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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Using Cofactors of elements of third column, evaluate $ \Delta = \begin{vmatrix} 1&x&yz \\ 1&y&zx \\ 1&z&xy \end{vmatrix} $

$\begin{array}{1 1} (x+y)(y+z)(z+x) \\ (x-y(z-y)(x-z) \\ (x-y)(y-z)(z-x) \\ (y-z)(y-x)(z-x) \end{array} $

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  • $\bigtriangleup$=Sum of the product of any row (or column) with their corresponding cofactors.
  • Given:
  • Using cofactors of elements of third column evaluate
$\bigtriangleup=\begin{vmatrix}1 & x & yz\\1 & y & zx\\1 & z & xy\end{vmatrix}$
we know that expanding the determinant $\bigtriangleup$ along $C_3$ we have,
Minor of$M_{13}=\begin{vmatrix}1 & y\\1 & z\end{vmatrix}=(z-y)$
Minor of$M_{23}=\begin{vmatrix}1 & x\\1 & z\end{vmatrix}=(z-x)$
Minor of$M_{33}=\begin{vmatrix}1 & x\\1 & y\end{vmatrix}=(y-x)$
Hence $a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}=yz(z-y)+zx(-)(z-x)+xy(y-x)$
Taking the common factor
We know $(a^2-b^2)=(a+b)(a-b)$
Therefore $\bigtriangleup=z(x+y)(x-y)+z^2(y-x)+z^2(y-x)+xy(y-x)$
Taking (x-y)as a common factor,
Taking the common factor x(z-x)
Hence $\bigtriangleup=(x-y)(y-z)(z-x)$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans

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