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Using Cofactors of elements of third column, evaluate delta = 1 1 1 x yz y zx z xy Ex 4.4 Q: 4

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).

 

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