Let A and B be matrices of order n. Prove that if
(I - AB) is invertible, (I - BA) is also invertible and
`(I-BA)^(-1) = I + B (I- AB)^(-1)A, ` where I be the identity matrix
of order n.

Explanation

Here, `I -BA = BIB^(-1)-BABB^(-1) = B(I-AB)B^(-1)` ...(i)
Hence, `abs(I-BA) = abs(B)abs(I-AB) abs(B^(-1)) = abs(B) abs(I-AB) 1/abs(B)`
`= abs(I-AB)`
If `abs(I-AB)ne 0, "then" abs(I-BA) ne 0`
i.e. if `(I -AB)` is invertible, then `(I - BA)` is also invertible.
Now, `(I-BA) [ I + B (I-AB)^(-1)A]`
`= (I- BA) + (I-BA) B (I-AB)^(-1)A ` [ using Eq. (i)]
`= (I- BA) +B (I-AB) B^(-1) B (I-AB)^(-1)A `
`=(I-BA)B(I-AB)(I-AB^(-1))A`
`= (I - BA) + BA= I`
Hence, `(I- BA)^(-1)=I + B(I-AB)^(-1) A.`