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Q : 4 Using Cofactors of elements of third column, evaluate $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

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Given determinant : $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

First finding Minors of the third column by the definition,

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 1 &y \\ 1 &z \end{vmatrix} =z-y$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &x \\ 1 &z \end{vmatrix} = z-x$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &x \\ 1 &y \end{vmatrix} =y-x$

Finding the Cofactors of the second row:

$A_{13}=$ Cofactor of $a_{13} = (-1)^{1+3}M_{13} = z-y$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = x-z$

$A_{33}=$ Cofactor of $a_{33} = (-1)^{3+3}M_{33} = y-x$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$

$= (z-y)yz + (x-z)zx +(y-x)xy$

$=yz^2-y^2z + zx^2-xz^2 + xy^2-x^2y$

$=z(x^2-y^2) + z^2(y-x) +xy(y-x)$

$= (x-y) \left [ zx+zy-z^2-xy \right ]$

$=(x-y)\left [ z(x-z) +y(z-x) \right ]$

$= (x-y)(z-x)[-z+y]$

$= (x-y)(y-z)(z-x)$

Thus, we have value of $\triangle = (x-y)(y-z)(z-x)$ .

Posted by

Divya Prakash Singh

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