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 the value of $k$ is

• A. 1
• B. 2
• C. -2
• D. 4

Answer 1

Answer: (D)

Using $a^2+b^2+c^2=0$, we can write $\Delta$ as
$\Delta =\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|=abc\left|\begin{array}{ccc}-a& a& a\\ b& -b& b\\ c& c& -c\end{array}\right|$
[taking $a,b,c$ common from $C_1,C_2,C_3$ respectively]
$=a^2{b}^2{c}^2\left|\begin{array}{ccc}-1& 1& 1\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|$
[taking $a,b,c$ common from $R_1,R_2,R_3$ respectively]
$=a^2{b}^2{c}^2\left|\begin{array}{ccc}0& 2& 1\\ 2& 0& 1\\ 0& 0& -1<br/>\end{array}\right|=-2a^2{b}^2{c}^2\left|\begin{array}{cc}2& 1\\ 0& -1<br/>\end{array}\right|=4a^2{b}^2{c}^2$
[applying $C_1\to C_1+C_3$ and $C_3\to C_2+C_3$ ]
Thus, $k=4$.