# Using Cofactors of elements of third column, evaluate

Question:

Using Cofactors of elements of third column, evaluate $\Delta=\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|$

Solution:

The given determinant is $\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right|$.

We have:

$\mathrm{M}_{13}=\left|\begin{array}{ll}1 & y \\ 1 & z\end{array}\right|=z-y$

$\mathrm{M}_{23}=\left|\begin{array}{ll}1 & x \\ 1 & z\end{array}\right|=z-x$

$\mathrm{M}_{33}=\left|\begin{array}{ll}1 & x \\ 1 & y\end{array}\right|=y-x$

$\therefore \mathrm{A}_{13}=$ cofactor of $a_{13}=(-1)^{1+3} \mathrm{M}_{13}=(z-y)$

$\mathrm{A}_{23}=$ cofactor of $\mathrm{a}_{23}=(-1)^{2+3} \mathrm{M}_{23}=-(z-x)=(x-z)$

$\mathrm{A}_{33}=$ cofactor of $a_{33}=(-1)^{3+3} \mathrm{M}_{33}=(y-x)$

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

\begin{aligned} \therefore \Delta &=a_{13} \mathrm{~A}_{13}+a_{23} \mathrm{~A}_{23}+a_{33} \mathrm{~A}_{33} \\ &=y z(z-y)+z x(x-z)+x y(y-x) \\ &=y z^{2}-y^{2} z+x^{2} z-x z^{2}+x y^{2}-x^{2} y \\ &=\left(x^{2} z-y^{2} z\right)+\left(y z^{2}-x z^{2}\right)+\left(x y^{2}-x^{2} y\right) \\ &=z\left(x^{2}-y^{2}\right)+z^{2}(y-x)+x y(y-x) \\ &=z(x-y)(x+y)+z^{2}(y-x)+x y(y-x) \\ &=(x-y)\left[z x+z y-z^{2}-x y\right] \\ &=(x-y)[z(x-z)+y(z-x)] \\ &=(x-y)(z-x)[-z+y] \\ &=(x-y)(y-z)(z-x) \end{aligned}

Hence, $\Delta=(x-y)(y-z)(z-x)$.