Question

Using properties of determinants prove the following :
$\left|\begin{array}{ccc}3x& -x+y& -x+z\\ x-y& 3y& z-y\\ x-z& y-z& 3z\end{array}\right|=3(x+y+z)(xy+yz+xz)$.

Answer 1

First take LHS: let $\Delta =\left|\begin{array}{ccc}3x& -x+y& -x+z\\ x-y& 3y& z-y\\ x-z& y-z& 3z\end{array}\right|$ [Applying $C_1\to C_1+C_2+C_3]$
$=\left|\begin{array}{ccc}x+y+z& -x+y& -x+z\\ x+y+z& 3y& z-y\\ x+y+z& y-z& 3z\end{array}\right|$
Takine x + y + z common from $C_1$
$=(x+y+z)\left|\begin{array}{ccc}1& -x+y& -x+z\\ 1& 3y& z-y\\ 1& y-z& 3z\end{array}\right|$
Applying $R_1\to R_1-R_3,R_2\to R_2-R_3$
$=(x+y+z)\left|\begin{array}{ccc}0& -x+y& -x+z\\ 0& 3y& z-y\\ 1& y-z& 3z\end{array}\right|$
Expanding along $C_1$
$=(x+y+z)(1\times \left|\begin{array}{cc}-x+z& -x-2z\\ 2y+z& -2z-y\end{array}\right|)$
$=(x+y+z)\left[\right(-x_z\left)\right(-2z-y)-(2y+z\left)\right(-x-2z\left)\right]$
$=(x+y+z)(3xy+3yz+3zx)$
$=3(x+y+z)(xy+yz+zx)$
LHS = RHS
Hence proved.