1.

If w is a complex cube root of unity. `|(a,b,c),(b,c,a),(c,a,b)| = -(a +b +c) (a + bk + ck^(2)) (a + bk^(2) + ck)`, then k equalsA. 1B. `-1`C. `omega`D. `-omega`

Answer» Correct Answer - C
We have,
`|(a,b,c),(b,c,a),(c,a,b)|`
`= |(a +b +c,b,c),(a +b +c,c,a),(a +b +c,a,b)| " " ["Applying " C_(1) rarr C_(1) + C_(2) + C_(3)]`
`= (a +b +c) |(1,b,c),(1,c,a),(1,a,b)|`
`= (a +b +c) |(1,b,c),(0,c -b,a -c),(0,a -b,b-c)| " " [("Applying" R_(2) rarr R_(2) - R_(1)),(" "R_(3) rarr R_(3) - R_(1))]`
`= (a +b +c) |(c -b,a -c),(a -b,b -c)|`
`= (a +b +c) (-b^(2) -c^(2) -a^(2) + ac + ab + bc - 2bc)`
`= -(a + b+c) (a^(2) +b^(2) + c^(2) -ab - bc - ca)`
`= - (a +b +c) (a + bomega + c omega^(2)) (a + b omega^(2) + c omega)`
`:. k = w`


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