Question

If $a,b,c$ are three complex numbers such that $a^2+b^2+c^2=0$ 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 value of $k$ is -

• A. $1$
• B. $2$
• C. $-2$
• D. $4$

Answer 1

The correct option is D $4$

$C_1\to a{C}_1$, $C_2\to b{C}_2$, $C_3\to c{C}_3$

$\frac{1}{abc}\left|\begin{array}{ccc}a(b^2+c^2)& a{b}^2& a{c}^2\\ a^{2}b& b(c^2+a^2)& b{c}^2\\ a^{2}c& b^{2}c& c(a^2+b^2)\end{array}\right|$

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

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

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

$=a^2{b}^2{c}^2[-1(0)-1(-1-1)+1(1+1\left)\right]$

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