Question

If $a,b,c$, are non zero complex numbers satisfying $a^2+b^2+c^2=0$ and
$\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+b^2& bc\\ ac& bc& a^2+b^2\end{array}\right|=k^2{a}^2{b}^2{c}^2,$ then $k$ is equal to

• A. $1$
• B. $3$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

Let $\Delta =\left|\begin{array}{ccc}{b}^2+c^2& ab& ac\\ ab& c^2+b^2& bc\\ ac& bc& a^2+b^2\end{array}\right|$
Multiply $C_1$ by $a$, $C_2$ by $b$ and $C_3$ by $c$ and hence divide by $abc.$$=\frac{1}{abc}\left|\begin{array}{ccc}a\left(b^2+c^2\right)& \left(a{b}^2\right)& a{c}^2\\ a^{2}b& b\left(c^2+b^2\right)& \left(b{c}^2\right)\\ a^{2}c& b^{2}c& c\left(a^2+b^2\right)\end{array}\right|$
Take out $a,b,c$ common fromi $R_1{R}_2$ and $R_3$ respectively.
$\therefore =\frac{abc}{abc}\left|\begin{array}{ccc}{b}^2+c^2& b^2& c^2\\ a^2& c^2+b^2& c^2\\ a^2& b^2& a^2+b^2\end{array}\right|$
Apply $C_1\to C_1-C_2-C_3$
$\Delta =\left|\begin{array}{ccc}0& b^2& c^2\\ -2c^2& c^2+b^2& c^2\\ -2b^2& b^2& a^2+b^2\end{array}\right|$
$=-2\left|\begin{array}{ccc}0& b^2& c^2\\ c^2& c^2+b^2& c^2\\ b^2& b^2& a^2+b^2\end{array}\right|$
Apply $C_2-C_1$ and $C_3-C_1$
$=-2\left|\begin{array}{ccc}0& b^2& c^2\\ c^2& a^2& 0\\ b^2& 0& a^2\end{array}\right|$
$=-2[-b^2(c^2{a}^2)+c^2(-a^2{b}^2{c}^2)]$
$=2a^2{b}^2{c}^2+2a^2{b}^2{c}^2=4a^2{b}^2{c}^2$
But $\Delta =k{a}^2{b}^2{c}^2\therefore k=4$