Question

If $a,b,c$ are non zero complex number satisfying $a^2+b^2+c^2=0$ 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

The correct option is D $4$

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

$b^2+c^2(c^2+a^2\times a^2+b^2-b^2{c}^2)-ab\left(\right(a^{3}b+a{b}^3)-ab{c}^2)$

$+ac(a{b}^{2}c-(a{c}^3+a^{3}c\left)\right)$

$b^2+c^2(a^2{c}^2+b^2{c}^2+a^4+a^2{b}^2-b^2{c}^2)-a^4{b}^2-a^2{b}^4$

$+a^2{b}^2{c}^2+a^2{b}^2{c}^2-a^2{c}^4-a^4{c}^2$

$a^2{b}^2{c}^2+a^4{b}^2+a^2{b}^4+a^2{c}^4+a^4{c}^2+a^2{b}^2{c}^2-a^4{b}^2-a^2{b}^4$

$+a^2{b}^2{c}^2+a^2{b}^2{c}^2-a^2{c}^4-a^4{c}^2$

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

$\therefore k=4$