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# Using properties of determinants, prove that: $\begin{vmatrix} 3a&-a+b&-a+c\\ -b+a&3b&-b+c\\ -c+a&-c+b&3c \end{vmatrix}= 3(a+b+c)(ab+bc+ca)$

Toolbox:
• If each element of a row(or a 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 column of a given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns.
• 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 $\Delta = \begin{vmatrix} 3a & -a+b& -a+c\\-b+a & 3b & -b+c\\-c+a&-c+b&3c\end{vmatrix}$
Let us apply $C_1\rightarrow C_1+C_2+C_3$
$\Delta = \begin{vmatrix} a+b+c & -a+b& -a+c\\a+b+c & 3b & -b+c\\a+b+c&-c+b&3c\end{vmatrix}$
Taking (a+b+c) as a common factor from $C_1$
$\Delta = \begin{vmatrix} 1 & -a+b& -a+c\\1 & 3b & -b+c\\1&-c+b&3c\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$ we have,
$\Delta =(a+b+c) \begin{vmatrix} 1 & -a+b& -a+c\\0 & 2b-a & a-b\\0&a-c&2c+a\end{vmatrix}$
Now expanding along $C_1$
$\Delta =(a+b+c) [1[(2b-a)(2c+a)-(a-b)(a-c)]-0+0]$
$\Delta =(a+b+c) [4bc+2ab-2ac-a^2-)a^2-ac-ab+bc)]$
$\;\;\; =(a+b+c) [4bc+2ab-2ac-a^2-a^2+ac+ab-bc)]$
$\;\;\;=(a+b+c)[3bc+3ab+3ac]$
$\;\;\;=3(a+b+c)[ab+bc+ac]$
Hence proved.

edited Mar 5, 2013