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DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
Using Cofactors of elements of third column, evaluate $\triangle=\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
✓

Answer

The given determinant is $\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
We have:
$M_{13} = \begin{vmatrix}1&y\\1&z\end{vmatrix}=z-y$
$M_{23} = \begin{vmatrix}1&x\\1&z\end{vmatrix}=z-x$
$M_{33} = \begin{vmatrix}1&x\\1&y\end{vmatrix}=y-x$
$\therefore A_{13} =$ cofactor of $a_{13} = (-1)^{1+3} M_{13} = (z - y)$
$A_{23} =$ cofactors of $a_{23} = (-1)^{2+3 }M_{23} = - (z - x) = (x - z)$
$A_{33} =$ cofactors of $a_{33} = (-1)^{3+3 }M_{33} = (y - x)$
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
$\therefore\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$= yz(z - y) + zx (x - z) + xy (y - x)$
$= yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y$
$=(x^2z - y^2z) + (yz^2 - xz^2) + (xy^2 - x^2y)$
$= z(x^2 - y^2) + z^2(y - x) + xy(y - x)$
$=z(x - y)(x + y) + z^2(y - x) + xy(y - x)$
$=(x- y)[zx + zy - z^2 - xy]$
$=(x - y)[z(x - z) + y(z - x)]$
$=(x - y)(z - x)[-z + y]$
$=(x - y)(y - z)(z - x)$

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