Question

Using Cofactors of elements of third column,evaluate$△=\begin{bmatrix}1&x&yz\\\ 1&y&zx\\\ 1&z&xy\end{bmatrix}$

Answer 1

The given determinant is$\begin{bmatrix}1&x&yz\\\ 1&y&zx\\\ 1&z&xy\end{bmatrix}$
We have:
$M_{13} =\begin{bmatrix}1&y\\\1&z\end{bmatrix}=z-y$
$M_{23} =\begin{bmatrix}1&x\\\1&z\end{bmatrix}=z-x$
$M_{33} =\begin{bmatrix}1&x\\\1&y\end{bmatrix}=y-z$
$∴A_{13}$ = cofactor of $a_{13} = (−1)^{1+3} M_{13} = (z − y)$
$A_{23}$ = cofactor of$a_{23} = (−1)^{2+3} M_{23} = − (z − x) = (x − z)$
$A_{33}$ = cofactor of $a_{33}$ $= (−1)^{3+3} M_{33} = (y − x)$
We know that ∆ is equal to the sum of the product of the elements of the second row
with their corresponding cofactors.
$∴∆=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}$
$=yz(z-y)+zx(x-z)+xy(y-x)$
$=yz^22-y^2z+x^2z-xz^2+xy^2-x^2y$
$=(x^2z-y^2z)+(yz^2-xz^2)+(xy^2-x^2y)$
$=z(x^2-y^2)+z^2(y-x)+xy(y-x)$
$=(x-y)[zx+zy-z^2-xy]$
$=(x-y)(z-x)[-z+y]$
$=(x-y)(y-z)(z-x)$
Hence, $∆=(x-y)(y-z)(z-x)$