A=([i,-i],[-i,i]),B=([1,-1],[-1,1])" ,then "A^(8)=quad " (Where "i^(2)=-1)

If A=[[i,-i],[-i, i ]] and B=[[1,-1],[-1,1]] ,then A^( 8 ) equals,(where i=sqrt(-1) ) 4B 128 B -128 B -64 B

If A=([i,-i],[-i,i]) , B=([1,-1],[-1,1]) , A^(8)=KB and K=a^(b) then a+b=

A=[(i,-i),(-i,i)],B=[(1,-1),(-1,1)]impliesA^(8)=

" If "A=[[i,-i],[-i,i]],B=[[1,-1],[-1,1]]" and if "A^8=alpha B,alpha in R" then the numerical value of "(alpha)/(32)" is "

if AB=[[i,-i],[-i,i]] and B=[[1,-1],[-1,1]] , then show that A^8=128B

If A={:[(i,i),(-i,i)] and B={:[(1,-1),(-1,1)] then show that A^(8)=128 B .

If A=({:(i,-i),(-i,i):})" and "B=({:(1,-1),(-1,1):}) , then show that A^8=128B .