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.
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.
For each of the following products, give its inverse.
\(\begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}0&1\\1&0 \end{bmatrix}\)
\( \begin{bmatrix}0&1\\-1&0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix}3&0\\0&1 \end{bmatrix} \)