Question

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

• A. $1$
• B. $3$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

Let $\Delta=\begin{vmatrix}b^{2}+c^{2}&ab&ac\\ ab&c^{2}+b^{2}&bc\\ ac&bc&a^{2}+b^{2}\end{vmatrix}$
Multiply $C_1$ by $a$, $C_2$ by $b$ and $C_3$ by $c$ and hence divide by $abc.$$=\frac{1}{abc}\begin{vmatrix}a\left(b^{2}+c^{2}\right)&\left(ab^{2}\right)&ac^{2}\\ a^{2}b&b\left(c^{2}+b^{2}\right)&\left(bc^{2}\right)\\ a^{2}c&b^{2}c&c\left(a^{2}+b^{2}\right)\end{vmatrix}$
Take out $a, b, c$ common fromi $R_{1} R_{2}$ and $R_{3}$ respectively.
$\therefore =\frac{abc}{abc}\begin{vmatrix}b^{2}+c^{2}&b^{2}&c^{2}\\ a^{2}&c^{2}+b^{2}&c^{2}\\ a^{2}&b^{2}&a^{2}+b^{2}\end{vmatrix}$
Apply $C_{1} \rightarrow C_{1} -C_{2}-C_{3}$
$\Delta=\begin{vmatrix}0&b^{2}&c^{2}\\ -2c^{2}&c^{2}+b^{2}&c^{2}\\ -2b^{2}&b^{2}&a^{2}+b^{2}\end{vmatrix}$
$=-2\begin{vmatrix}0&b^{2}&c^{2}\\ c^{2}&c^{2}+b^{2}&c^{2}\\ b^{2}&b^{2}&a^{2}+b^{2}\end{vmatrix}$
Apply $C_{2} - C_{1}$ and $C_{3}-C_{1}$
$=-2\begin{vmatrix}0&b^{2}&c^{2}\\ c^{2}&a^{2}&0\\ b^{2}&0&a^{2}\end{vmatrix}$
$= -2[-b^2 (c^2a^2) + c^2 (-a^2b^2c^2)]$
$= 2a^2b^2c^2 + 2a^2b^2c^2 = 4a^2b^2c^2$
But $\Delta = ka^2b^2c^2 \therefore k = 4$