InterviewSolution
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InterviewSolution
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By using properties of determinants. Show that:(i) `|a-b-c2a2a2bb-c-a2b2c2cc-a-b|=(a+b+c)^3` (ii) `|x+y+2z x y z y+z+2x y z x z+x+2y|=2(x+y+z)^3` |
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Answer» `|{:(a-b-c,2a,2a),(ab,b-c-a,2a),(2c,2c,c-a-b):}|=|{:(-(a+b+c),0,2a),(a+b+v,-(a+b+c),ab),(0,a+b+c,c-a-b):}|` `(C_(1)toC_(1)-C_(2),C_(2)toC_(2)-C_(3))` `=(a+b+c)^(2)|{:(-1,0,2a),(0,-1,2b+2a),(0,1,c-a-b):}|` `(R_(2)toR_(2)+R_(1))` `=(a+b+c)^(2).(-1)|{:(-1,2b+2a),(1,c-a-b):}|` (Expanding along `C_(1)`) `=(a+b+c)^(2)(-1)(-c+a+b-2a-ab)` `=(a+b+c)^(2)(-1)(-a-b-c)` `=(a+b+c)^(2)(a+b+c)` `=(a+b+c)=R.H.S.` `|{:(x+y+2s,x,y),(z,y+z+2z,y),(z,x,z+x+2y):}|=|{:(2x+2y+2z,x,y),(2x+2y+2z,y+z+2x,y),(2x+2y+2z,z,z+x+2y):}|` `(C_(1)toC_(1)+C_(2)+C_(3))` `=(2x+2y+2z)|{:(1,x,y),(1,y+z+2x,u),(1,x,z+x+2y):}|` `=2(x+y+z)|{:(1,x,y),(1,y+z+2x,0),(0,0,x+y+z):}|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `=2(x+y+z).1|{:(x+y+z,0),(0,x+y+z):}|` (Expanding along `C_(1)`) `=2(x+y+z)^(3)=R.H.S` |
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