# Inverse of Matrix Product

## Theorem

Let $\mathbf {A, B}$ be square matrices of order $n$

Let $\mathbf I$ be the $n \times n$ unit matrix.

Let $\mathbf A$ and $\mathbf B$ be invertible.

Then the matrix product $\mathbf {AB}$ is also invertible, and:

$\paren {\mathbf A \mathbf B}^{-1} = \mathbf B^{-1} \mathbf A^{-1}$

## Proof

We are given that $\mathbf A$ and $\mathbf B$ are invertible.

From Product of Matrices is Invertible iff Matrices are Invertible, $\mathbf A \mathbf B$ is also invertible.

By the definition of inverse matrix:

$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$

and

$\mathbf B \mathbf B^{-1} = \mathbf B^{-1} \mathbf B = \mathbf I$

Now, observe that:

 $\ds \paren {\mathbf A \mathbf B} \paren {\mathbf B^{-1} \mathbf A^{-1} }$ $=$ $\ds \paren {\mathbf A \paren {\mathbf B \mathbf B^{-1} } } \mathbf A^{-1}$ Matrix Multiplication is Associative $\ds$ $=$ $\ds \paren {\mathbf A \mathbf I} \mathbf A^{-1}$ $\ds$ $=$ $\ds \mathbf A \mathbf A^{-1}$ Definition of Identity Element $\ds$ $=$ $\ds \mathbf I$

Similarly:

 $\ds \paren {\mathbf B^{-1} \mathbf A^{-1} } \paren {\mathbf A \mathbf B}$ $=$ $\ds \paren {\mathbf B^{-1} \paren {\mathbf A^{-1} \mathbf A} } \mathbf B$ Matrix Multiplication is Associative $\ds$ $=$ $\ds \paren {\mathbf B^{-1} \mathbf I} \mathbf B$ $\ds$ $=$ $\ds \mathbf B^{-1} \mathbf B$ Definition of Identity Element $\ds$ $=$ $\ds \mathbf I$

The result follows from the definition of inverse.

$\blacksquare$