Suppose A and B are n Times n matrices assume that A is inve

Suppose A and B are n Times n matrices, assume that A is invertible (B may not be invertible) and also assume that AB = BA. Prove that A^-1 B = BA^-1.

Solution

As A is invertible,

AA^-1 = I

Multiply both the sides of above equation by B,

(AA^-1) B = IB

A (A^-1B) = IB .. (By associative property of matrix multiplication)

A (A^-1B) = B .. (By definition of identity)

Divide both the sides by A,

A^-1B = B / A .... (1)

AA^-1 = I

Multiply both the sides of above equation by B,

B (AA^-1) = BI

(BA) A^-1 = BI .. (By associative property of matrix multiplication)

(AB) A^-1 = BI .. (As AB = BA)

A(BA^-1) = BI .. (By associative property of matrix multiplication)

A(BA^-1) = B ..(By definition of identity)

Divide both the side by A,

BA^-1 = B / A ... (2)

So, from equations (1) and (2), we get

A^-1 B = BA^-1 = B / A

Hence, A^-1 B = BA^-1