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:
M13 = $\begin{vmatrix}1&y\\1&z\end{vmatrix}=z-y$
M23 = $\begin{vmatrix}1&x\\1&z\end{vmatrix}=z-x$
M33 = $\begin{vmatrix}1&x\\1&y\end{vmatrix}=y-x$
$\therefore$ A13 = cofactor of a13 = (-1)1+3 M13 = (z - y)
A23 = cofactors of a23 = (-1)2+3 M23 = - (z - x) = (x - z)
A33 = cofactors of a33 = (-1)3+3 M33 = (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.
$\therefore\triangle$ = a13A13 + a23A23 + a33A33
= yz(z - y) + zx (x - z) + xy (y - x)
= yz2 - y2z + x2z - xz2 + xy2 - x2y
=(x2z - y2z) + (yz2 - xz2) + (xy2 - x2y)
= z(x2 - y2) + z2(y - x) + xy(y - x)
=z(x - y)(x + y) + z2(y - x) + xy(y - x)
=(x- y)[zx + zy - z2 - 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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