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If `x , y , z`are not all zero such that`a x+y+z=0`,`x+b y+z=0`,`x+y+c z=0`then prove that`1/(1-a)+1/(1-b)+1/(1-c)=1` |
Answer» `D= | (a,1,1),(1,b,1),(1,1,c)| = 0` `R_2 -> R_2 - R_1` `R_3 -> R_3 - cR_1` `D= |(a,1,1),((1-a),b-1,0,),(1-ac,1-c,0)|` `=> (1-a)(1-c)- (b-1)(1-ac) = 0` `= 1-a-c+ac- (b-1-abc+ac) = 0` `= 1 - a-c+ac-b+1+abc-ac= 0` `= 2-a-b-c+abc= 0 ` `abc = a+b+c - 2` to prove :`1/(1-a) + 1/(1-b) + 1/(1-c) = 1` LHS :`1/(1-a) + 1/(1-b) + 1/(1-c) = ((1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b))/((1-a)(1-b)(1-c))` `= (3-2(a+b+c) + (ab+bc+ac))/(1- (a+b+c) + (ab + bc + ac) - ab)` `= 1` |
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