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If A1=[0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0],...

If `A_1=[0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0],A_2=[0 0 0i0 0-i0 0i0 0-i0 0 0],t h e nA_i A_k+A_k A_i` is equal to `2lifi=k` b. `Oifi!=k` c. `2lifi!=k` d. `O` always

A

`2I` if `i=k`

B

`O` if `i ne k`

C

`2I` if `i ne k`

D

`O` always

Text Solution

Verified by Experts

The correct Answer is:
A, B
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`
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