Question
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}=0$ 

Answer

$\text{Let}\ \triangle=\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}$
$=\triangle=(-1)^3\begin{vmatrix}0&-a&b\\a&0&c\\-b&-c&0\end{vmatrix}$ [Taking (-1) common from each row]
Interchanging rows and columns in the determinants on R.H.S.,
$\triangle=-\begin{vmatrix}0&a&-b\\-a&0&-c\\b&c&0\end{vmatrix}\ \Rightarrow\triangle=-\triangle\ \Rightarrow\triangle+\triangle=0$
$\Rightarrow 2\triangle=0\ \Rightarrow\triangle=0$ Proved.

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