Question

If $\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|=K{a}^2{b}^2{c}^2$, then $K=$

• A. $-4$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

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

[Taking common $a$, $b$ and $c$ from row-$1$, row-$2$ and row-$3$ respectively.

$=abc\left|\begin{array}{ccc}-a& b& c\\ a& -b& c\\ a& b& -c\end{array}\right|$

[Taking common $a$, $b$ and $c$ from column-$1$, column-$2$ and column-$3$ respectively.

$=(abc{)}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$

[$R_1^{\prime }=R_1+R_2$]

$=(abc{)}^2\left|\begin{array}{ccc}0& 0& 2\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$

$=4a^2{b}^2{c}^2$

Comparing it with the given problem we get, $K=4$