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

$\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|$

Cofactors of elements of third row are

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}x& yz\\ y& zx\end{array}\right|$

$\Rightarrow C_{31}=x^{2}z-y^{2}z$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& yz\\ 1& zx\end{array}\right|$

$\Rightarrow C_{32}=-(zx-yz)=yz-zx$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& x\\ 1& y\end{array}\right|$

$\Rightarrow C_{33}=y-x$

Now, $\Delta =1C_{31}+z{C}_{32}+xy{C}_{33}$

$\Delta =x^{2}z-y^{2}z+z(yz-zx)+xy(y-x)$

$\Rightarrow \Delta =x^{2}z-y^{2}z+y{z}^2-x{z}^2+x{y}^2-x^{2}y$