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Q.
If $a^2 + b^2 + c^2 = 0$ and $\begin{vmatrix}b^{2}+c^{2}&ab&ac\\ ab&c^{2}+a^{2}&bc\\ ac&bc&a^{2}+b^{2}\end{vmatrix} = ka^{2}b^{2}c^{2} $ then k is equal to
Determinants
Solution:
Given determinant = $\begin{vmatrix}-a^{2}&ab&ac\\ ab&-b^{2}&bc\\ ac&bc&-c^{2}\end{vmatrix} $
$ \left(\because a^{2} +b^{2}+c^{2}= 0 \right)$
$ =abc \begin{vmatrix}-a&b&c\\ a&-b&c\\ a&b&-c\end{vmatrix}$
(taking out a from $R_1$, b from $R_2$ and c from $R_3$)
$=\left(abc\right)\left(abc\right) \begin{vmatrix}-1&1&1\\ 1&-1&1\\ 1&1&-1\end{vmatrix} $ ,
operate $ C_{2} \to C_{2} +C_{1}$ and $ C_{3} \to C_{3}+C_{2} $
$=a^{2}b^{2}c^{2}\begin{vmatrix}-1&0&2\\ 1&0&0\\ 1&2&0\end{vmatrix}=4a^{2} b^{2}c^{2} $
Hence, $k = 4 $