Let A and B be square matrices of the same order such that `A^(2)=I` and `B^(2)=I`, then which of the following is CORRECT ?

Options

A. IF A and B are inverse to each other, then `A=B`.
B. If `AB=BA`, then there exists matrix `C=(AB+BA)/2` such that `C^(2)=C`.
C. If `AB=BA`, then there exists matrix `D=AB-BA` such that `D^(n)=O` for some `n in N`.
D. If `AB=BA` then `(A+B)^(5)=16 (A+B)`.

Explanation

We have `A^(2)=I` and `B^(2)=I`
`:. A=A^(-1)` and `B=B^(-1)`
If A and B are inverse to each other, then `A=B^(-1)=B`.
`C=(AB+BA)/2`
`implies C^(2)=((AB)^(2)+(BA)^(2)+AB.BA+BA.AB)/4`
Now, `ABBA=AIA=A A=I`
Similarly, `BA AB=I`
Also `(AB)^(2)=ABAB`
`=A AB B" "("if "AB=BA)`
`=A^(2)B^(2)`
`=I`
Similarly `(BA)^(2)=I`
`:. C^(2)=I`
Thus, if `AB=BA`, then `C^(2)=I`
`D=AB-BA`
`implies D^(2)=(AB)^(2)+(BA)^(2)-ABxxBA-BAxxAB`
`=I+I-I-I" "("if "AB=BA)`
`=O`
Thus, if `AB=BA`, then `D^(n)=O`.
`(A+B)^(5)=A^(5)+5A^(4)B+10A^(3)B^(2)+10A^(2)B^(3)+5AB^(4)+B^(5)`
`=A+5B+10A+10B+5A+B`
`=16(A+B)`