Using the properties of determinants ,show that..

1 x+y x2+y2

1 y+z y2+z2

1 z+x z2+x2

=(x-y)(y-z)(z-x)

(((x2,y2,z2 are x square,y square and z square respectively)))

(meritnation exprt please ans .....)

Question

Using the properties of determinants ,show that
1 x+y x2+y2
1 y+z y2+z2
1 z+x z2+x2
=(x-y)(y-z)(z-x)
(((x2,y2,z2 are x square,y square and z square
respectively)))
(meritnation exprt please ans )

Santosh Kumar · Accepted Answer

To show

$| \begin{matrix} 1 & x + y & x^{2} + y^{2} \\ 1 & y + z & y^{2} + z^{2} \\ 1 & z + x & z^{2} + x^{2} \end{matrix} | = (x - y) (y - z) (z - x)$

Taking LHS

$Δ = | \begin{matrix} 1 & x + y & x^{2} + y^{2} \\ 1 & y + z & y^{2} + z^{2} \\ 1 & z + x & z^{2} + x^{2} \end{matrix} |$

Applying R₁ → R₁ – R₃; R₂ → R₂ – R₃, we get

$Δ = | \begin{matrix} 0 & y - z & y^{2} - z^{2} \\ 0 & y - x & y^{2} - x^{2} \\ 1 & z + x & z^{2} + x^{2} \end{matrix} | = | \begin{matrix} 0 & y - z & (y + z) (y - z) \\ 0 & y - x & (y + x) (y - x) \\ 1 & z + x & z^{2} + x^{2} \end{matrix} | = (y - z) (y - x) | \begin{matrix} 0 & 1 & y + z \\ 0 & 1 & y + x \\ 1 & z + x & z^{2} + x^{2} \end{matrix} | = (y - z) (y - x) [y + x - y - z] = (y - z) (y - x) (x - z) = (x - y) (y - z) (z - x) = R.H.S$

4

Using the properties of determinants ,show that.. 1 x+y x2+y2 1 y+z y2+z2 1 z+x z2+x2 =(x-y)(y-z)(z-x) (((x2,y2,z2 are x square,y square and z square respectively))) (meritnation exprt please ans .....)

Using the properties of determinants ,show that..

1 x+y x2+y2

1 y+z y2+z2

1 z+x z2+x2

=(x-y)(y-z)(z-x)

(((x2,y2,z2 are x square,y square and z square respectively)))

(meritnation exprt please ans .....)