# Evaluate $\begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x & y \end{vmatrix}$

Toolbox:
• If each element of a row (or a column) of a determinant is multiplied by a constant k,then its value get multiplied by k.
• By this property we can take out any common factor from any one row or any one column.
• 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 $\Delta= \begin{vmatrix} x & y & x+y\\ y & x+y & x\\ x+y & x & y \end{vmatrix}$
Let us apply $R_1\rightarrow R_1+R_2+R_3$
$\Delta= \begin{vmatrix} 2x+2y & 2x+2y & 2x+2y\\ y & x+y & x\\ x+y & x & y \end{vmatrix}$
Taking 2(x+y) as the common factor ,we get
$\Delta=2(x+y) \begin{vmatrix} 1 & 1& 1\\ y & x+y & x\\ x+y & x & y \end{vmatrix}$
Apply $C_2\rightarrow C_2-C_1$ and $C_2-C_3$
$\Delta=2(x+y) \begin{vmatrix} 1 & 0& 0\\ y & x & y\\ x+y & -y & x-y \end{vmatrix}$
Now expanding along $R_1$ we get,
$\Delta=1[x(x-y)+y^2]=(x^2-xy+y^2)2(x+y)$
$\;\;\;\;=2(x-y)(x^2-xy+y^2)$
$\;\;\;\;=2(x^3-x^2y+xy^2+x^2y-xy^2+y^3]$
$\;\;\;\;=2(x^3+y^3).$

edited Mar 2, 2013