If A, B, C are square matrices of the same order such that AB = BA = I and AC = CA = I then B = C.
We have
AB = BA = I ...(1)
and AC = CA = I ...(2)
Now B = BI
= B(AC) [∵ of (2)]
= (BA) C [∵ of associative properties of multiplication]
= IC [∵ of (1)]
= C
∴ B = C
Hence the result.
Note : AB = BA = I ⇒ B is inverse of A
Again AC = CA = 1 ⇒ C is inverse of A
Also B = C
∴ inverse of a matrix A, if it exists, is unique.