Question

Using cofactors of elements of third column, evaluate
$\Delta =\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$

Answer 1

Given, $\Delta =\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$
Cofactors of the elements of third column are
$A_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}1& y\\ 1& z\end{array}\right|=1(z-y)=z-y{A}_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& x\\ 1& z\end{array}\right|=-1(z-x)=x-zand{A}_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& x\\ 1& y\end{array}\right|=1(y-x)=y-x$
Now, expansion of $\Delta$ using cofactors of elements of third column is given by
$\Delta =a_{13}{A}_{13}+a_{23}{A}_{23}+a_{33}{A}_{33}==yz(z-y)+zx(x-z)+xy(y-x)=y{z}^2-y^{2}z+z{x}^2-z^{2}x+x{y}^2-x^{2}y=x^2(z-y)+x(y^2-z^2)+yz(z-y)=(z-y)\{x^2-x(y+z)+yz\}=(z-y)\{x^2-xy-xz+yz\}=(z-y)\left[x\right(x-y)-z(x-y\left)\right]=(y-z)(x-y)(z-x)=(x-y)(y-z)(z-x)$