`[[i,0,0],[0,i,0],[0,0,i]]` then `A^(4n+1)` =.... `n in N`

If A=[(i,0,0),(0,i,0),(0,0,i)] then for ninN,A^(4n+1)=

If A = [(i,0),(0,i)] then A^(4n) (n in N ) equals :

If A=[{:(i,0),(0,i):}], n inN then A^(4n)= ...... (where I is imaginary complex number and i^(2)=-1)

If A=[[i,0],[0,i]];i=sqrt(-1), then A^(n) is equal to

If {:A=[(i,0),(0,i)]:},n in N" then " A^(4n) equal

If A=[[cos theta , sin theta],[sin theta,-costheta]], B = [[1,0],[-1,1]], C=ABA^(T), then A^(T) C^(n) A, n in I^(+) equals to a. [[-n,1],[1,0]] b. [[1,-n],[0,1]] c. [[0,1],[1,-n]] d. [[1,0],[-n,1]]

If A=[(i,0), (0,i)],\ n in N , then A^(4n) equals [(0,i), (i,0)] (b) [(0 ,0) ,(0, 0)] (c) [(1, 0) ,(0, 1)] (d) [(0,i) ,(i,0)]

if A=[[i,0] , [0,i]] where i=sqrt(-1) and x epsilon N then A^(4x) equals to:

if A=[[i,0] , [0,i]] where i=sqrt(-1) and x in N then A^(4x) equals to: