Inverse of Matrix Product

Theorem

Let $A,B$ : Invertible matrices of the same size

Then $AB$ is invertible and

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

Remark

This theorem can be extended to:

Theorem

A product of any number of invertible matrices is invertible, and the inverse of the product is the product of the inverses in the reverse order.

Proof

Let $A,B$ : Invertible matrices of the same size

Then

$$\begin{array}{rl}(AB)(B^{-1}{A}^{-1})& =A(B{B}^{-1})A^{-1}\\ & =AI{A}^{-1}\\ & =A{A}^{-1}\\ & =I\end{array}$$

and

$$\begin{array}{rl}(B^{-1}{A}^{-1})(AB)& =B^{-1}(A^{-1}A)B\\ & =B^{-1}IB\\ & =B^{-1}B\\ & =I\end{array}$$

By definition of inverse matrix, if $CD=DC=I$, then $C^{-1}=D$

Since we have that $(AB)(B^{-1}{A}^{-1})=(B^{-1}{A}^{-1})(AB)=I$, we conclude that $(AB{)}^{-1}=B^{-1}{A}^{-1}$