Question

Using the properties of determinant and without expanding , prove that:
$\left|\begin{array}{ccc}x& a& x+a\\ y& b& y+b\\ z& c& z+c\end{array}\right|=0$

Answer 1

$\left|\begin{array}{ccc}x& a& x+a\\ y& b& y+b\\ z& c& z+c\end{array}\right|=0\left|\begin{array}{ccc}x& a& x\\ y& b& y\\ z& c& z\end{array}\right|+\left|\begin{array}{ccc}x& a& a\\ y& b& b\\ z& c& c\end{array}\right|$

$C_1\to C_1-C_3$ for first Determinant

$C_2\to C_2-C_3$ for second Determinant

$\left|\begin{array}{ccc}0& a& x\\ 0& b& y\\ 0& c& z\end{array}\right|+\left|\begin{array}{ccc}x& 0& a\\ y& 0& b\\ z& 0& c\end{array}\right|=0$