Let `alpha` and `beta` be the roots of the equation `x^(2) + x + 1 = 0`.
Then, for `y ne 0` in R.
`|{:(y+1, alpha,beta), (alpha, y+beta, 1),(beta, 1, y+alpha):}|` is

Options

A. `y(y^(2) -1)`
B. `y(y^(2) -3)`
C. `y^(3)-1`
D. `y^(3)`

Answer

B

Explanation

Given, quadratic equation is `x^(2) + x +1 = 0` having roots `alpha, beta`.
Then, `alpha + beta = -1 " and "alphabeta = 1`
Now, given determinant
`Delta = [{:(y+1,alpha,beta), (alpha, y+beta, 1),(beta, 1, y+alpha):}]`
On applying `R_(1) to R_(1) + R_(2) + R_(3)`, we get
`Delta = [{:(y+1 + alpha +beta,y+1 + alpha + beta,y+1+alpha +beta), (alpha, y+beta, 1),(beta, 1, y+alpha):}]`
`= [{:(y," "y,y), (alpha, y+beta, 1),(beta," "1, y+alpha):}] " "[because alpha +beta =-1]`
On applying `C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3)-C_(1)`, we get
`Delta = [{:(y," "0," "0), (alpha, y+beta-alpha, 1-alpha),(beta," "1-beta, y+alpha-beta):}]`
`=y[(y+(beta-alpha))(y-(beta-alpha))-(1-alpha)(1-beta)] " " ["expanding along "R_(1)]`
`=y[y^(2)-(beta-alpha)^(2) - (1-alpha-beta+alphabeta)]`
`=y[y^(2) -beta^(2)-alpha^(2)+2alpha beta-1+(alpha + beta)-alphabeta]`
`= y [y^(2)-(alpha + beta)^(2) + 2alpha beta +2alpha beta -1 + (alpha + beta)-alphabeta]`
`=y[y^(2)-1+3-1-1] = y^(3) " " [because alpha +beta = -1 " and "alphabeta = 1]`