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 if |
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Answer» As `1 + omega +omega^(2) =0 ` `D= |{:((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))):}|` `=|{:(-(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)` `rArr x=y=z" " "(":. x+y+z ne 0")"` |
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