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` is

Question

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` is

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` is
A. `0`
B. `2`
C. `-1`
D. None of these

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`.

4 views Answered 2023-02-05 15:29