# Using the property of determinants and without expanding, prove that $\begin{vmatrix} -a^2&ab&ac \\ ba&-b^2&bc \\ ca&cb&-c^2 \end{vmatrix} = 4a^2b^2c^2$

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.
• By this property we can take out any common factor from any one row or any one column of the determinant.
• 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 $\bigtriangleup=\begin{vmatrix}-a^2 & ab & ac\\ba & -b^2 & bc\\ca & cb & -c^2\end{vmatrix}$

We can take the common factors a, b and c respectively from $R_1,R_2$ and $R_3$

Therefore$\bigtriangleup=abc\begin{vmatrix}-a & b & c\\a & -b & c\\a & b & -c\end{vmatrix}$

Again taking a,b,c as common factors from $R_1,R_2$ and $R_3$ respectively,

$\bigtriangleup=a^2b^2c^2\begin{vmatrix}-1 & 1 & 1\\1 & -1 & 1\\1 & 1 &-1\end{vmatrix}$

By applying $R_2\rightarrow R_2+R_1$ and $R_3\rightarrow R_3+R_1$,we get,

$\bigtriangleup=a^2b^2c^2\begin{vmatrix}-1 & 1 & 1\\0 & 0 & 2\\0 & 2 &0\end{vmatrix}$

Now by expanding we get $a^2b^2c^2(-1)\begin{vmatrix}0 & 2\\2 & 0\end{vmatrix}$

$\bigtriangleup=a^2b^2c^2(-1)(-4)=4a^2b^2c^2.$

Hence proved.

edited Feb 24, 2013