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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.

Quick Quiz

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