If `A_1=[(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (1, 0, 0, 0)],A_2=[(0, 0, 0,i),(0, 0,-i, 0), (0,i,0, 0),(-i,0, 0, 0)],t h e nA_i A_k+A_k A_i` is equal to
a. `2Iifi=k`
b. `Oifi!=k`
c. `2lifi!=k`
d. `O` always

Options

A. `2I` if `i=k`
B. `O` if `i ne k`
C. `2I` if `i ne k`
D. `O` always

Answer

A, B

Explanation

Let `i=k=1` (say). Then,
`A_(r)A_(k)=A_(k)A_(i)=A_(1)A_(1)`
`A_(i)A_(k)=A_(1)A_(1)=[(0,0,0,0),(0,0,1,0),(0,1,0,0),(1,0,0,0)]xx[(0,0,0,1),(0,0,1,0),(0,1,0,0),(1,0,0,0)]`
`=[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]=I`
`A_(2)A_(2)=[(0,0,0,i),(0,0,-i,0),(0,i,0,0),(-i,0,0,0)]xx[(0,0,0,i),(0,0,-i,0),(0,i,0,0),(-i,0,0,0)]`
`=[(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)]=I`
`:. A_(i)A_(k)+A_(k)A_(i)=I+I=2I`
If `i ne k` let `i=1` and `k=2`, then
`A_(i)A_(k)=A_(1)A_(2)=[(0,0,0,1),(0,0,1,0),(0,1,0,0),(1,0,0,0)]xx[(0,0,0,i),(0,0,-i,0),(0,i,0,0),(-i,0,0,0)]`
`=[(-i,0,0,0),(0,i,0,0),(0,0,-i,0),(0,0,0,i)]`
Also, `A_(2)A_(1)=[(0,0,0,i),(0,0,-i,0),(0,i,0,0),(-i,0,0,0)]xx[(0,0,0,1),(0,0,1,0),(0,1,0,0),(1,0,0,0)]`
`=[(i,0,0,0),(0,-i,0,0),(0,0,i,0),(0,0,0,-i)]`
`implies A_(1)A_(2)+A_(2)A_(1)=O`