Common Errors with Matrix Inverses

Why Familiar Scalar Rules Can Fail

With real numbers, taking a reciprocal reverses multiplication: \((ab)^{-1}=a^{-1}b^{-1}\), because scalar multiplication commutes. Matrices do not generally commute, so the same-looking rule can fail even when every inverse exists. Addition presents a different problem: there is no general rule that distributes matrix inversion across a sum. These distinctions matter when simplifying matrix equations, because an expression that looks familiar can change the answer or may not even be defined.

In this tutorial, all matrices in inverse expressions are square matrices of the same size with real entries. An inverse of a matrix \(\mathbf{A}\) is a matrix \(\mathbf{A}^{-1}\) satisfying both \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}\) and \(\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}\). A matrix is called invertible when such an inverse exists. In particular, before manipulating an inverse, check that the matrix inside it is invertible. The fact that two matrices are individually invertible does not guarantee that their sum is invertible.

The Inverse of a Sum Does Not Distribute

The proposed addition rule is tempting because it resembles the distributive laws used in scalar arithmetic. A quick way to test a proposed inverse is to multiply it by the original matrix. If the product is not the identity matrix, the proposal is not an inverse. The following example makes the failure exact and easy to check.

There is also a separate issue with the sum rule: sometimes the inverse on its left side does not exist at all, even when the inverses on its right side do exist. For example, take \(\mathbf{A}=\mathbf{I}_2\) and \(\mathbf{B}=-\mathbf{I}_2\). Each is invertible, and \(\mathbf{B}^{-1}=-\mathbf{I}_2\), but \(\mathbf{A}+\mathbf{B}=\mathbf{O}\). The zero matrix has no inverse, because multiplying it by any matrix gives the zero matrix, not \(\mathbf{I}_2\). Thus an expression such as \((\mathbf{A}+\mathbf{B})^{-1}\) must be checked for existence before attempting to simplify it.

The Product Rule Reverses the Order

For products, there is a valid general formula, but the order is essential. When \(\mathbf{A}\) and \(\mathbf{B}\) are invertible, the inverse of \(\mathbf{A}\mathbf{B}\) is \(\mathbf{B}^{-1}\mathbf{A}^{-1}\), not generally \(\mathbf{A}^{-1}\mathbf{B}^{-1}\). The reason is visible in the multiplication: the factor nearest \(\mathbf{A}\mathbf{B}\) must cancel first.

Proof. Set \(\mathbf{C}=\mathbf{B}^{-1}\mathbf{A}^{-1}\). Using associativity of matrix multiplication and the defining inverse identities for \(\mathbf{A}\) and \(\mathbf{B}\), we obtain

and, in the other order,

Therefore \(\mathbf{C}\) is a two-sided inverse of \(\mathbf{A}\mathbf{B}\), so \(\mathbf{A}\mathbf{B}\) is invertible and its inverse is \(\mathbf{C}\). The inverse is unique: if \(\mathbf{C}\) and \(\mathbf{D}\) were both inverses of the same matrix \(\mathbf{M}\), then \(\mathbf{C}=\mathbf{C}(\mathbf{M}\mathbf{D})=(\mathbf{C}\mathbf{M})\mathbf{D}=\mathbf{D}\). Hence the stated formula follows. \(\square\)

This proof also explains why simply reversing the order is not an optional convention. To cancel the adjacent factors in \((\mathbf{A}\mathbf{B})(\mathbf{B}^{-1}\mathbf{A}^{-1})\), we first use \(\mathbf{B}\mathbf{B}^{-1}=\mathbf{I}_n\), then \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\). If the inverse factors are written in the other order, those cancellations are no longer adjacent.

A Valid Rule for Scalar Multiples

Not every apparently simple inverse rule is false. If \(c\) is a nonzero real scalar and \(\mathbf{A}\) is invertible, then the inverse of \(c\mathbf{A}\) is \(\frac{1}{c}\mathbf{A}^{-1}\). The requirement \(c\ne0\) is necessary: if \(c=0\), then \(c\mathbf{A}\) is the zero matrix and has no inverse.

Proof. Let \(\mathbf{C}=\frac{1}{c}\mathbf{A}^{-1}\). Since scalar multiplication can be regrouped with matrix multiplication,

and

Thus \(\mathbf{C}\) is a two-sided inverse of \(c\mathbf{A}\), which proves the formula. \(\square\)

Common Mistakes and a Reliable Check

Mistake: distributing an inverse across a sum. A wrong calculation might write \((\mathbf{A}+\mathbf{B})^{-1}=\mathbf{A}^{-1}+\mathbf{B}^{-1}\), then substitute \(\mathbf{A}=\mathbf{B}=\mathbf{I}_2\) and conclude that \((2\mathbf{I}_2)^{-1}=2\mathbf{I}_2\). The correction is to multiply back: \((2\mathbf{I}_2)(2\mathbf{I}_2)=4\mathbf{I}_2\ne\mathbf{I}_2\). The actual inverse is \(\tfrac12\mathbf{I}_2\).

Mistake: reversing the product rule in the wrong direction. Writing \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{A}^{-1}\mathbf{B}^{-1}\) ignores that matrix multiplication may not commute. The correction is to reverse the factors, then verify by multiplication: \((\mathbf{A}\mathbf{B})(\mathbf{B}^{-1}\mathbf{A}^{-1})=\mathbf{I}_n\) and \((\mathbf{B}^{-1}\mathbf{A}^{-1})(\mathbf{A}\mathbf{B})=\mathbf{I}_n\).

When checking an inverse identity, use this short sequence of questions:

• Are the matrices square and of compatible size?
• Does each inverse appearing in the expression exist?
• For an inverse of a product, are the factors in reversed order?
• Does multiplying the proposed inverse on both sides give the identity?

The order check is especially important when there are three or more factors. Repeatedly applying the product rule gives \((\mathbf{A}\mathbf{B}\mathbf{C})^{-1}=\mathbf{C}^{-1}\mathbf{B}^{-1}\mathbf{A}^{-1}\) when all three matrices are invertible. The factors reverse their full order; they are not individually inverted while left in place. By contrast, a sum has no analogous distribution rule, and the sum itself must be checked for invertibility.