InterviewSolution
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InterviewSolution
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if `omega!=1` is cube root of unity and x+y+z`!=`0 then `|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|`=0 ifA. `x^(2)+y^(2)+z^(2)=0`B. `x+yomega+zomega^(2)=0` or `x=y=z`C. `x ne y ne z ne 0`D. x=2y=3z |
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Answer» Correct Answer - B `(b)` As `1+omega+omega^(2)=0` `D=|{:(-(x)/(omega^(2)),-y,-(z)/(omega)),(-y,-(z)/(omega),-(x)/(omega^(2))),(-(z)/(omega),-(x)/(omega^(2)),-y):}|=x^(3_+y^(3)+z^(3)-3xyz` `=(1)/(2)(x+y+z){(x-y)^(2)+(y-z)^(2)+(z-x)^(2)}` `=(x+y+z)(x+yomega+zomega^(2))(x+yomega^(2)+zomega)` The determinant varnishes if `x=y=z` or `x+yomega+zomega^(2)=0` |
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