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

1

Explanation

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`