Let A, B be two matrices different from identify matrix such that `AB=BA` and `A^(n)-B^(n)` is invertible for some positive integer n. If `A^(n)-B^(n)=A^(n+1)-B^(n+1)=A^(n+1)-B^(n+2)`, then

Options

A. `I-A` is non-singular
B. `I-B` is non-singular
C. `I-A` is singular
D. `I-B` is singular

Explanation

`A^(n+2)-B^(n+2)=(A+B) (A^(n+1)-B^(n+1))-AB(A^(n)-B^(n))`
`implies A^(n)-B^(n)=(A+B) (A^(n)-B^(n))-AB(A^(n)-B^(n))`
`implies I=A+B-AB" "[ :' A^(n)-B^(n)" is invertible"]`
`implies (I-A) (I-B)=O`
Now, `A, B, ne I`.
`:. I-A ne O` and `I-B ne O`.
If (I-A) is non-singular, then `(I-A)^(-1)` exists.
`:. (I-A)^(-1) (I-A) (I-B)=O`
`implies I-B=O`, this is contradiction the given condition.
Thus, `(I-A)` is singular.
With similar reasons, `(I-B)` is also singular.