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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 toA. `[[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» Correct Answer - D `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]]` |
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