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# By using the properties of determinants show that $(i) \begin{vmatrix} a-b-c&2a&2a \\ 2b&b-c-a&2b \\ 2c&2c&c-a-b \end{vmatrix} = (a+b+c)^3$

Note: This is part 1 of a 2 part question, split as 2 separate questions here.

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 given 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.
(i)let  $\begin{vmatrix}a-b-c& 2a & 2a\\2b & b-c-a & 2b\\2c &2c & c-a-b\end{vmatrix}$

By adding $R_1+R_2+R_3$ we apply $R_1\rightarrow R_1+R_2+R_3$

Therefore $\bigtriangleup=\begin{vmatrix}a+b+c& a+b+c & a+b+c\\2b & b-c-a & 2b\\2c &2c & c-a-b\end{vmatrix}$

Take (a+b+c) as the common factor from $R_1$

Therefore $\bigtriangleup=(a+b+c)\begin{vmatrix}1& 1 & 1\\2b & b-c-a & 2b\\2c &2c & c-a-b\end{vmatrix}$

By applying $C_1\rightarrow C_1-C_2$ and $4C_2\rightarrow C_2-C_3$ we get

Therefore $\bigtriangleup=(a+b+c)\begin{vmatrix}0& 0 & 1\\(b+c+a) & -(a+b+c) & 2b\\0 &(a+b+c) & c-a-b\end{vmatrix}$

Take (a+b+c) as common factors in $C_1$ and $C_2$

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

$\bigtriangleup=(a+b+c)^3[1(1-(0)]$

$\quad=(a+b+c)^3$

Hence $\begin{vmatrix}a-b-c & 2a & 2a\\2b & b-c-a & 2b\\2c & 2c & c-a-b\end{vmatrix}=(a+b+c)^3$

Hence proved.

edited Mar 11, 2013