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For a real number `alpha,`if the system `[1alphaalpha^2alpha1alphaalpha^2alpha1][x y z]=[1-1 1]`of linear equations, has infinitely many solutions, then `1+alpha+alpha^2=` |
Answer» Correct Answer - 1 Since the system of equations hasss infinitely many solutions `|{:(1,,alpha,,alpha^(2)),(alpha,,1,,alpha),(alpha^(2),,alpha,,1):}|=0` `rArr 1(1-alpha^(2)) -alpha(alpha-alpha^(3)) +alpha^(2)(alpha^(2)-alpha^(2))=0` `rArr (1-alpha^(2)) -alpha^(2) +alpha^(4)=0` `rArr (alpha^(2) -1)^(2)=0` `rArr alpha= +-1` For `alpha=1` we get `x+y+z=1 " and " x+y+z=-1` Hence system has no solution . Fora `alpha=-1` all three equations become x-y+z=1 which represents coincident planes. `:. 1+alpha+alpha^(2) =1` |
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