C_1$ and $C_3\to C_3+C_2$
$=a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 0& 2\\ 1& 0& 0\\ 1& 2& 0\end{array}\right|=4a^2{b}^2{c}^2$
Hence, $k=4$