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:


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

Proof
By hypothesis, $\mathbf A$ and $\mathbf B$ are invertible.

Thus, by the definition of inverse matrix:


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

and


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

Now, observe that:

Similarly:

The result follows from the definition of inverse.

Also see

 * Transpose of Matrix Product