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, \(∆=(x-y)(y-z)(z-x)\)