Why the Order Reverses
For numbers, the order of factors does not affect their product. Matrix multiplication is different: even when \(\mathbf{A}\mathbf{B}\) and \(\mathbf{B}\mathbf{A}\) are both defined, they need not be equal. When finding the inverse of a product, order matters for the same reason. If we multiply two invertible matrices first, the inverse we obtain must undo the second factor first, then undo the first.
Throughout this tutorial, the matrices are square matrices with real entries. For an \(n\times n\) matrix \(\mathbf{A}\), an inverse is an \(n\times n\) matrix \(\mathbf{A}^{-1}\) such that both \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\) and \(\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\). A matrix is invertible if it has such an inverse. The examples use small matrices so that each multiplication can be checked directly.
Proof. Set \(\mathbf{C}=\mathbf{B}^{-1}\mathbf{A}^{-1}\). Associativity of matrix multiplication and the defining inverse identities give \[ (\mathbf{A}\mathbf{B})\mathbf{C} =\mathbf{A}\mathbf{B}\mathbf{B}^{-1}\mathbf{A}^{-1} =\mathbf{A}\mathbf{I}_n\mathbf{A}^{-1} =\mathbf{I}_n. \] Multiplying in the other order gives \[ \mathbf{C}(\mathbf{A}\mathbf{B}) =\mathbf{B}^{-1}\mathbf{A}^{-1}\mathbf{A}\mathbf{B} =\mathbf{B}^{-1}\mathbf{I}_n\mathbf{B} =\mathbf{I}_n. \] Thus \(\mathbf{C}\) is a two-sided inverse of \(\mathbf{A}\mathbf{B}\), which proves both that the product is invertible and that its inverse has the stated form. \(\square\)
The theorem supplies the order rule, but an actual calculation still requires the inverses of the factors and careful multiplication. In each example, we will check the proposed inverse by multiplying it by the product on both the left and the right. Checking both orders makes the verification explicit and avoids relying on an unverified candidate.
Compute the Factors, Then Reverse Them
Worked Example: Two Triangular Factors
Let \[ \mathbf{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}2&0\\1&1\end{bmatrix}. \] Find \((\mathbf{A}\mathbf{B})^{-1}\) and verify it in both orders.
First, the inverse of \(\mathbf{A}\) is \(\begin{bmatrix}1&-2\\0&1\end{bmatrix}\). Direct multiplication checks this candidate:
For \(\mathbf{B}\), the candidate inverse is \(\begin{bmatrix}\tfrac12&0\\-\tfrac12&1\end{bmatrix}\). The two checks are \[ \begin{bmatrix}2&0\\1&1\end{bmatrix} \begin{bmatrix}\tfrac12&0\\-\tfrac12&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad \begin{bmatrix}\tfrac12&0\\-\tfrac12&1\end{bmatrix} \begin{bmatrix}2&0\\1&1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] So both factors are invertible. Multiply the factors in the given order:
The inverse theorem says to reverse the inverse factors: \[ (\mathbf{A}\mathbf{B})^{-1} =\mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{bmatrix}\tfrac12&0\\-\tfrac12&1\end{bmatrix} \begin{bmatrix}1&-2\\0&1\end{bmatrix} = \begin{bmatrix}\tfrac12&-1\\-\tfrac12&2\end{bmatrix}. \] Now check the product followed by its proposed inverse, entry by entry:
In the reverse multiplication order, \[ \begin{bmatrix}\tfrac12&-1\\-\tfrac12&2\end{bmatrix} \begin{bmatrix}4&2\\1&1\end{bmatrix} = \begin{bmatrix}2-1&1-1\\-2+2&-1+2\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] Both checks give the identity, confirming the answer.
Order Is Not a Cosmetic Detail
Worked Example: Detecting the Wrong Order
Consider \[ \mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}1&0\\2&1\end{bmatrix}. \] Find the inverse of the product and show what goes wrong if the inverse factors are not reversed.
Multiplying each factor by its proposed inverse in either order gives the identity: \[ \mathbf{A}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}, \qquad \mathbf{B}^{-1}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}. \] For instance, the products in the original factor order are \[ \mathbf{A}\mathbf{A}^{-1} =\begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad \mathbf{B}\mathbf{B}^{-1} =\begin{bmatrix}1&0\\0&1\end{bmatrix}; \] reversing each pair gives the same identity matrices. First calculate the product: \[ \mathbf{A}\mathbf{B} = \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&0\\2&1\end{bmatrix} = \begin{bmatrix}1+2&0+1\\0+2&0+1\end{bmatrix} = \begin{bmatrix}3&1\\2&1\end{bmatrix}. \] The correct inverse is \[ \mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{bmatrix}1&0\\-2&1\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix} = \begin{bmatrix}1&-1\\-2&3\end{bmatrix}. \]
Check on the right: \[ \begin{bmatrix}3&1\\2&1\end{bmatrix} \begin{bmatrix}1&-1\\-2&3\end{bmatrix} = \begin{bmatrix}3-2&-3+3\\2-2&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] Check on the left: \[ \begin{bmatrix}1&-1\\-2&3\end{bmatrix} \begin{bmatrix}3&1\\2&1\end{bmatrix} = \begin{bmatrix}3-2&1-1\\-6+6&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] Thus the reversed order works on both sides.
A tempting but incorrect calculation is to keep the factor order and use \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). In this example, \[ \mathbf{A}^{-1}\mathbf{B}^{-1} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{bmatrix}1&0\\-2&1\end{bmatrix} = \begin{bmatrix}3&-1\\-2&1\end{bmatrix}. \] It is not the inverse, as multiplying the product by this matrix shows: \[ \begin{bmatrix}3&1\\2&1\end{bmatrix} \begin{bmatrix}3&-1\\-2&1\end{bmatrix} = \begin{bmatrix}7&-2\\4&-1\end{bmatrix} \ne \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] The failure is concrete: the factors do not commute, and the unreversed candidate does not undo their combined action.
A Product with Fractional Entries
Worked Example: A Nontrivial Inverse in Each Factor
Take \[ \mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}, \qquad \mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}. \] Find the inverse of \(\mathbf{A}\mathbf{B}\) and verify both multiplication orders.
The factor inverses are \[ \mathbf{A}^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}, \qquad \mathbf{B}^{-1}=\begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix}. \] Their checks include \[ \mathbf{A}\mathbf{A}^{-1} =\begin{bmatrix}2-1&-2+2\\1-1&-1+2\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}, \quad \mathbf{A}^{-1}\mathbf{A} =\begin{bmatrix}2-1&1-1\\-2+2&-1+2\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}, \] and \[ \mathbf{B}\mathbf{B}^{-1} =\begin{bmatrix}1&-\tfrac12+\tfrac12\\0&1\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}, \quad \mathbf{B}^{-1}\mathbf{B} =\begin{bmatrix}1&1-1\\0&1\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}. \] Next, \[ \mathbf{A}\mathbf{B} = \begin{bmatrix}2\cdot1+1\cdot0&2\cdot1+1\cdot2\\ 1\cdot1+1\cdot0&1\cdot1+1\cdot2\end{bmatrix} = \begin{bmatrix}2&4\\1&3\end{bmatrix}. \] Reverse the factor order when multiplying the inverses: \[ (\mathbf{A}\mathbf{B})^{-1} =\mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix} \begin{bmatrix}1&-1\\-1&2\end{bmatrix} = \begin{bmatrix}\tfrac32&-2\\-\tfrac12&1\end{bmatrix}. \]
The check on the right is \[ \begin{bmatrix}2&4\\1&3\end{bmatrix} \begin{bmatrix}\tfrac32&-2\\-\tfrac12&1\end{bmatrix} = \begin{bmatrix}3-2&-4+4\\\tfrac32-\tfrac32&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] The check on the left is \[ \begin{bmatrix}\tfrac32&-2\\-\tfrac12&1\end{bmatrix} \begin{bmatrix}2&4\\1&3\end{bmatrix} = \begin{bmatrix}3-2&6-6\\-1+1&-2+3\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}. \] Both products confirm the computed inverse.
A Reliable Order Check
For two invertible square matrices, there are two useful routes. One can multiply \(\mathbf{A}\mathbf{B}\) first and then find the inverse of that matrix, or use the theorem and compute \(\mathbf{B}^{-1}\mathbf{A}^{-1}\). The second route is often shorter, but it is especially easy to make an order error. A disciplined calculation separates the tasks: check the factor inverses, compute the product in its stated order, reverse the inverse factors, and multiply to verify.
Confirm that both factors are square matrices of the same size and that each has a two-sided inverse.
Use the given order \(\mathbf{A}\mathbf{B}\); do not silently switch it to \(\mathbf{B}\mathbf{A}\).
Form \(\mathbf{B}^{-1}\mathbf{A}^{-1}\), not \(\mathbf{A}^{-1}\mathbf{B}^{-1}\).
Multiply the product by the candidate on the right and then on the left. Each result must be the identity matrix.
The theorem proves why the reversed expression works, while the checks catch arithmetic mistakes in a particular calculation. If either factor is not invertible, the theorem's hypotheses are not met, so this rule cannot be used to conclude that the product has the stated inverse. The method here concerns products of two invertible square matrices; for a longer product, the same reversal principle can be applied repeatedly, but the calculations should still preserve the order at every step.
Common Pitfall: Treating Matrix Products Like Number Products
The most common mistake is to write \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{A}^{-1}\mathbf{B}^{-1}\) because the original factors appear as \(\mathbf{A}\) then \(\mathbf{B}\). The second worked example disproves that calculation: its proposed unreversed product gives \(\begin{bmatrix}7&-2\\4&-1\end{bmatrix}\) when multiplied by \(\mathbf{A}\mathbf{B}\), rather than the identity. The correction is to reverse the factors and then verify the result.
A second pitfall is to check only one side and lose track of what has actually been shown. For a proposed inverse \(\mathbf{C}\), the defining check is that \((\mathbf{A}\mathbf{B})\mathbf{C}=\mathbf{I}_n\) and \(\mathbf{C}(\mathbf{A}\mathbf{B})=\mathbf{I}_n\). The theorem guarantees both for the correctly ordered candidate; carrying out both checks in an example also confirms that the entries were calculated correctly.
Check Your Understanding
Use the reverse-order rule, show the matrix products needed to support your result, and check each proposed product inverse in both orders.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\3&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}2&1\\0&1\end{bmatrix}\), find \((\mathbf{A}\mathbf{B})^{-1}\).
- State the inverse of a product \(\mathbf{A}\mathbf{B}\) when both factors are invertible, and explain why its factors appear in reversed order.
- For \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\), calculate both \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) and \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). Which one is the inverse of \(\mathbf{A}\mathbf{B}\)?
- Why must the inverse theorem's hypothesis that both factors are invertible be stated?
- What two matrix equations should be checked to verify a proposed inverse of a square matrix?
