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if `a+b+c=0` then `Delta=|{:(a-x,c,b),(c,b-x,a),(b,a,c-x):}|=0` isA. `x=0`B. `x=sqrt(a^(2)+b^(2)+c^(2))`C. `x=-sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))`D. `x=-sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))` |
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Answer» Correct Answer - A::B::C Applying `R_(1)toR_(1)R_(2)+R_(3)` and taking `a+b+c-x` common from the first row the `(a+b+c-x)|{:(1,0,0),(c,b-x,a),(b,a,c-x):}|=0` Applying `C_(2)toC_(2)-C_(1)` and `C_(3)toC_(3)-C_(5)` then `(a+b+c-x)|{:(1,0,0),(c,b-x-c,a-c),(b,a-b,c-x-b):}|=0` `impliesx^(x^(2)-a^(2)-b^(2)-c^(2)+ab+bc+ac)=0` `impliesx=0+-sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))` `becausea+b+c=0` |
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