Question

Prove that$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+z+2y\end{array}\right|=2{(x+y+z)}^3$

Answer 1

L.H.S.

$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|$

$C_1=C_1+C_2+C_3$

$=\left|\begin{array}{ccc}2(x+y+z)& x& y\\ 2(x+y+z)& y+z+2x& y\\ 2(x+y+z)& x& z+x+2y\end{array}\right|$

Taking out common $2(x+y+z)$ from $C_1$

$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 1& y+z+2x& y\\ 1& x& z+x+2y\end{array}\right|$

$R_2=R_2-R_1$

$R_3=R_3-R_1$

$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& x+y+z& 0\\ 0& 0& x+y+z\end{array}\right|$

Taking out common $(x+y+z)$ from both $R_2$ and $R_1$

$=2(x+y+z{)}^3\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

Expanding along $C_1$

$=2(x+y+z{)}^3$