InterviewSolution
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InterviewSolution
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If `w ne 1` is a cube root of unity and `Delta=|{:(x+w^(2),w,1),(w,w^(2),1+x),(1,x+w,w^(2)):}|=0`, then value of `x` isA. `0`B. `2`C. `-1`D. None of these |
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Answer» Correct Answer - A `(a)` Applying `C_(1)toC_(1)+C_(2)+C_(3)`. We get `Delta=|{:(x+w^(2)+w+1,w,1),(w+w^(2)+1+x,w^(2),1+x),(1+x+w+w^(2),x+w,w^(2)):}|` `=|{:(x,w,1),(x,w^(2),1+x),(x,x+w,w^(2)):}|` [using `1+w+w^(2)=0`] `Delta` is clearly equal to `0` for `x=0`. |
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