# Evaluate $\begin{vmatrix} 1 & x & y\\ 1 & x+y & y\\ 1 & x & x+y \end{vmatrix}$

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value 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 $\Delta = \begin{vmatrix} 1 & x & y\\1 & x+y & y\\1& x& x+y\end{vmatrix}$
Apply R2 → R2 - R1 and R3 → R3 - R1
$\Delta=\begin{vmatrix}1 & x & y\\0 & y & -x\\0 & 0 & x\end{vmatrix}$
Now expanding $C_1$ we have
$\Delta=1(yx+0)-0+0=xy$
Hence $\Delta=xy$.
edited Mar 2, 2013