### Inverse of a product

Let $$A, B\in\mathbb{F}^{n\times n}$$ where $$\mathbb{F}$$ denotes a field and $$n$$ is a positive integer. Let $$C = AB$$.

If $$A$$ and $$B$$ are both invertible, then $$C$$ is invertible and $$C^{-1}$$ is given by $$B^{-1}A^{-1}$$. (This is not a typo. The inverse of $$B$$ does come before the inverse of $$A$$.) Indeed, if $$D = B^{-1}A^{-1}$$, then $D C = (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}(AB^{-1})) = B^{-1}((A^{-1}A)B^{-1}) = B^{-1} (I_nB) = B^{-1}B = I_n$ and $C D = (AB)(B^{-1}A^{-1}) = A(B(B^{-1}A^{-1})) = A((BB^{-1})A^{-1}) = A(I_n A^{-1}) = AA^{-1} = I_n.$

More generally, if $$A_1,\ldots,A_k$$ are $$n\times n$$ invertible matrices, then $$(A_1\cdots A_k)^{-1}= A_k^{-1}\cdots A_1^{-1}.$$

A less obvious result is the following: If $$C$$ is invertible, then $$A$$ and $$B$$ are both invertible.

Proving the above result is not entirely trivial without other tools.

### Example

Let $$A= \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1\end{bmatrix}.$$ Thus, $$A = \begin{bmatrix} 1 & 3\\1 & 2\end{bmatrix}$$. Then $A^{-1}= \begin{bmatrix} 1 & 3 \\ 0 & 1\end{bmatrix}^{-1} \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}^{-1} \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}^{-1} = \begin{bmatrix} 1 & -3 \\ 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}.$

Thus, $$A^{-1} = \begin{bmatrix} -2 & 3\\1 & -1\end{bmatrix}$$.

In the above, the individual matrix inverses are easy to obtain because they are inverses of elementary matrices.

## Exercises

For each of the following products, give its inverse.

1. $$\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}0&1\\1&0 \end{bmatrix}$$

2. $$\begin{bmatrix}0&1\\-1&0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix}3&0\\0&1 \end{bmatrix}$$