InterviewSolution
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InterviewSolution
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`|(x,x^2,y2),(y,y^2,2x),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+2x)` |
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Answer» `L.H.S. = |[x,x^2 ,yz] ,[y , y^2, zx],[z,z^2,xy]|` Operating `R_1->R_1- R_2 and R_2->R_2-R_3` `=|[x-y,x^2-y^2,yz-zx],[y-z,y^2-z^2,zx-xy],[z,z^2,xy]|` `=|[x-y,(x-y)(x+y),yz-zx],[y-z,(y-z)(y+z),zx-xy],[z,z^2,xy]|` `=(x-y)(y-z)|[1,x+y,-z],[1,y+z,-x],[z ,z^2,xy]|` Now, operating `R_1->R_1-R_2` `=(x-y)(y-z)|[0,x-z,x-z],[1,y+z,-x],[z ,z^2,xy]|` `=(x-y)(y-z)(z-x)|[0,-1,-1],[1,y+z,-x],[z ,z^2,xy]|` `=(x-y)(y-z)(z-x)[1(xy+zx)-1(z^2-yz-z^2)]` `=(x-y)(y-z)(z-x)(xy+yz+zx)= R.H.S.` |
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