" 5.If "A=[[i,0],[0,i]],[n in N,i=sqrt(-1)," then prove that "A^(4n)" is equal to "[[1,0],[0,1]]]

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

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 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:

If i=sqrt(-1) , then (i^(n)+i^(-n), n in Z) is equal to

If i=sqrt(-1) , then {i^(n)+i^(-n), n in Z} is equal to

If i=sqrt(-1) , then (i^(n)+i^(-n), n in Z) 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]]