If-A-and-B-are-invertible-matrices-then-AB-1-B-1-A-1-A-1-B-1-proove-

Question Number 156610 by jlewis last updated on 13/Oct/21

If A and B are invertible matrices, then :

{(AB)}^{- 1} = B^{- 1} A^{- 1} \neq A^{- 1} B^{- 1}

proove .

Answered by physicstutes last updated on 14/Oct/21

Consider the theorems {A A}^{- 1} = I

and {B B}^{- 1} = I

now assume : {(A B)}^{- 1} = B^{- 1} A^{- 1}

pre - multiply both sides by A B

\Rightarrow A B {(A B)}^{- 1} = {A B B}^{- 1} A^{- 1}

\Rightarrow I = {A I A}^{- 1}

\Rightarrow I = {A A}^{- 1}

\Rightarrow I = I which is true!

Also consider :

{(A B)}^{- 1} = A^{- 1} B^{- 1}

pre - multiply both sides by A B

\Rightarrow (A B) {(A B)}^{- 1} = {A B A}^{- 1} B^{- 1}

\Rightarrow I \neq {A B A}^{- 1} B^{- 1}

Answered by TheSupreme last updated on 13/Oct/21

B^{- 1} A^{- 1} A B = B^{- 1} (A^{- 1} A) B = B^{- 1} B = I

t h e i n e q u a l i t y c a n b e d e m o n s t r a t e w i t h n o n c o m m u t a t i v i t y o f m a t r i x m o l t i p l i c a t i o n