Question

Let $a,b,c,x,y,zϵR$, and
${\Delta }_1=\left|\begin{array}{ccc}{(a+x)}^2& {(a+y)}^2& {(a+z)}^2\\ {(b+x)}^2& {(b+y)}^2& {(b+z)}^2\\ {(c+x)}^2& {(c+y)}^2& {(c+z)}^2\end{array}\right|$
and ${\Delta }_2=\left|\begin{array}{ccc}{(1-ax)}^2& {(1-ay)}^2& {(1-az)}^2\\ {(1-bx)}^2& {(1-by)}^2& {(1-bz)}^2\\ {(1-cx)}^2& {(1-cy)}^2& {(1-cz)}^2\end{array}\right|$
find ${\Delta }_1-{\Delta }_2$

Answer 1

${\Delta }_1=\left|\begin{array}{ccc}{(a+x)}^2& {(a+y)}^2& {(a+z)}^2\\ {(b+x)}^2& {(b+y)}^2& {(b+z)}^2\\ {(c+x)}^2& {(c+y)}^2& {(c+z)}^2\end{array}\right|$
and ${\Delta }_2=\left|\begin{array}{ccc}{(1-ax)}^2& {(1-ay)}^2& {(1-az)}^2\\ {(1-bx)}^2& {(1-by)}^2& {(1-bz)}^2\\ {(1-cx)}^2& {(1-cy)}^2& {(1-cz)}^2\end{array}\right|$
$\Rightarrow {\Delta }_1=\left|\begin{array}{ccc}{a}^2+2ax+x^2& a^2+2ay+y^2& a^2+2az+z^2\\ b^2+2bx+x^2& b^2+2by+y^2& b^2+2bz+z^2\\ c^2+2cx+x^2& c^2+2cy+y^2& c^2+2cz+z^2\end{array}\right|$
$\to \left|\begin{array}{ccc}{a}^2& 2a& 1\\ b^2& 2b& 1\\ c^2& 2c& 1\end{array}\right|\times \left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$ (row by row)
$\Rightarrow {\Delta }_2=\left|\begin{array}{ccc}1-2ax+a^2{x}^2& 1-2ay+a^2{y}^2& 1-2az+a^2{z}^2\\ 1-2bx+b^2{x}^2& 1-2by+b^2{y}^2& 1-2bz+b^2{z}^2\\ 1-2cx+c^2{x}^2& 1-2cy+c^2{y}^2& 1-2cz+c^2{z}^2\end{array}\right|$
$\to \left|\begin{array}{ccc}1& -2a& a^2\\ 1& -2b& b^2\\ 1& -2c& c^2\end{array}\right|\times \left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$ (row by row)
by interchanging columns ${\Delta }_1={\Delta }_2$
$\therefore {\Delta }_1-{\Delta }_2=0$