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

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
Let $\bigtriangleup=\begin{vmatrix}0 & a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}$

Let $R_1\rightarrow CR_1$,by applying this property

$\bigtriangleup=\frac{1}{C}\begin{vmatrix}0 & ac & -bc\\-a & 0 & -c\\b & c & 0\end{vmatrix}$

Similarly let $R_1\rightarrow R_1-bR_2$,by applying this property

$\bigtriangleup=\frac{1}{C}\begin{vmatrix}ab & ac & 0\\-a & 0 & -c\\b & c & 0\end{vmatrix}$

By taking 'a' as the common factor from $R_1$

$\bigtriangleup=\frac{a}{C}\begin{vmatrix}b & c & 0\\-a & 0 & -c\\b & c & 0\end{vmatrix}$

We can see $R_1$ and $R_3$ are identical.

Hence the value of the determinant is 0.

Therefore $\begin{vmatrix}0 & a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}$=0.

edited Feb 24, 2013