Question

Using properties of determinants, prove that

`|((x+y)^2,zx,zy),(zx,(z+y)^2,xy),(zy,xy,(z+x)^2)|=2xyz(x+y+z)^3`

 

Answer 1

`|((x+y)^2,zx,zy),(zx,(z+y)^2,xy),(zy,xy,(z+x)^2)|=2xyz(x+y+z)^3`

L.H.S.

Multipiying R₁, R₂ and R₃ by z, x, y respectively

`=1/(xyz)|(z(x+y)^2,z^2x,z^2y),(x^2z,x(z+y)^2,x^2y),(y^2z,xy^2,y(z+x)^2)|`

take common z, x, y from C₁, C₂, & C₃

`=(xyz)/(xyz)|((x+y)^2,z^2,z^2),(x^2,(z+y)^2,x^2),(y^2,y^2,(z+x)^2)|`

C₁ → C₁ - C₃ and C₂  C₂ - C₃

taking common x+y+z from C₁ & C₂

`=(x+y+z)^2|((x+y+z),0,z^2),(0,z+y-x,x^2),(y-z-x,y-z-x,(z+x)^2)|`

R₃ → R₃ - (R₁ + R₂)

`=(x+y+z)^2|(x+y+z,0,z^2),(0,z+y-x,x^2),(-2x,-2zx,2xz)|`

C₁ → zC₁, C₂ → xC₃

`=(x+y+z)^2/(xz)=|(z(x+y-z),0,z^2),(0,x(z+y-x),x^2),(-2xz,-2zx,2xz)|`

C₁ → C₁ + C₃   C₂ → C₂ + C₃

 

 

`=(x+y+x^2)/(xz)|(z(x+y),z^2,z^2),(x^2,x(z+y),x^2),(0,0,2xz)|`

taking z and x common from R₁ & R₂

`=(x+y+x)^2/(xz)xxzx|(x+y,z,z),(x,z+y,x),(0,0,2xz)|`

expansion along R₃

= (x+y+z)² × 2xz ((x + y) (z + y) – xz)

= (x+y+z)² × 2xz (xz + xy + yz + y2 - xz)

= (x+y+z)² × 2xz (xy + yz + y²)

= 2xyz (x + y + z)³