Question

If $a,b,c$ are three complex numbers such that $a^2+b^2+c^2=0$ and
$\Delta =\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+a^2& bc\\ ac& bc& a^2+b^2\end{array}\right|=k{a}^2{b}^2{c}^2$
then value of $k$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $4$

Answer 1

The correct option is D $4$

Using $a^2+b^2+c^2=0$ , we can write $\Delta$ as

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

$[Takinga,b,ccommonfrom{C}_1,C_2,C_{3}respectively]$

$a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|=a^2{b}^2{c}^2\left|\begin{array}{ccc}0& 2& 1\\ 0& 0& 1\\ 2& 0& -1\end{array}\right|$

$[Using{C}_1\to C_1+C_{2}and{C}_2\to C_2+C_3]$

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

Thus $k=4$