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:


 * $\left({\mathbf{AB}}\right)^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$

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