If `A=[[(-1+isqrt(3))/(2i),(-1-isqrt(3))/(2i)],[(1+isqrt(3))/(2i),(1-isqrt(3))/(2i)]]`, `I = sqrt(-1) and f (x) = x^(2) + 2, `
then `f(A)` equals to

Options

A. `[[1,0],[0,1]]`
B. `((3-isqrt(3))/2)[[1,0],[0,1]]`
C. `((5-isqrt(3))/2)[[1,0],[0,1]]`
D. `(2+isqrt(3))[[1,0],[0,1]]`

Answer

D

Explanation

`because omega = (-1+isqrt(3))/2 and omega ^(2) =(-1-isqrt(3))/2` ltvrgt Also, `omega ^(3) = 1 and omega + omega ^(2) = -1 `
Thus, ` A = [[-iomega,-iomega^(2)],[iomega^(2),iomega ]]`
` A^(2) = [[-iomega,-iomega^(2)],[iomega^(2),iomega ]][[-iomega,-iomega^(2)],[iomega^(2),iomega ]]= [[-omega^(2)+omega,0],[0,-omega^(2)+omega]]`
Now, `f (A) =A^(2) + 2I= [[-omega^(2)+omega,0],[0,-omega^(2)+omega]]+ [[2,0],[0,2]]`
`= [[-omega^(2)+omega+2,0],[0,-omega^(2)+omega+2]] `
`=(-omega^(2)+omega+2)[[1,0],[0,1]]= (2 + isqrt(3)) [[1,0],[0,1]]`